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

$$ \begin{eqnarray} cos\,75^\circ &=& cos(45+30)^\circ\\[6pt] &=& cos\,45^\circ\,cos\,30^\circ- sin\,45^\circ\,sin\,30^\circ\\[6pt] &=& \small\frac{1}{\sqrt{2}}\times\frac{\sqrt{3}}{2}\normalsize - \small\frac{1}{\sqrt{2}}\times\frac{1}{2}\normalsize \\[6pt] &=& \small\frac{\sqrt{3}-1}{2\sqrt{2}}\normalsize \end{eqnarray} $$

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

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

$$ \begin{eqnarray} sin\,15^\circ &=& sin(45-30)^\circ\\[6pt] &=& sin\,45^\circ\,cos\,30^\circ- cos\,45^\circ\,sin\,30^\circ\\[6pt] &=& \small\frac{1}{\sqrt{2}}\times\frac{\sqrt{3}}{2}\normalsize - \small\frac{1}{\sqrt{2}}\times\frac{1}{2}\normalsize \\[6pt] &=& \small\frac{\sqrt{3}-1}{2\sqrt{2}}\normalsize \end{eqnarray} $$

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

There is only one trig identity for \(sin\,2x\) so we need to know the value of \(cos\,x.\)

The question tells us that \(x\) is an acute angle, so we can imagine it inside a right-angled triangle with opposite side \(5\) units and hypotenuse \(13\) units. Then we will use Pythagoras 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 now we know that \(cos\,x=\frac{12}{13}\) and we're ready to find \(sin\,2x.\)

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

The question tells us that \(2x\) is an acute angle, so we can imagine it inside a right-angled triangle with opposite side \(4\) units and adjacent side \(3\) units. The hypotenuse is obviously \(5\) units, so \(cos\,2x=\frac{3}{5}.\)

Now we use the trig identity that links \(cos\,2x\) and \(cos\,x.\)

$$ \begin{eqnarray} cos\,2x &=& 2\,cos^{2}\,x-1 \\[6pt] \small\frac{3}{5}\normalsize &=& 2\,cos^{2}\,x-1 \\[6pt] \small\frac{3}{5}\normalsize &=& 2\,cos^{2}\,x-\small\frac{5}{5}\normalsize \\[6pt] \small\frac{3}{5}\normalsize +\small\frac{5}{5}\normalsize &=& 2\,cos^{2}\,x \\[6pt] \small\frac{8}{5}\normalsize &=& 2\,cos^{2}\,x \\[6pt] \small\frac{4}{5}\normalsize &=& cos^{2}\,x \\[6pt] cos\,x &=& \small\sqrt{\frac{4}{5}}\normalsize \\[6pt] cos\,x &=& \small\frac{2}{\sqrt{5}}\normalsize \end{eqnarray} $$

You will notice that \(tan\,x\) and \(tan\,2x\) are absent from the list of trig identities at the top of this page. In fact, the only identity involving \(tan\,x\) that you are expected to know is from Nat 5: \(tan\,x=\large\frac{sin\,x}{cos\,x}\normalsize.\) We will use that here.

We will use one of the identities for \(cos\,2x\) to find \(sin^{2}\,x.\)

$$ \begin{eqnarray} cos\,2x &=& 1-2\,sin^{2}\,x \\[6pt] \small\frac{12}{13}\normalsize &=& \small\frac{13}{13}-2\,sin^{2}\,x \\[6pt] 2\,sin^{2}\,x &=& \small\frac{13}{13}-\small\frac{12}{13}\normalsize \\[6pt] 2\,sin^{2}\,x &=& \small\frac{1}{13} \\[6pt] sin^{2}\,x &=& \small\frac{1}{26}\normalsize \end{eqnarray} $$

Now to find \(cos^{2}\,x.\) One way to do this is to use this Nat 5 trig identity:

$$ \begin{eqnarray} sin^{2}\,x+cos^{2}\,x &=& 1 \\[6pt] \small\frac{1}{26}\normalsize+cos^{2}\,x &=& 1 \\[6pt] cos^{2}\,x &=& \small\frac{25}{26}\normalsize \end{eqnarray} $$

