Higher Maths
Trigonometry

We use the expansion of \(\text{cos}(A+B)\) with \(A=45^\circ\) and \(B=30^{\circ}\small,\) because they are angles whose exact values we have learned – or should have learned!

$$ \begin{eqnarray} \text{cos}\,75^\circ &=& \text{cos}(45+30)^\circ \\[8pt] &=& \text{cos}\,45^\circ\,\text{cos}\,30^\circ-\text{sin}\,45^\circ\,\text{sin}\,30^\circ \\[8pt] &=& \small{\dfrac{1}{\sqrt{2}}}\times\small{\dfrac{\sqrt{3}}{2}}-\small{\dfrac{1}{\sqrt{2}}}\times\small{\dfrac{1}{2}} \\[8pt] &=& \small\frac{\sqrt{3}-\normalsize{1}}{2\sqrt{2}}\normalsize \end{eqnarray} $$

\(195^\circ\) is a third quadrant angle, so its sine is negative. The related acute angle is \(15^{\circ}.\)

So we use the expansion of \(\text{sin}(A-B)\) with \(A=45^\circ\) and \(B=30^{\circ}\small.\)

$$ \begin{eqnarray} \text{sin}\,15^\circ &=& \text{sin}(45-30)^\circ\\[8pt] &=& \text{sin}\,45^\circ\,\text{cos}\,30^\circ-\text{cos}\,45^\circ\,\text{sin}\,30^\circ\\[8pt] &=& \small\frac{1}{\sqrt{2}}\times\frac{\sqrt{3}}{2}\normalsize - \small\frac{1}{\sqrt{2}}\times\frac{1}{2}\normalsize \\[8pt] &=& \small\frac{\sqrt{3}-1}{2\sqrt{2}}\normalsize \end{eqnarray} $$

So \(\text{sin}\,195^\circ=-\,\text{sin}\,15^\circ=\large\frac{1\,-\,\sqrt{3}}{2\sqrt{2}}\)

Note 1: \(A=60^\circ\small,\) \(B=45^{\circ}\) would also have worked here.

Note 2: We advise learning the exact trig values to save valuable time in your exams.

This solution uses the double-angle identity: \(\text{sin}\,2x=2\,\text{sin}\,x\,\text{cos}\,x\small.\)

We have been given the value of \(\text{sin}\,x\small,\) so we just need to find \(\text{cos}\,x\small.\)

The question tells us that \(x\) is an acute angle, so it is small enough to fit inside a right-angled triangle.

\(\text{sin}\,x=\frac{5}{13}\small,\) so we can use an opposite side of \(5\) units and a hypotenuse of \(13\) units.

We use Pythagoras' Theorem to find the adjacent side, \(a\) units:

$$ \begin{eqnarray} a^2 &=& 13^2-5^2 \\[6pt] &=& 169-25\\[6pt] &=& 144\\[6pt] a&=& \sqrt{144}\\[6pt] &=& 12 \end{eqnarray} $$

So \(\text{cos}\,x=\frac{12}{13}\) and now we can find \(\text{sin}\,2x\small.\)

$$ \begin{eqnarray} \text{sin}\,2x &=& 2\,\text{sin}\,x\,\text{cos}\,x \\[6pt] &=& 2\small\left(\frac{5}{13}\right)\left(\frac{12}{13}\right)\normalsize \\[6pt] &=& \small\frac{120}{169}\normalsize \end{eqnarray} $$

(a)  The question tells us that \(x\lt \frac{\pi}{4}\) radians. That's \(45^\circ\small,\) if you're still thinking in degrees.

So \(2x\lt \frac{\pi}{2}\small.\) In other words, \(2x\) is an acute angle. So \(2x\) is small enough to fit inside a right-angled triangle with opposite side \(4\) units and adjacent side \(3\) units. The hypotenuse is therefore \(5\) units, so \(\text{cos}\,2x=\frac{3}{5}\small.\)

(b)  There are three identities for \(\text{cos}\,2x\) on the formulae list . We choose the identity that involves \(\text{cos}\,x\small.\)

$$ \begin{gather} \text{cos}\,2x=2\,\text{cos}^{2}\,x-1 \\[8pt] \small\frac{3}{5}\normalsize=2\,\text{cos}^{2}\,x-1 \\[8pt] \small\frac{3}{5}\normalsize=2\,\text{cos}^{2}\,x-\small\frac{5}{5}\normalsize \\[8pt] 2\,\text{cos}^{2}\,x=\small\frac{3}{5}\normalsize +\small\frac{5}{5}\normalsize \\[8pt] 2\,\text{cos}^{2}\,x=\small\frac{8}{5}\normalsize \\[8pt] \text{cos}^{2}\,x=\small\frac{4}{5}\normalsize \\[8pt] \text{cos}\,x=\small\sqrt{\frac{4}{5}}\normalsize \\[8pt] \text{cos}\,x=\small\frac{2}{\sqrt{5}}\normalsize \end{gather} $$

Note: We don't need to worry about the negative square root of \(\frac{4}{5}\small,\) as \(x\) is an acute angle, so \(\text{cos}\,x\gt 0\small.\)

You will notice that \(\text{tan}\,x\) and \(\text{tan}\,2x\) are absent from the trig identities on the formula sheet .

