Mathematics 9709 · Pure Mathematics 2

Further trigonometric identities

Proving an identity is not the same as solving. You rewrite one side until it turns into the other.
Never move terms across the $\equiv$ sign.
Start on the messier side, open it up with the compound and double angle formulae, then tidy until it matches.

Work on one side only, usually the busier one. Three moves handle almost every proof: open up compound/double angles into $\sin$ and $\cos$ , choose the $\cos2A$ form that cancels the awkward term, and split the fraction $\tfrac{a+b}{c}=\tfrac{a}{c}+\tfrac{b}{c}$ to bring out simple ratios.

Expand the messy side, then simplify

Open up any compound or double angle into $\sin$ and $\cos$ .
Choose the $\cos2A$ form that cancels the awkward term (leaves a clean square).
Split the fraction $\tfrac{a+b}{c}=\tfrac{a}{c}+\tfrac{b}{c}$ to bring out simple ratios like $\cot$ and $\tan$ .

Worked examples

Worked example

Prove $\dfrac{1+\cos2x}{\sin2x}=\cot x$

Choose the cos form that makes a clean square on top: $1+\cos2x=1+(2\cos^2x-1)=2\cos^2x$ .
Expand the bottom with the double angle: $\sin2x=2\sin x\cos x$ , giving $\dfrac{2\cos^2x}{2\sin x\cos x}$ .
Cancel $2\cos x$ top and bottom: $=\dfrac{\cos x}{\sin x}=\cot x$ . ✓

Answer

\dfrac{1+\cos2x}{\sin2x}\equiv\cot x

Worked example

Real question: prove $\operatorname{cosec}2\theta-\cot2\theta\equiv\tan\theta$

9709/32 M/J 2021 Q6(a). First put both reciprocals over the one bottom, $\sin2\theta$ . Then a double angle finishes it.

Write both terms over the same bottom: $\operatorname{cosec}2\theta-\cot2\theta=\dfrac{1}{\sin2\theta}-\dfrac{\cos2\theta}{\sin2\theta}=\dfrac{1-\cos2\theta}{\sin2\theta}$ .
On top, pick the form that makes a clean square: $1-\cos2\theta=1-(1-2\sin^2\theta)=2\sin^2\theta$ . On the bottom, $\sin2\theta=2\sin\theta\cos\theta$ .
So $\dfrac{2\sin^2\theta}{2\sin\theta\cos\theta}=\dfrac{\sin\theta}{\cos\theta}=\tan\theta$ . ✓

Answer

\operatorname{cosec}2\theta-\cot2\theta\equiv\tan\theta

Why examiners always add a part (b)

The proof is never the real question

Part (a) proves $\operatorname{cosec}2\theta-\cot2\theta\equiv\tan\theta$ . Part (b) of that same paper then asks you to integrate $\operatorname{cosec}2\theta-\cot2\theta$ . That looks impossible. But (a) lets you swap the whole mess for $\tan\theta$ , and $\displaystyle\int_{\pi/4}^{\pi/3}\tan\theta\,d\theta=\big[-\ln\cos\theta\big]_{\pi/4}^{\pi/3}=\tfrac12\ln2$ . (9709/32 Jun 2021 Q6)

That is why proofs are on this paper: they let you swap a hard thing to integrate for an easy one. When you see “prove this identity, hence integrate”, use the proved form in the integral.

Where the marks go

Common mistake

Picking the

\cos2A

form at random sends a proof to a dead end. If the line has

\sin^2

, use

1-2\sin^2A

; if it has

\cos^2

, use

2\cos^2A-1

Common mistake

Proofs are marked on the working. Show every step and the fraction split, even when the last line looks obvious. And never move terms across the

\equiv

Now you try

Show that $\cos^4\theta-\sin^4\theta\equiv\cos2\theta$ . (9709-style)

Your turn— tap to reveal the worked answer (9709-style)

The left side is a difference of two squares: $\cos^4\theta-\sin^4\theta=(\cos^2\theta-\sin^2\theta)(\cos^2\theta+\sin^2\theta)$ .

The second bracket is $1$ , and the first is exactly the $\cos2\theta$ form:

Answer

=\cos^2\theta-\sin^2\theta=\cos2\theta\ \checkmark

PreviousDouble angle formulae NextExpressing a sin θ + b cos θ as R sin(θ ± α) or R cos(θ ∓ α)

Further trigonometric identities

Proving an identity is not the same as solving. You rewrite one side until it turns into the other.
Never move terms across the $\equiv$ sign.
Start on the messier side, open it up with the compound and double angle formulae, then tidy until it matches.

Expand the messy side, then simplify

Open up any compound or double angle into $\sin$ and $\cos$ .
Choose the $\cos2A$ form that cancels the awkward term (leaves a clean square).
Split the fraction $\tfrac{a+b}{c}=\tfrac{a}{c}+\tfrac{b}{c}$ to bring out simple ratios like $\cot$ and $\tan$ .

Worked examples

Worked example

Prove $\dfrac{1+\cos2x}{\sin2x}=\cot x$

Choose the cos form that makes a clean square on top: $1+\cos2x=1+(2\cos^2x-1)=2\cos^2x$ .
Expand the bottom with the double angle: $\sin2x=2\sin x\cos x$ , giving $\dfrac{2\cos^2x}{2\sin x\cos x}$ .
Cancel $2\cos x$ top and bottom: $=\dfrac{\cos x}{\sin x}=\cot x$ . ✓

Answer

\dfrac{1+\cos2x}{\sin2x}\equiv\cot x

Worked example

Real question: prove $\operatorname{cosec}2\theta-\cot2\theta\equiv\tan\theta$

9709/32 M/J 2021 Q6(a). First put both reciprocals over the one bottom, $\sin2\theta$ . Then a double angle finishes it.

Write both terms over the same bottom: $\operatorname{cosec}2\theta-\cot2\theta=\dfrac{1}{\sin2\theta}-\dfrac{\cos2\theta}{\sin2\theta}=\dfrac{1-\cos2\theta}{\sin2\theta}$ .
On top, pick the form that makes a clean square: $1-\cos2\theta=1-(1-2\sin^2\theta)=2\sin^2\theta$ . On the bottom, $\sin2\theta=2\sin\theta\cos\theta$ .
So $\dfrac{2\sin^2\theta}{2\sin\theta\cos\theta}=\dfrac{\sin\theta}{\cos\theta}=\tan\theta$ . ✓

Answer

\operatorname{cosec}2\theta-\cot2\theta\equiv\tan\theta

Why examiners always add a part (b)

The proof is never the real question

That is why proofs are on this paper: they let you swap a hard thing to integrate for an easy one. When you see “prove this identity, hence integrate”, use the proved form in the integral.

Where the marks go

Common mistake

Picking the

\cos2A

form at random sends a proof to a dead end. If the line has

\sin^2

, use

1-2\sin^2A

; if it has

\cos^2

, use

2\cos^2A-1

Common mistake

Proofs are marked on the working. Show every step and the fraction split, even when the last line looks obvious. And never move terms across the

\equiv

Now you try

Show that $\cos^4\theta-\sin^4\theta\equiv\cos2\theta$ . (9709-style)

Your turn— tap to reveal the worked answer (9709-style)

The left side is a difference of two squares: $\cos^4\theta-\sin^4\theta=(\cos^2\theta-\sin^2\theta)(\cos^2\theta+\sin^2\theta)$ .

The second bracket is $1$ , and the first is exactly the $\cos2\theta$ form:

Answer

=\cos^2\theta-\sin^2\theta=\cos2\theta\ \checkmark