Further Trig Identities | CAIE 9709 Pure Mathematics 2 Notes

Pure Mathematics 2

Mathematics9709

Rewrite one side into the other

A proof must work for every angle in the original domain. Numerical substitutions can check a result but cannot prove an identity. Start with the more complicated side and choose a rewrite suggested by the target.

Reciprocal names often simplify into sine and cosine. A squared pair suggests a Pythagorean identity; a doubled input suggests a double-angle formula.

Worked example

Prove that 2 sin θ cosec 2θ ≡ sec θ.

(9709/22/M/J/24 Q7(a)) Start with the left-hand side. It requires $\sin2\theta\ne0$ , so both $\sin\theta$ and $\cos\theta$ are non-zero.

\begin{gathered}2\sin\theta\cosec2\theta =\frac{2\sin\theta}{\sin2\theta}\\ =\frac{2\sin\theta}{2\sin\theta\cos\theta}\\ =\frac1{\cos\theta}=\sec\theta\end{gathered}

Cancelling sine is valid on that original domain. The simplified expression is also defined at some extra angles, such as zero, but those are not newly allowed values of the original expression.

Your turnPractice

Prove $\sin x+\cos x\cot x\equiv\cosec x$ , stating the original restriction.

Show worked answer

\begin{gathered}\sin x+\frac{\cos^2x}{\sin x}\\ =\frac{\sin^2x+\cos^2x}{\sin x}\\ =\frac1{\sin x}=\cosec x\end{gathered}

Require $\sin x\ne0$ throughout.

Combine fractions before applying the identity

For fractions with different denominators, use a common denominator. Here the product of a sum and a difference gives a difference of squares.

Worked example

Simplify 1/(1 + cos x) + 1/(1 − cos x).

The original denominators require $\cos x\ne1,-1$ .

\begin{gathered}\frac1{1+\cos x}+\frac1{1-\cos x}\\=\frac2{1-\cos^2x}\\=\frac2{\sin^2x}\\=2\cosec^2x\end{gathered}

The numerator loses its cosine terms; the denominator becomes a squared sine. Both sides exclude the same multiples of $\pi$ .

Your turnPractice

Prove $(1-\cos2x)/\sin2x\equiv\tan x$ . Is the original expression defined at $x=0$ ?

Show worked answer

\begin{gathered}\frac{1-\cos2x}{\sin2x}\\=\frac{2\sin^2x}{2\sin x\cos x}\\=\tan x\end{gathered}

The original requires $\sin2x\ne0$ . It is not defined at zero, even though $\tan0=0$ .

Rewrite one side into the other

Reciprocal names often simplify into sine and cosine. A squared pair suggests a Pythagorean identity; a doubled input suggests a double-angle formula.

Worked example

Prove that 2 sin θ cosec 2θ ≡ sec θ.

(9709/22/M/J/24 Q7(a)) Start with the left-hand side. It requires $\sin2\theta\ne0$ , so both $\sin\theta$ and $\cos\theta$ are non-zero.

\begin{gathered}2\sin\theta\cosec2\theta =\frac{2\sin\theta}{\sin2\theta}\\ =\frac{2\sin\theta}{2\sin\theta\cos\theta}\\ =\frac1{\cos\theta}=\sec\theta\end{gathered}

Cancelling sine is valid on that original domain. The simplified expression is also defined at some extra angles, such as zero, but those are not newly allowed values of the original expression.

Your turnPractice

Prove $\sin x+\cos x\cot x\equiv\cosec x$ , stating the original restriction.

Show worked answer

\begin{gathered}\sin x+\frac{\cos^2x}{\sin x}\\ =\frac{\sin^2x+\cos^2x}{\sin x}\\ =\frac1{\sin x}=\cosec x\end{gathered}

Require $\sin x\ne0$ throughout.

Combine fractions before applying the identity

For fractions with different denominators, use a common denominator. Here the product of a sum and a difference gives a difference of squares.

Worked example

Simplify 1/(1 + cos x) + 1/(1 − cos x).

The original denominators require $\cos x\ne1,-1$ .

\begin{gathered}\frac1{1+\cos x}+\frac1{1-\cos x}\\=\frac2{1-\cos^2x}\\=\frac2{\sin^2x}\\=2\cosec^2x\end{gathered}

The numerator loses its cosine terms; the denominator becomes a squared sine. Both sides exclude the same multiples of $\pi$ .

Your turnPractice

Prove $(1-\cos2x)/\sin2x\equiv\tan x$ . Is the original expression defined at $x=0$ ?

Show worked answer

\begin{gathered}\frac{1-\cos2x}{\sin2x}\\=\frac{2\sin^2x}{2\sin x\cos x}\\=\tan x\end{gathered}

The original requires $\sin2x\ne0$ . It is not defined at zero, even though $\tan0=0$ .