The only identity involving \(\text{tan}\,x\) that you need to know is \(\text{tan}\,x=\large\frac{\text{sin}\,x}{\text{cos}\,x}\) from Nat 5. We will use that here.

First, we use the relevant identity for \(\text{cos}\,2x\) to find \(\text{sin}^{2}\,x.\)

$$ \begin{gather} \text{cos}\,2x=1-2\,\text{sin}^{2}\,x \\[7pt] \small\frac{12}{13}\normalsize=\small\frac{13}{13}\normalsize-2\,\text{sin}^{2}\,x \\[7pt] 2\,\text{sin}^{2}\,x=\small\frac{13}{13}-\small\frac{12}{13}\normalsize \\[7pt] 2\,\text{sin}^{2}\,x= \small\frac{1}{13} \\[7pt] \text{sin}^{2}\,x=\small\frac{1}{26}\normalsize \end{gather} $$

Now to find \(\text{cos}^{2}\,x.\) We could use another double angle formula, but it is easier to use this Nat 5 trig identity:

$$ \begin{gather} \text{sin}^{2}\,x+\text{cos}^{2}\,x=1 \\[7pt] \small\frac{1}{26}\normalsize+\text{cos}^{2}\,x=\small\frac{26}{26} \\[7pt] \text{cos}^{2}\,x=\small\frac{25}{26}\normalsize \end{gather} $$

Now, \(\text{tan}^2\,x=\large\frac{\text{sin}^2\,x}{\text{cos}^2\,x}\normalsize=\large\frac{\Large{{}^{1}\!/\!_{26}}}{\Large{{}^{25}\!/\!_{26}}}\normalsize= \large\frac{1}{25}\small.\)

So \(\text{tan}\,x=\sqrt{\!\large{\frac{1}{25}}}=\large\frac{1}{5}\small.\)

(a)  \(\text{sin}\,x=\large\frac{3}{5}\small,\) so we can imagine \(x^\circ\) inside a right-angled triangle with opposite side \(3\) units and hypotenuse \(5\) units. This is a \(3,4,5\) triangle, so the adjacent side is \(4\) units, giving us \(\text{cos}\,x=\large\frac{4}{5}\small.\)

(b)  We use the only relevant identity:

$$ \begin{eqnarray} \text{sin}\,2x &=& 2\,\text{sin}\,x\,\text{cos}\,x \\[6pt] &=& 2\small\left(\frac{3}{5}\right)\left(\frac{4}{5}\right)\normalsize \\[6pt] &=& \small\frac{24}{25}\normalsize \end{eqnarray} $$

(c)  We could use any of the three identities for \(\text{cos}\,2x\) but we know that \(x\lt 45^\circ\) so \(2x\) is still an acute angle. Therefore, it is easier to find the adjacent side in a right-angled triangle containing \(2x^{\circ}\small.\)

Using Pythagoras with the answer to the previous part:

$$ \begin{eqnarray} a^2 &=& 25^2-24^2 \\[6pt] &=& 625-576\\[6pt] &=& 49\\[6pt] a&=& \sqrt{49}\\[6pt] &=& 7 \end{eqnarray} $$

So \(\text{cos}\,2x=\large\frac{7}{25}\small.\)

(d)  This and part (e) will use the fact that \(3x=2x+x\) together with the relevant addition formulae.

$$ \begin{eqnarray} \text{cos}\,3x &=& \text{cos}\,(2x+x) \\[6pt] &=& \text{cos}\,2x\,\text{cos}\,x - \text{sin}\,2x\,\text{sin}\,x\\[6pt] &=& \small\left(\frac{7}{25}\right)\small\left(\frac{4}{5}\right) - \small\left(\frac{24}{25}\right)\small\left(\frac{3}{5}\right)\\[6pt] &=& \small\frac{28-72}{125}\\[6pt] &=& -\!\small\frac{44}{125} \end{eqnarray} $$

(e)  This part uses the similar identity for \(\text{sin}\,3x\small.\)

$$ \begin{eqnarray} \text{sin}\,3x &=& \text{sin}\,(2x+x) \\[6pt] &=& \text{sin}\,2x\,\text{cos}\,x+\text{cos}\,2x\,\text{sin}\,x\\[6pt] &=& \small\left(\frac{24}{25}\right)\small\left(\frac{4}{5}\right) + \small\left(\frac{7}{25}\right)\small\left(\frac{3}{5}\right)\\[6pt] &=& \small\frac{96+21}{125}\\[6pt] &=& \small\frac{117}{125} \end{eqnarray} $$

This equation uses radians rather than degrees. We would recommend that, if possible, you work in radians throughout. However, that takes practice and it may not suit every student. Be guided by your teacher in school.

We can use the trig identity that links \(\text{cos}\,2x\) and \(\text{sin}\,x\) to turn this into a quadratic in \(\text{sin}\,x\small.\)

$$ \begin{gather} \text{cos}\,2x=\text{sin}\,x \\[8pt] 1-2\,\text{sin}^{2}\,x=\text{sin}\,x \\[8pt] 2\,\text{sin}^{2}\,x+\text{sin}\,x -1=0 \\[8pt] (2\,\text{sin}\,x-1)(\text{sin}\,x+1)=0 \\[8pt] 2\,\text{sin}\,x-1=0\:\:\:\small\textsf{or}\normalsize\:\:\:\text{sin}\,x+1=0 \\[6pt] \text{sin}\,x=\small\frac{1}{2}\normalsize\:\:\:\small\textsf{or}\normalsize\:\:\:\text{sin}\,x=-1 \\[6pt] x=\small\frac{\pi}{6}\small{,}\:\:\pi-\small\frac{\pi}{6}\normalsize\:\:\:\small\textsf{or}\normalsize\:\:\:x=\small\frac{3\pi}{2}\\[6pt] x=\small\frac{\pi}{6},\:\:\frac{5\pi}{6},\:\:\small\frac{3\pi}{2} \end{gather} $$

This trig equation involves only the double angle \(2x\small.\) There is no single angle in the equation.