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

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

(a)  Clearly, \(x^\circ\) is in a \(3,4,5\) triangle, so \(cos\,x=\large\frac{4}{5}\small.\)

(b)  We use the only relevant identity:

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

(c)  We could use any of the three identities for \(cos\,2x\) but, as \(2x\) is still an acute angle, it is easier to simply find the adjacent side in the right-angled triangle containing \(2x^{\circ}\small.\)

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

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

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

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

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

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

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

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

So there are 3 solutions in the given interval: \(x=\large\frac{\pi}{6}\normalsize,\ \large\frac{5\pi}{6}\normalsize,\ \large\frac{3\pi}{2}\normalsize\)

This trigonometric equation involves the double angle \(2x\). We need to find all solutions, in degrees, in the range \(0\leq x \lt 360,\) so \(0\leq 2x \lt 720.\)

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

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

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

However, we need to take care to find all the values of \(2x\) in the range \(0\leq 2x \lt 720^\circ.\) 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 \(240+360=600.\) These are both in range.

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

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

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

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

Because \(cos\,x\) could be either positive or negative, there are solutions 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.

$$ \begin{eqnarray} sin\,2x+6\,cos\,x &=& 0 \\[6pt] 2\,sin\,x\,cos\,x+6\,cos\,x &=& 0 \\[6pt] (cos\,x)(2\,sin\,x+6) &=& 0 \\[6pt] cos\,x=0\:\:\ &\small\textsf{or}\normalsize&\:\,2\,sin\,x+6=0 \\[6pt] x=90^\circ, 270^\circ\:\:\ &\small\textsf{or}\normalsize&\:\,sin\,x=-3 \\[6pt] \:\:&\,&\:\, \small\textsf{(no solutions)}\normalsize\\[6pt] \end{eqnarray} $$

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

The interval \(0\leq x \lt 360\) corresponds to \(0\leq 2x \lt 720\small,\) so we need to find all solutions in this interval.

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

This trigonometric equation involves the triple angle \(3x\). We need to find all solutions, in degrees, in the range \(0\leq x \lt 180,\) so \(0\leq 3x \lt 540\) and \(-60\leq 3x-60 \lt 480.\)

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

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

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

However, we need to take care to find all the values of \(3x-60\) in the range \(-60\leq 3x-60 \lt 480.\) 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 \(3x-60=-30^\circ,210^\circ,330^\circ.\)

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

Dividing by \(3,\) we obtain \(x=10^\circ,90^\circ,130^\circ.\)

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 bring all terms to the left, factorise and solve using National 5 methods.

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

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 possible to get into an awful muddle if you don't follow a correct process very precisely. We recommend the same method that is detailed in the HSN notes, where you will find several other examples like this.

First, expand \(k\,cos(x-a)\) using the relevant addition formula and then rearrange it to compare it with the given expression.

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

So now we know that \(k\,cos\,a=\sqrt{3}\) and \(k\,sin\,a=1.\) The next step is to work out which quadrant \(a\) is in. Here, both \(sin\,a\) and \(cos\,a\) are positive, so \(a\) is in the first quadrant.

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

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

There is an easy formula to find \(k\) 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:

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

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

Now we can state our final answer:

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

First, expand \(k\,cos(x-a)\) using the relevant addition formula and then rearrange it to compare it with the given expression.

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

So now we know that \(k\,cos\,a=2\) and \(k\,sin\,a=-3.\) The next step is to work out which quadrant \(a\) is in. Here, \(cos\,a\gt 0\) which limits \(a\) to the first (A) or fourth (C) quadrant. But \(sin\,a\lt 0,\) 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:

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

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

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:

\(2\,cos\,x^\circ-3sin\,x^\circ=\sqrt{13}\ cos(x-303.7)^\circ\) (correct to 1 d.p.)

First, expand \(k\,cos(x+a)\) using the relevant addition formula and then rearrange it to compare it with the given expression. This example needs a little more rearrangement than the previous two examples.

