Mathematics 9709 · Pure Mathematics 2

Double angle formulae

A double angle is just a compound angle with both angles the same.
Put $B=A$ into $\sin(A+B)$ and $\cos(A+B)$ and you get three new identities directly.
That includes the most useful one, $\cos2A$ , which comes in three different forms.

$\sin2A=2\sin A\cos A$ , $\tan2A=\dfrac{2\tan A}{1-\tan^2A}$ , and the three-form $\cos2A=\cos^2A-\sin^2A=2\cos^2A-1=1-2\sin^2A$ . Pick the $\cos2A$ form that matches the ratio already in the equation. Then the whole equation is in one ratio.

Three formulae, and the three forms of cos 2A

$\sin2A=2\sin A\cos A,\qquad \tan2A=\dfrac{2\tan A}{1-\tan^2A}$

$\cos2A=\cos^2A-\sin^2A=2\cos^2A-1=1-2\sin^2A$

All three forms come from the first one by putting in $\sin^2A+\cos^2A=1$ .
Picking the right one is the main skill. Know $\cos A$ , use $2\cos^2A-1$ ; know $\sin A$ , use $1-2\sin^2A$ .
A wrong form still works. It just takes more algebra.

Worked examples

Worked example

$\sin x=-\tfrac35$ with $180^\circ<x<270^\circ$ : find $\sin2x$ and $\cos2x$

First find $\cos x$ . In the third quadrant cosine is also negative, so $\cos x=-\tfrac45$ (the $3,4,5$ triangle, both legs negative here).
$\sin2x=2\sin x\cos x=2\left(-\tfrac35\right)\!\left(-\tfrac45\right)=\tfrac{24}{25}$ (two negatives multiply to a positive).
$\cos2x=\cos^2x-\sin^2x=\left(-\tfrac45\right)^2-\left(-\tfrac35\right)^2=\tfrac{16-9}{25}=\tfrac{7}{25}$ .

Answer

\sin2x=\tfrac{24}{25},\quad \cos2x=\tfrac{7}{25}

Worked example

Solve $\cos2x+3\sin x=2$ for $0\le x\le2\pi$

The other ratio in the equation is sine, so reach for the form with no cosine in it: $\cos2x=1-2\sin^2x$ .
Substitute and tidy: $1-2\sin^2x+3\sin x=2\Rightarrow 2\sin^2x-3\sin x+1=0$ .
Factorise the quadratic in $\sin x$ : $(2\sin x-1)(\sin x-1)=0$ , so $\sin x=\tfrac12$ or $\sin x=1$ .
Read off in radians on $[0,2\pi]$ : $\sin x=\tfrac12\Rightarrow x=\tfrac{\pi}{6},\tfrac{5\pi}{6}$ ; $\sin x=1\Rightarrow x=\tfrac{\pi}{2}$ .

Answer

x=\tfrac{\pi}{6},\ \tfrac{\pi}{2},\ \tfrac{5\pi}{6}

Why the right form is faster

Why picking the right form helps, with a real case

Every $\cos2A$ form is correct, so why does the choice matter? Because each form is already written in one ratio. Say the rest of the equation is in $\sin$ . Pick $1-2\sin^2A$ and the whole equation is in $\sin$ : a clean quadratic. Pick $\cos^2A-\sin^2A$ instead and you carry both ratios. Now you must swap one with an extra line of $\cos^2=1-\sin^2$ . Same answer, but more chances to slip up.

Where the marks go

Common mistake

Picking the wrong

\cos2A

form is allowed but slow. Match the form to the ratio you were given, so the equation drops to one ratio in a single step.

Common mistake

When you square to find a missing ratio, the quadrant decides the sign. In

180^\circ<x<270^\circ

both

\sin x

and

\cos x

are negative, so don't just take the positive root.