For that reason, we do not use any of the double angle formulae.

We need to find all solutions, in degrees, in the interval \(0\leqslant x \lt 360\small,\) so \(0\leqslant 2x \lt 720\small.\)

\(\text{cos}\,2x=-\frac{1}{2}\) and the related acute angle \(\alpha=\text{cos}^{-1}\,\left(\frac{1}{2}\right)=60^\circ\small.\)

Because \(\text{cos}\,2x \lt 0\small,\) there are solutions in the second (S) and third (T) quadrants.

So \(2x=180-\alpha=120^\circ\) and \(2x=180+\alpha=240^\circ\small.\)

However, we need to take care to find all the values of \(2x\) in the interval \(0\leqslant 2x \lt 720^\circ\small.\) We do this by adding or subtracting multiples of \(360^\circ\) until we have found them all. It is important to note that this must be done before dividing by \(2\) to get the values of \(x\) themselves.

\(120+360=480\) and \(240+360=600\small.\) These are both within the interval.

So \(2x=120^\circ\small,\normalsize\ 240^\circ\small,\normalsize\ 480^\circ\small,\normalsize\ 600^\circ\small.\)

Halving, \(x=60^\circ\small,\normalsize\ 120^\circ\small,\normalsize\ 240^\circ\small,\normalsize\ 300^\circ\small.\)

This is a relatively simple equation. We start by rearranging it to make \(\text{cos}\,x\) the subject:

$$ \begin{gather} 2\,\text{cos}^2\,x=1 \\[6pt] \text{cos}^2\,x=\small\frac12 \\[6pt] \text{cos}\,x=\pm\small\frac{1}{\sqrt{2}} \end{gather} $$

It is very important not to forget the \(\pm\) above, or we 'lose' half of the solutions!

The related acute angle \(\alpha=\text{cos}^{-1}\,\left(\large\frac{1}{\sqrt{2}}\normalsize\right)=45^\circ\small.\)

Because \(\text{cos}\,x\) could be either positive or negative, there are solutions in all four quadrants.

So the four solutions are:
\(x=45^\circ\)
\(x=180-45=135^\circ\)
\(x=180+45=225^\circ\)
\(x=360-45=315^\circ\)

(a)  This equation involves both double angle and single angle, so our first job is to use the relevant identity to eliminate the double angle. Then we factorise and solve, as usual.

$$ \begin{gather} \text{sin}\,2x+6\,\text{cos}\,x=0\\[7pt] 2\,\text{sin}\,x\,\text{cos}\,x+6\,\text{cos}\,x=0\\[7pt] (\text{cos}\,x)(2\,\text{sin}\,x+6)=0\\[7pt] \text{cos}\,x=0\:\:\:\small\textsf{or}\normalsize\:\:\:2\,\text{sin}\,x+6=0\\[7pt] x=90^\circ\small,\normalsize\:270^\circ\:\:\:\small\textsf{or}\normalsize\:\:\:\text{sin}\,x=-3\\[7pt] \end{gather} $$

\(\text{sin}\,x=-3\) has no solutions as \(-1\lt\text{sin}\,x\lt 1\small.\)

So the only solutions are \(x=90^\circ\) and \(x=270^\circ\small.\)

(b)  This equation is very closely related to part (a), but with \(x\) replaced by \(2x\small.\)

So we don't need to solve from scratch again. We can just use the fact that \(90^\circ\) and \(270^\circ\) are solutions for \(2x\small.\)

The interval \(0\leqslant x \lt 360\) means that we need to solve for \(0\leqslant 2x \lt 720\small,\) so we need to remember to add \(360\) to find all solutions in this interval.

$$ \begin{gather} \text{sin}\,4x+6\,\text{cos}\,2x=0\\[7pt] \text{sin}\,2(2x)+6\,\text{cos}\,2x=0\\[7pt] 2x = 90^\circ\small,\normalsize\:270^\circ\small,\normalsize\:90^\circ\!+\!360^\circ\small,\normalsize\:270^\circ\!+\!360^\circ\\[7pt] 2x = 90^\circ\small,\normalsize\:270^\circ\small,\normalsize\:450^\circ\small,\normalsize\:630^\circ \\[7pt] x = 45^\circ\small,\normalsize\:135^\circ\small,\normalsize\:225^\circ\small,\normalsize\:315^\circ\\[7pt] \end{gather} $$

This equation involves the triple angle \(3x\small.\) We need to find all solutions, in degrees, in the interval \(0\leqslant x \lt 180\small.\)

Our first job is to inspect the equation and adjust this interval: \(0\leqslant 3x \lt 540\small\) so \(-60\leqslant 3x-60 \lt 480\small.\)

Now, \(\text{sin}\,(3x-60)=-\frac{1}{2}\) and the related acute angle \(\alpha=sin^{-1}\left(\frac{1}{2}\right)=30^\circ\small.\)

Because \(\text{sin}\,(3x-60)\lt 0\small,\) there are solutions in the third (T) and fourth (C) quadrants.

