Equations with Compound Angles | CAIE 9709 Pure Mathematics 2 Notes

Pure Mathematics 2

Mathematics9709

Expand when different angle brackets occur

When both sides involve $x$ with different shifts, expand into $\sin x$ and $\cos x$ . A relation such as $a\cos x=b\sin x$ can then become a tangent equation, after checking cosine is not zero.

Worked example

Solve sin(60° − x) = 2 sin x for 0° < x < 360°.

\begin{gathered}\frac{\sqrt3}{2}\cos x\\=2\sin x+\frac12\sin x\\\sqrt3\cos x=5\sin x\end{gathered}

If cosine were zero, sine would be 1 or −1, so this equation would fail. Dividing by cosine is therefore safe.

\tan x=\frac{\sqrt3}{5}

Answer

x\approx19.1^\circ,\ 199.1^\circ

Tangent repeats every $180^\circ$ . Check the unrounded values in the original equation.

Your turnPractice

Solve $\cos(x+30^\circ)=2\sin x$ for $0^\circ\le x\le360^\circ$ .

Show worked answer

\begin{gathered}\frac{\sqrt3}{2}\cos x\\=2\sin x+\frac12\sin x\end{gathered}

This gives the same tangent equation, hence $x\approx19.1^\circ,199.1^\circ$ . Neither endpoint satisfies the original equation.

Keep the restrictions when clearing a denominator

Worked example

Solve tan(x + 45°) = 6 tan x for 0° ≤ x < 180°.

Let $t=\tan x$ . The original equation excludes $x=90^\circ$ and $x=45^\circ$ . The addition formula gives a quotient, so keep its denominator non-zero.

\begin{gathered}\frac{t+1}{1-t}=6t,\qquad t\ne1\\ 6t^2-5t+1=0\\(3t-1)(2t-1)=0\\t=\frac13,\ \frac12\end{gathered}

The calculator roots give convenient factors. Use inverse tangent for each root; adding $180^\circ$ would leave the interval.

Answer

x\approx18.4^\circ,\ 26.6^\circ

Neither is excluded. Substitute the unrounded angles into the original equation to check.

The discriminant $b^2-4ac$ tests whether a quadratic has real roots. If it is negative, the trigonometric equation has no real solutions through that quadratic either.

Your turnPractice

Solve $\tan(x+45^\circ)=3\tan x$ for $0^\circ\le x<180^\circ$ .

Show worked answer

\begin{gathered}\frac{t+1}{1-t}=3t\\3t^2-2t+1=0\end{gathered}

The discriminant is $(-2)^2-4(3)(1)=-8$ , so there are no real solutions. Clearing the denominator does not make the excluded angles valid.

Expand when different angle brackets occur

When both sides involve $x$ with different shifts, expand into $\sin x$ and $\cos x$ . A relation such as $a\cos x=b\sin x$ can then become a tangent equation, after checking cosine is not zero.

Worked example

Solve sin(60° − x) = 2 sin x for 0° < x < 360°.

\begin{gathered}\frac{\sqrt3}{2}\cos x\\=2\sin x+\frac12\sin x\\\sqrt3\cos x=5\sin x\end{gathered}

If cosine were zero, sine would be 1 or −1, so this equation would fail. Dividing by cosine is therefore safe.

\tan x=\frac{\sqrt3}{5}

Answer

x\approx19.1^\circ,\ 199.1^\circ

Tangent repeats every $180^\circ$ . Check the unrounded values in the original equation.

Your turnPractice

Solve $\cos(x+30^\circ)=2\sin x$ for $0^\circ\le x\le360^\circ$ .

Show worked answer

\begin{gathered}\frac{\sqrt3}{2}\cos x\\=2\sin x+\frac12\sin x\end{gathered}

This gives the same tangent equation, hence $x\approx19.1^\circ,199.1^\circ$ . Neither endpoint satisfies the original equation.

Keep the restrictions when clearing a denominator

Worked example

Solve tan(x + 45°) = 6 tan x for 0° ≤ x < 180°.

Let $t=\tan x$ . The original equation excludes $x=90^\circ$ and $x=45^\circ$ . The addition formula gives a quotient, so keep its denominator non-zero.

\begin{gathered}\frac{t+1}{1-t}=6t,\qquad t\ne1\\ 6t^2-5t+1=0\\(3t-1)(2t-1)=0\\t=\frac13,\ \frac12\end{gathered}

The calculator roots give convenient factors. Use inverse tangent for each root; adding $180^\circ$ would leave the interval.

Answer

x\approx18.4^\circ,\ 26.6^\circ

Neither is excluded. Substitute the unrounded angles into the original equation to check.

The discriminant $b^2-4ac$ tests whether a quadratic has real roots. If it is negative, the trigonometric equation has no real solutions through that quadratic either.

Your turnPractice

Solve $\tan(x+45^\circ)=3\tan x$ for $0^\circ\le x<180^\circ$ .

Show worked answer

\begin{gathered}\frac{t+1}{1-t}=3t\\3t^2-2t+1=0\end{gathered}

The discriminant is $(-2)^2-4(3)(1)=-8$ , so there are no real solutions. Clearing the denominator does not make the excluded angles valid.