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

So now we know that \(-k\,sin\,a=5\) and \(k\,cos\,a=-2.\) The next step is to work out which quadrant \(a\) is in. Here, both \(cos\,a\) and \(sin\,a\) are negative, which means that \(a\) can only be in the third (T) quadrant.

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

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

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

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:

\(5\,sin\,x^\circ-2\,cos\,x^\circ=\sqrt{29}\ cos(x+248.2)^\circ\) (correct to 1 d.p.)

Note that this solution should use radians throughout.

First, expand \(k\,sin(x+a)\) using the relevant addition formula and then rearrange it to compare it with the given expression.

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

So now we know that \(k\,cos\,a=1\) and \(k\,sin\,a=-\sqrt{3}.\) The next step is to work out which quadrant \(a\) is in. As \(cos\,a\gt 0,\) the first (A) or fourth (C) quadrants are possibilities. But \(sin\,a\lt 0\) 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:

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

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

Finally, we find \(k\):

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

So our final answer is:

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

Note that this solution should use radians throughout.

First, expand \(k\,sin(x-a)\) using the relevant addition formula and then rearrange it to compare it with the given expression. This example needs extreme care over the \(+\) and \(-\) signs.

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

So now we know that \(k\,cos\,a=-2\) and \(-k\,sin\,a=7.\) The next step is to work out which quadrant \(a\) is in. As both \(cos\,a\) and \(sin\,a\) are negative, \(a\) must be in the third (T) quadrant.

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

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

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

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:

\(-2sin\,x+7\,cos\,x=\sqrt{53}\,sin(x-4.434)\) (correct to 3 d.p.)

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

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

\(sin\,a\) and \(cos\,a\) are both positive, so \(a\) is in the first quadrant. So \(a=tan^{-1}\,\frac{1}{3}=18.4^\circ\) (to 1 d.p.)

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

\(3\,cos\,x^\circ+sin\,x^\circ=\sqrt{10}\,cos(x-18.4)^\circ\) (to 1 d.p.)

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

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

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

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

There is no way to solve equations of this type without the wave function. So we will express \(-3\,cos\,x^\circ+2\,sin\,x\) in the form \(k\,cos(x-a)^\circ\) where \(k\gt 0\) and \(0\lt a \lt 360.\)

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

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

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

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

\(-3\,cos\,x^\circ+2\,sin\,x^\circ=\sqrt{13}\,cos(x-146.3)^\circ\) (to 1 d.p.)

Now we are ready to start solving the equation:

$$ \begin{eqnarray} -3\,cos\,x^\circ+2\,sin\,x^\circ &=& -1 \\[6pt] \sqrt{13}\,cos(x-146.3)^\circ &=& -1 \\[6pt] cos(x-146.3)^\circ &=& \frac{-1}{\sqrt{13}} \end{eqnarray} $$

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

\(cos(x-146.3)^\circ \lt 0\) 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 range, 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, note that \(400.2^\circ\) is out of the required range, so we subtract \(360^\circ\) to obtain the correct solution. \(400.2-360=40.2^\circ\) (to 1 d.p.)

Solutions: \(x=40.2^\circ\) or \(x=252.4^{\circ}\)

This final example ramps things up to just about the maximum level of difficulty. It has double angles and is in radians.

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

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

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

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

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

\(5\,sin\,2x-4\,cos\,2x=\sqrt{41}\,cos(2x-2.246)\) (to 3 d.p.)

Now we can solve the equation:

$$ \begin{eqnarray} 5\,sin\,2x-4\,cos\,2x &=& 3 \\[6pt] \sqrt{41}\,cos(2x-2.246) &=& 3 \\[6pt] cos(2x-2.246) &=& \frac{3}{\sqrt{41}} \end{eqnarray} $$

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

\(cos(2x-2.246) \gt 0\) 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 \(0\leq x \lt 2\pi,\) which corresponds to \(0\leq 2x \lt 4\pi, \) 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:

\(2x= 1.163, 3.329, 7.446, 9.612\)

Only now should we halve these to obtain the final solutions for \(x\). Because 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 2 d.p.

\(x=0.58, 1.66, 3.72, 4.81\)

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.

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