So \(3x-60=180+\alpha=210^\circ\) and \(3x-60=360-\alpha=330^\circ\small.\)

However, we need to take care to find all the values of \(3x-60\) in the interval \(-60\leqslant 3x-60 \lt 480\small.\) We do this by adding or subtracting multiples of \(360^\circ\) until we have found them all. It is important to note that this must be done before adding \(60\) and dividing by \(3\) to find the values of \(x\) themselves.

Only \(330-360=-30\) is within the required interval.

So \(3x-60=-30^\circ\small,\normalsize\:210^\circ\small,\normalsize\:330^\circ\small.\)

Adding \(60\small,\) we obtain \(3x=30^\circ\small,\normalsize\:270^\circ\small,\normalsize\:390^\circ\small.\)

Dividing by \(3\) gives us \(x=10^\circ\small,\normalsize\:90^\circ\small,\normalsize\:130^\circ\small.\)

This equation contains both a double angle \(2x\small\) and a single angle \(x\small,\) so our first job is to use the relevant identity to eliminate the double angle, giving us a quadratic equation in \(\text{sin}\,x\small.\)

Then we bring all terms to the left, factorise and solve each linear equation using Nat 5 methods.

$$ \begin{gather} \text{sin}\,x+2 = 3\,\text{cos}\,2x \\[7pt] \text{sin}\,x+2 = 3(1-2\,\text{sin}^{2}\,x) \\[7pt] \text{sin}\,x+2 = 3-6\,\text{sin}^{2}\,x \\[7pt] 6\,\text{sin}^{2}\,x + \text{sin}\,x -1 = 0 \\[7pt] (3\,\text{sin}\,x-1)(2\,\text{sin}\,x+1) = 0 \\[7pt] 3\,\text{sin}\,x-1=0\:\ \small\textsf{or}\normalsize\:\:2\,\text{sin}\,x+1=0\\[7pt] 3\,\text{sin}\,x=1\:\ \small\textsf{or}\normalsize\:\:2\,\text{sin}\,x=-1\\[7pt] \text{sin}\,x=\small\frac{1}{3}\normalsize\:\:\small\textsf{or}\normalsize\:\:\text{sin}\,x=-\small\frac{1}{2}\\[7pt] x^\circ=19.5\small,\normalsize\:180\!-\!19.5\:\:\small\textsf{or}\normalsize\:\:180\!+\!30\small,\normalsize\:360\!-\!30\\[7pt] x=19.5^\circ\small,\normalsize\:160.5^\circ\small,\normalsize\:210^\circ\small,\normalsize\:330^\circ \\[6pt] \end{gather} $$

For 'wave function' questions like this, it is important for you to find a method that you are comfortable with and stick to it exactly. It is easy to get into an awful muddle if you don't follow a correct process precisely. We recommend the same method that is detailed in the HSN notes , where you will find more worked examples.

Step 1: Expand \(\boldsymbol{k\,\textbf{cos}\,(x-a)}\)

We start by using the relevant addition formula to expand \(k\,\text{cos}\,(x-a)\small.\) Then we reorder the factors within each term to compare our expansion with the given expression.

$$ \begin{eqnarray} && \!\!\!\!k\,\text{cos}\,(x-a)\\[6pt] &=& k\,\text{cos}\,x\,\text{cos}\,a+k\,\text{sin}\,x\,\text{sin}\,a\\[6pt] &=& \underbrace{k\,\text{cos}\,a}_{\large{\sqrt{3}}}\,\text{cos}\,x+ \underbrace{k\,\text{sin}\,a}_{\large{1}}\,\text{sin}\,x \end{eqnarray} $$

So, by comparing with the expression in the question, we can see that \(k\,\text{cos}\,a=\sqrt{3}\) and \(k\,\text{sin}\,a=1\small.\)

Step 2: Work out which quadrant \(\boldsymbol{a}\) is in

From the previous working, both \(\text{cos}\,a\) and \(\text{sin}\,a\) are positive, so \(a\) is in the first quadrant.

Step 3: Find \(\boldsymbol{a}\)

This part works by dividing the questions that we obtained in Step 1, making use of a Nat 5 trig identity.

$$ \begin{eqnarray} \frac{k\,\text{sin}\,a}{k\,\text{cos}\,a} &=& \frac{1}{\sqrt{3}} \\[6pt] \text{tan}\,a &=& \frac{1}{\sqrt{3}} \\[6pt] a &=& 30^\circ \\[6pt] \end{eqnarray} $$

In this example, Step 2 told us that \(a\) is an acute angle, so there is no need to think about the quadrant diagram, but in other examples, that will be necessary.

Step 4: Find \(\boldsymbol{k}\)

There is an easy formula to find \(k\small,\) but in this example, we will explain in full the theory behind it. Then in the further examples below, we can use the shortcut.

