Using Identities in Equations | CAIE 9709 Pure Mathematics 2 Notes

Pure Mathematics 2

Mathematics9709

Use the proved result with its exact coefficient

Hence asks you to use the preceding result. If a proof gives $2\sin\theta\cosec2\theta=\sec\theta$ , then one copy of $\sin\theta\cosec2\theta$ is $\tfrac12\sec\theta$ , not a whole secant.

Worked example

Solve tan²θ + 7 sin θ cosec 2θ = 8 for −π < θ < π.

(9709/22/M/J/24 Q7(b)) Keep $\sin2\theta\ne0$ . Use the proved identity and $\tan^2\theta=\sec^2\theta-1$ .

\begin{gathered}\sec^2\theta-1+\frac72\sec\theta=8\\u=\sec\theta\\2u^2+7u-18=0\\u=\frac{-7\pm\sqrt{193}}4\end{gathered}

The polynomial solver gives awkward roots, so use the quadratic formula for supported exact working. Both have magnitude greater than 1; keep both. Convert each using $\cos\theta=1/u$ and find both signs of the angle in the interval.

Answer

\theta\approx\pm0.952,\ \pm1.76

These are radians, to 3 significant figures. None makes $\sin2\theta$ zero. Keep full root and angle values on the calculator until the final line.

Your turnPractice

Solve $\tan\theta+\cot\theta=4$ for $0<\theta<\pi$ , giving exact answers.

Show worked answer

\begin{gathered}\tan\theta+\cot\theta\\=\frac{\sin^2\theta+\cos^2\theta}{\sin\theta\cos\theta}\\=\frac2{\sin2\theta}\\\sin2\theta=\frac12\\0<2\theta<2\pi\end{gathered}

Answer

\theta=\frac\pi{12},\ \frac{5\pi}{12}

The original requires both sine and cosine non-zero; both solutions pass. Alternatively, a quadratic in tangent gives $2\pm\sqrt3$ .

Choose a rewrite that matches the other terms

Inspect the whole equation before choosing an identity. An unsquared ratio often identifies the useful quadratic variable; a product may factor without any division. A graph intersection gives an equation by setting the two outputs equal.

Your turnPractice

Find the intersections of $y=1+2\cos2x$ and $y=4\cos^2x-\sin x$ for $0\le x\le2\pi$ .

Show worked answer

\begin{gathered}1+2(2\cos^2x-1)\\=4\cos^2x-\sin x\\1-2=-\sin x\\\sin x=1\\x=\frac\pi2,\qquad y=-1\end{gathered}

Answer

\left(\frac\pi2,-1\right)

Both graph formulae give $y=-1$ at this angle. The equal cosine-squared terms cancel, so no quadratic solve is needed.

Use the proved result with its exact coefficient

Hence asks you to use the preceding result. If a proof gives $2\sin\theta\cosec2\theta=\sec\theta$ , then one copy of $\sin\theta\cosec2\theta$ is $\tfrac12\sec\theta$ , not a whole secant.

Worked example

Solve tan²θ + 7 sin θ cosec 2θ = 8 for −π < θ < π.

(9709/22/M/J/24 Q7(b)) Keep $\sin2\theta\ne0$ . Use the proved identity and $\tan^2\theta=\sec^2\theta-1$ .

\begin{gathered}\sec^2\theta-1+\frac72\sec\theta=8\\u=\sec\theta\\2u^2+7u-18=0\\u=\frac{-7\pm\sqrt{193}}4\end{gathered}

Answer

\theta\approx\pm0.952,\ \pm1.76

These are radians, to 3 significant figures. None makes $\sin2\theta$ zero. Keep full root and angle values on the calculator until the final line.

Your turnPractice

Solve $\tan\theta+\cot\theta=4$ for $0<\theta<\pi$ , giving exact answers.

Show worked answer

\begin{gathered}\tan\theta+\cot\theta\\=\frac{\sin^2\theta+\cos^2\theta}{\sin\theta\cos\theta}\\=\frac2{\sin2\theta}\\\sin2\theta=\frac12\\0<2\theta<2\pi\end{gathered}

Answer

\theta=\frac\pi{12},\ \frac{5\pi}{12}

The original requires both sine and cosine non-zero; both solutions pass. Alternatively, a quadratic in tangent gives $2\pm\sqrt3$ .

Choose a rewrite that matches the other terms

Your turnPractice

Find the intersections of $y=1+2\cos2x$ and $y=4\cos^2x-\sin x$ for $0\le x\le2\pi$ .

Show worked answer

\begin{gathered}1+2(2\cos^2x-1)\\=4\cos^2x-\sin x\\1-2=-\sin x\\\sin x=1\\x=\frac\pi2,\qquad y=-1\end{gathered}

Answer

\left(\frac\pi2,-1\right)

Both graph formulae give $y=-1$ at this angle. The equal cosine-squared terms cancel, so no quadratic solve is needed.