Start with one of the Nat 5 trig identities and follow these steps:

$$ \begin{gather} \text{cos}^{2}\,a+\text{sin}^{2}\,a=1\\[6pt] k^2\left(\text{cos}^{2}\,a+\text{sin}^{2}\,a\right)=k^2\\[6pt] k^2\,\text{cos}^{2}\,a+k^2\,\text{sin}^{2}\,a=k^2\\[6pt] \left(k\,\text{cos}\,a\right)^2+\left(k\,\text{sin}\,a\right)^2=k^2\\[6pt] \left(\sqrt{3}\right)^2+1^2=k^2\\[6pt] 3+1=k^2\\[6pt] k^2=4\\[6pt] k=2 \end{gather} $$

In future examples, we'll skip this detail and just use the Pythagoras-like relationship directly.

We now know \(a\) and \(k\small,\) so we can write our final answer:

\(\sqrt{3}\,\text{cos}\,x^\circ+\text{sin}\,x^\circ=2\,\text{cos}\,(x-30)^\circ\)

First, expand \(k\,\text{cos}\,(x-a)\) using the relevant addition formula. Then reorder to compare it with the given expression.

$$ \begin{eqnarray} && \!\!\!\!\!\!k\,\text{cos}\,(x-a) \\[6pt] &=& k\,\text{cos}\,x\,\text{cos}\,a+ k\,\text{sin}\,x\,\text{sin}\,a \\[6pt] &=& \underbrace{k\,\text{cos}\,a}_{\large{2}}\,\text{cos}\,x+ \underbrace{k\,\text{sin}\,a}_{\large{-3}}\,\text{sin}\,x \end{eqnarray} $$

So now we know that \(k\,\text{cos}\,a=2\) and \(k\,\text{sin}\,a=-3\small.\)

The next step is to work out which quadrant \(a\) is in. Here, \(\text{cos}\,a\gt 0\) which limits \(a\) to the first (A) or fourth (C) quadrant. But \(\text{sin}\,a\lt 0,\) which means that \(a\) must be in the third (T) or fourth (C) quadrant. The only quadrant that meets both conditions is the fourth (C) quadrant.

Next, we find \(a\) by dividing the two equations:

$$ \begin{eqnarray} \frac{k\,\text{sin}\,a}{k\,\text{cos}\,a} &=& \frac{-3}{2} \\[6pt] \text{tan}\,a &=& -\!\frac{3}{2} \\[6pt] \end{eqnarray} $$

The related acute angle is \(\text{tan}^{-1}\,\large\frac{3}{2}\normalsize=56.3^\circ\) (accurate to 1 d.p.) but \(a\) is in the fourth quadrant, so \(a=360-56.3=303.7^\circ\small.\)

Finally, we find \(k\):

$$ \begin{eqnarray} k^2 &=& 2^2+(-3)^2 \\[6pt] &=& 4+9 \\[6pt] &=& 13 \\[6pt] k &=& \sqrt{13} \end{eqnarray} $$

So our final answer is:

\(2\,\text{cos}\,x^\circ-3\,\text{sin}\,x^\circ=\sqrt{13}\ \text{cos}\,(x-303.7)^\circ\small.\)

First, expand \(k\,\text{cos}\,(x+a)\) using the relevant addition formula. Then reorder the terms to compare the expansion with the given expression. This example needs a little more rearrangement than the previous two examples.

$$ \begin{eqnarray} && \!\!\!\!\!\!k\,\text{cos}\,(x+a) \\[6pt] &=& k\,\text{cos}\,x\,\text{cos}\,a-k\,\text{sin}\,x\,\text{sin}\,a \\[6pt] &=& k\,\text{cos}\,a\,\text{cos}\,x-k\,\text{sin}\,a\,\text{sin}\,x \\[6pt] &=& \!-k\,\text{sin}\,a\,\text{sin}\,x+k\,\text{cos}\,a\,\text{cos}\,x \\[6pt] &=& \underbrace{-k\,\text{sin}\,a}_{\large{5}}\,\text{sin}\,x+\underbrace{k\,\text{cos}\,a}_{\large{-2}}\,\text{cos}\,x \end{eqnarray} $$

So now we know that \(-k\,\text{sin}\,a=5\) and \(k\,\text{cos}\,a=-2\small.\)

The next step is to work out which quadrant \(a\) is in. Both \(\text{cos}\,a\) and \(\text{sin}\,a\) are negative, which means that \(a\) can only be in the third (T) quadrant.

Next, we find \(a\) by dividing the two equations. Care is needed with the negative signs.

$$ \begin{eqnarray} \frac{-k\,\text{sin}\,a}{k\,\text{cos}\,a} &=& \frac{5}{-2}\normalsize \\[6pt] \frac{k\,\text{sin}\,a}{k\,\text{cos}\,a} &=& \frac{5}{2}\normalsize \\[6pt] \text{tan}\,a &=& \frac{5}{2} \\[6pt] \end{eqnarray} $$

The related acute angle is \(\text{tan}^{-1}\,\large\frac{5}{2}\normalsize=68.2^\circ\) (to 1 d.p.) but \(a\) is in the third quadrant, so \(a=180+68.2=248.2^\circ\small.\)

Finally, we find \(k\):

$$ \begin{eqnarray} k^2 &=& (-2)^2+5^2 \\[6pt] &=& 4+25 \\[6pt] &=& 29 \\[6pt] k &=& \sqrt{29} \end{eqnarray} $$

So our final answer is:

\(5\,\text{sin}\,x^\circ-2\,\text{cos}\,x^\circ=\sqrt{29}\ \text{cos}\,(x+248.2)^\circ\)

This equation uses radians rather than degrees. We would recommend that, if possible, you work in radians throughout. However, that takes practice and it may not suit every student. Be guided by your teacher in school.

First, expand \(k\,\text{sin}\,(x+a)\) using the relevant addition formula. The reorder the terms to compare your expansion with the given expression.

$$ \begin{eqnarray} && \!\!\!\!\!\!k\,\text{sin}\,(x+a) \\[6pt] &=& k\,\text{sin}\,x\,\text{cos}\,a+k\,\text{cos}\,x\,\text{sin}\,a \\[6pt] &=& k\,\text{cos}\,a\,\text{sin}\,x+k\,\text{sin}\,a\,\text{cos}\,x \\[6pt] &=& \underbrace{k\,\text{cos}\,a}_{\large{1}}\,\text{sin}\,x+\underbrace{k\,\text{sin}\,a}_{-\sqrt{3}}\,cos\,x \end{eqnarray} $$

So now we know that \(k\,\text{cos}\,a=1\) and \(k\,\text{sin}\,a=-\sqrt{3}\small.\)

The next step is to work out which quadrant \(a\) is in. As \(\text{cos}\,a\gt 0\small,\) the first (A) or fourth (C) quadrants are possibilities. But \(\text{sin}\,a\lt 0\) rules out the first (A) quadrant. So \(a\) can only be in the fourth (C) quadrant.

Next, we find \(a\) by dividing the two equations:

$$ \begin{eqnarray} \frac{k\,\text{sin}\,a}{k\,\text{cos}\,a} &=& \frac{-\sqrt{3}}{1} \\[6pt] \text{tan}\,a &=& -\!\sqrt{3} \\[6pt] \end{eqnarray} $$

The related acute angle is \(\text{tan}^{-1}\,\sqrt{3}=\large\frac{\pi}{3}\normalsize\) but \(a\) is in the fourth quadrant, so \(a=2\pi-\large\frac{\pi}{3}\normalsize=\large\frac{5\pi}{3}\small.\)

Finally, we find \(k\):

$$ \begin{eqnarray} k^2 &=& 1^2+\left(-\sqrt{3}\right)^2 \\[6pt] &=& 1+3 \\[6pt] &=& 4 \\[6pt] k &=& 2 \end{eqnarray} $$

So our final answer is:

\(\text{sin}\,x-\sqrt{3}\,\text{cos}\,x=2\,\text{sin}\left(x+\large\frac{5\pi}{3}\normalsize\right)\)

This equation uses radians rather than degrees. We would recommend that, if possible, you work in radians throughout. However, that takes practice and it may not suit every student. Be guided by your teacher in school.

First, expand \(k\,\text{sin}(x-a)\) using the relevant addition formula. Then reorder the factors to enable comparison of your expansion with the given expression. This example needs extreme care over the \(+\) and \(-\) signs.

$$ \begin{eqnarray} && \!\!\!\!\!\!k\,\text{sin}\,(x-a) \\[6pt] &=& k\,\text{sin}\,x\,\text{cos}\,a-k\,\text{cos}\,x\,\text{sin}\,a \\[6pt] &=& k\,\text{cos}\,a\,\text{sin}\,x-k\,\text{sin}\,a\,\text{cos}\,x \\[6pt] &=& \underbrace{k\,\text{cos}\,a}_{\large{-2}}\,\text{sin}\,x+(\underbrace{-k\,\text{sin}\,a}_{\large{7}})\,\text{cos}\,x \end{eqnarray} $$

So now we know that \(k\,\text{cos}\,a=-2\) and \(-k\,\text{sin}\,a=7\small.\)

The next step is to work out which quadrant \(a\) is in. As both \(\text{cos}\,a\) and \(\text{sin}\,a\) are negative, \(a\) must be in the third (T) quadrant.

Next, we find \(a\) by dividing the two equations:

$$ \begin{eqnarray} \frac{-k\,\text{sin}\,a}{k\,\text{cos}\,a} &=& \frac{7}{-2} \\[6pt] \frac{k\,\text{sin}\,a}{k\,\text{cos}\,a} &=& \frac{7}{2} \\[6pt] \text{tan}\,a &=& \frac{7}{2} \\[6pt] \end{eqnarray} $$

The related acute angle is \(\text{tan}^{-1}\,\large\frac{7}{2}\normalsize=1.292\) (to 3 d.p.) but \(a\) is in the third quadrant, so \(a=\pi+1.292=4.434\small.\)

Finally, we find \(k\):

$$ \begin{eqnarray} k^2 &=& (-2)^2+7^2 \\[6pt] &=& 4+49 \\[6pt] &=& 53 \\[6pt] k &=& \sqrt{53} \end{eqnarray} $$

So our final answer is:

\(-2\,\text{sin}\,x+7\,\text{cos}\,x=\sqrt{53}\,\text{sin}\,(x-4.434)\)

(a)  This part is a relatively simple 'wave function' question, which we will present below without any commentary:

$$ \begin{eqnarray} && \!\!\!\!\!\!k\,\text{cos}\,(x-a) \\[6pt] &=& k\,\text{cos}\,x\,\text{cos}\,a+k\,\text{sin}\,x\,\text{sin}\,a \\[6pt] &=& k\,\text{cos}\,a\,\text{cos}\,x+k\,\text{sin}\,a\,\text{sin}\,x \\[6pt] &=& \underbrace{k\,\text{cos}\,a}_{\large{3}}\,\text{sin}\,x+\underbrace{k\,\text{sin}\,a}_{\large{1}}\,\text{sin}\,x \end{eqnarray} $$

$$ \begin{eqnarray} \frac{k\,\text{sin}\,a}{k\,\text{cos}\,a} &=& \frac{1}{3} \\[6pt] \text{tan}\,a &=& \frac{1}{3} \\[6pt] \end{eqnarray} $$

\(\text{sin}\,a\gt 0\) and \(\text{cos}\,a\gt 0\small,\) so \(a\) is acute. So \(a=\text{tan}^{-1}\,\large\frac{1}{3}\normalsize=18.4^\circ\) (to 1 d.p.)

$$ \begin{eqnarray} k^2 &=& 3^2+1^2 \\[6pt] &=& 10 \\[6pt] k &=& \sqrt{10} \end{eqnarray} $$

\(\therefore\:3\,\text{cos}\,x^\circ+\text{sin}\,x^\circ=\sqrt{10}\,\text{cos}\,(x-18.4)^\circ\)

(b)  The maximum value is \(\sqrt{10}\) and (correct to 1 d.p.) occurs when:

$$ \begin{gather} \text{cos}\,(x-18.4)^\circ=1\\[6pt] x-18.4=\text{cos}^{-1}\,1\\[6pt] x-18.4=0\\[6pt] x=18.4^\circ \end{gather} $$

(c)  The minimum value is \(-\sqrt{10}\) and (correct to 1 d.p.) occurs when:

$$ \begin{gather} \text{cos}\,(x-18.4)^\circ=-1\\[6pt] x-18.4=\text{cos}^{-1}\,\left(-1\right)\\[6pt] x-18.4=180\\[6pt] x=180+18.4\\[6pt] x=198.4^\circ \end{gather} $$

Other methods exist, but a good way of solving equations of this type is use the wave function.

Let's express \(-3\,\text{cos}\,x^\circ+2\,\text{sin}\,x\) in the form \(k\,\text{cos}\,(x-a)^\circ\) where \(k\gt 0\) and \(0\leqslant a \lt 360\small.\)

$$ \begin{eqnarray} && \!\!\!\!\!\!k\,\text{cos}\,(x-a) \\[6pt] &=& k\,\text{cos}\,x\,\text{cos}\,a+k\,\text{sin}\,x\,\text{sin}\,a \\[6pt] &=& k\,\text{cos}\,a\,\text{cos}\,x+k\,\text{sin}\,a\,\text{sin}\,x \\[6pt] &=& \underbrace{k\,\text{cos}\,a}_{\large{-3}}\,\text{cos}\,x+\underbrace{k\,\text{sin}\,a}_{\large{2}}\,\text{sin}\,x \end{eqnarray} $$

$$ \begin{eqnarray} \therefore\:\frac{k\,\text{sin}\,a}{k\,\text{cos}\,a} &=& \frac{2}{-3} \\[6pt] \text{tan}\,a &=& -\!\frac{2}{3} \\[6pt] \end{eqnarray} $$

\(\text{sin}\,a\gt 0\) but \(\text{cos}\,a\lt 0\small,\) so \(a\) is in the second (S) quadrant. The related acute angle is \(\text{tan}^{-1}\,\large\frac{2}{3}\normalsize=33.7^\circ\) (to 1 d.p.) so \(a=180-33.7=146.3^\circ\small.\)

$$ \begin{eqnarray} k^2 &=& (-3)^2+2^2 \\[6pt] &=& 9+4 \\[6pt] &=& 13 \\[6pt] k &=& \sqrt{13} \end{eqnarray} $$

\(\therefore\:-3\,\text{cos}\,x^\circ+2\,\text{sin}\,x^\circ=\sqrt{13}\,\text{cos}\,(x-146.3)^\circ\)

Now we are ready to start solving the equation:

$$ \begin{gather} -3\,\text{cos}\,x^\circ+2\,\text{sin}\,x^\circ=-1\\[7pt] \sqrt{13}\,\text{cos}\,(x-146.3)^\circ=-1\\[7pt] \text{cos}\,(x-146.3)^\circ=-\frac{1}{\sqrt{13}} \end{gather} $$

The related acute angle is \(\text{cos}^{-1}\,\large\frac{1}{\sqrt{13}}\normalsize=73.9^\circ\) (accurate to 1 d.p.)

\(\text{cos}\,(x-146.3)^\circ \lt 0\) so it has solutions in the second (S) and third (T) quadrants.

In the second quadrant:

$$ \begin{eqnarray} x-146.3 &=& 180-73.9 \\[6pt] x &=& 180-73.9+146.3 \\[6pt] &=& 252.4^\circ \end{eqnarray} $$

\(252.4^\circ\) is in the required interval, so it doesn't need any adjustment.

Now, in the third quadrant:

$$ \begin{eqnarray} x-146.3 &=& 180+73.9 \\[6pt] x &=& 180+73.9+146.3 \\[6pt] &=& 400.2^\circ \end{eqnarray} $$

However, \(400.2^\circ\) is outwith the required interval, so we subtract \(360^\circ\) to obtain the correct solution: \(400.2-360=40.2^\circ\small.\)

Solutions: \(x=40.2^\circ\small,\normalsize\:x=252.4^{\circ}\)

This example is about as difficult as the wave function can get! It has double angles and is in radians.

First, we will express \(5\,\text{sin}\,2x-4\,\text{cos}\,2x\) in the form \(k\,\text{cos}\,(2x-a)\) where \(k\gt 0\) and \(0\lt a \lt 2\pi\small.\)

$$ \begin{eqnarray} && \!\!\!\!\!\!k\,\text{cos}\,(2x-a) \\[6pt] &=& k\,\text{cos}\,2x\,\text{cos}\,a+k\,\text{sin}\,2x\,\text{sin}\,a \\[6pt] &=& k\,\text{cos}\,a\,\text{cos}\,2x+k\,\text{sin}\,a\,\text{sin}\,2x \\[6pt] &=& \underbrace{k\,\text{sin}\,a}_{\large{5}}\,\text{sin}\,2x+\underbrace{k\,\text{cos}\,a}_{\large{-4}}\,\text{cos}\,2x \end{eqnarray} $$

$$ \begin{eqnarray} \frac{k\,\text{sin}\,a}{k\,\text{cos}\,a} &=& \frac{5}{-4} \\[6pt] \text{tan}\,a &=& -\!\frac{5}{4} \\[6pt] \end{eqnarray} $$

\(\text{sin}\,a\gt 0\) but \(\text{cos}\,a\lt 0,\) so \(a\) is in the second (S) quadrant. The related acute angle is \(\text{tan}^{-1}\,\large\frac{5}{4}\normalsize=0.896\) (to 3 d.p.) so \(a=\pi-0.896=2.246\small.\)

$$ \begin{eqnarray} k^2 &=& 5^2+(-4)^2 \\[6pt] &=& 25+16 \\[6pt] &=& 41 \\[6pt] k &=& \sqrt{41} \end{eqnarray} $$

\(\therefore\:5\,\text{sin}\,2x-4\,\text{cos}\,2x=\sqrt{41}\,\text{cos}\,(2x-2.246)\)

Now we can start solving the equation:

$$ \begin{gather} 5\,\text{sin}\,2x-4\,\text{cos}\,2x=3\\[7pt] \sqrt{41}\,\text{cos}\,(2x-2.246)=3\\[7pt] \text{cos}\,(2x-2.246)=\frac{3}{\sqrt{41}} \end{gather} $$

The related acute angle is \(\text{cos}^{-1}\,\large\frac{3}{\sqrt{41}}\normalsize=1.083\) (to 3 d.p.)

\(\text{cos}\,(2x-2.246) \gt 0\) so it has solutions in the first (A) and fourth (C) quadrants.

In the first quadrant:

$$ \begin{eqnarray} 2x-2.246 &=& 1.083 \\[6pt] 2x &=& 1.083+2.246 \\[6pt] 2x &=& 3.329 \end{eqnarray} $$

We need to find all solutions in the range \(0\leqslant x\lt 2\pi,\) which corresponds to \(0\leqslant 2x\lt 4\pi, \) so there is another solution \(2x=3.329+2\pi=9.612.\)

Now, in the fourth quadrant:

$$ \begin{eqnarray} 2x-2.246 &=& 2\pi-1.083 \\[6pt] 2x &=& 2\pi-1.083+2.246 \\[6pt] 2x &=& 7.446 \end{eqnarray} $$

Note that \(7.446\) is above \(2\pi,\) so there is another solution \(2x=7.446-2\pi=1.163.\)

Putting this together, we have these four values:

\(2x= 1.163\small,\normalsize\:3.329\small,\normalsize\:7.446\small,\normalsize\:9.612\)

Only now should we halve these to obtain the final solutions for \(x\).

Because, after halving, there is a possibility that the third decimal places include a rounding error, we should drop the last decimal place and give our solutions for \(x\) correct to only two decimal places.

\(x=0.58\small,\normalsize\:1.66\small,\normalsize\:3.72\small,\normalsize\:4.81\)

Here, the right hand side is much simpler than the left hand side.

So it makes sense to start with the LHS and use known identities to simplify it.

$$ \begin{eqnarray} \textsf{LHS} &=& \small\frac{\text{sin}\,2x}{2\,\text{cos}\,x}\normalsize-\text{sin}\,x\,\text{cos}^2\,x \\[9pt] &=& \small\frac{2\,\text{sin}\,x\,\text{cos}\,x}{2\,\text{cos}\,x}\normalsize-\text{sin}\,x\,\text{cos}^2\,x \\[9pt] &=& \small\frac{\cancel{2}\,\text{sin}\,x\,\cancel{\text{cos}\,x}}{\cancel{2}\,\cancel{\text{cos}\,x}}\normalsize-\text{sin}\,x\,\text{cos}^2\,x \\[9pt] &=& \text{sin}\,x - \text{sin}\,x\,\text{cos}^2\,x \\[9pt] &=& (\text{sin}\,x)(1 - \text{cos}^2\,x) \\[9pt] &=& (\text{sin}\,x)(\text{sin}^2\,x) \\[9pt] &=& \text{sin}^3\,x \\[9pt] &=& \textsf{RHS} \end{eqnarray} $$