Now you try

Show that $\cot^2\theta+2\cos2\theta=4$ can be written as $4\sin^4\theta+3\sin^2\theta-1=0$ . (9709/33 Oct/Nov 2023 Q6)

Your turn— tap to reveal the worked answer (9709/33 Oct/Nov 2023 Q6)

Write $\cot^2\theta=\dfrac{\cos^2\theta}{\sin^2\theta}=\dfrac{1-\sin^2\theta}{\sin^2\theta}$ and use the sine form $\cos2\theta=1-2\sin^2\theta$ , then multiply through by $\sin^2\theta$ :

$(1-\sin^2\theta)+\sin^2\theta(2-4\sin^2\theta)=4\sin^2\theta$

Answer

4\sin^4\theta+3\sin^2\theta-1=0\ \checkmark

PreviousCompound angle formulae NextFurther trigonometric identities

Double angle formulae

A double angle is just a compound angle with both angles the same.
Put $B=A$ into $\sin(A+B)$ and $\cos(A+B)$ and you get three new identities directly.
That includes the most useful one, $\cos2A$ , which comes in three different forms.

Three formulae, and the three forms of cos 2A

$\sin2A=2\sin A\cos A,\qquad \tan2A=\dfrac{2\tan A}{1-\tan^2A}$

$\cos2A=\cos^2A-\sin^2A=2\cos^2A-1=1-2\sin^2A$

All three forms come from the first one by putting in $\sin^2A+\cos^2A=1$ .
Picking the right one is the main skill. Know $\cos A$ , use $2\cos^2A-1$ ; know $\sin A$ , use $1-2\sin^2A$ .
A wrong form still works. It just takes more algebra.

Worked examples

Worked example

$\sin x=-\tfrac35$ with $180^\circ<x<270^\circ$ : find $\sin2x$ and $\cos2x$

First find $\cos x$ . In the third quadrant cosine is also negative, so $\cos x=-\tfrac45$ (the $3,4,5$ triangle, both legs negative here).
$\sin2x=2\sin x\cos x=2\left(-\tfrac35\right)\!\left(-\tfrac45\right)=\tfrac{24}{25}$ (two negatives multiply to a positive).
$\cos2x=\cos^2x-\sin^2x=\left(-\tfrac45\right)^2-\left(-\tfrac35\right)^2=\tfrac{16-9}{25}=\tfrac{7}{25}$ .

Answer

\sin2x=\tfrac{24}{25},\quad \cos2x=\tfrac{7}{25}

Worked example

Solve $\cos2x+3\sin x=2$ for $0\le x\le2\pi$

The other ratio in the equation is sine, so reach for the form with no cosine in it: $\cos2x=1-2\sin^2x$ .
Substitute and tidy: $1-2\sin^2x+3\sin x=2\Rightarrow 2\sin^2x-3\sin x+1=0$ .
Factorise the quadratic in $\sin x$ : $(2\sin x-1)(\sin x-1)=0$ , so $\sin x=\tfrac12$ or $\sin x=1$ .
Read off in radians on $[0,2\pi]$ : $\sin x=\tfrac12\Rightarrow x=\tfrac{\pi}{6},\tfrac{5\pi}{6}$ ; $\sin x=1\Rightarrow x=\tfrac{\pi}{2}$ .

Answer

x=\tfrac{\pi}{6},\ \tfrac{\pi}{2},\ \tfrac{5\pi}{6}

Why the right form is faster

Why picking the right form helps, with a real case

Where the marks go

Common mistake

Picking the wrong

\cos2A

form is allowed but slow. Match the form to the ratio you were given, so the equation drops to one ratio in a single step.

Common mistake

When you square to find a missing ratio, the quadrant decides the sign. In

180^\circ<x<270^\circ

both

\sin x

and

\cos x

are negative, so don't just take the positive root.

Now you try

Show that $\cot^2\theta+2\cos2\theta=4$ can be written as $4\sin^4\theta+3\sin^2\theta-1=0$ . (9709/33 Oct/Nov 2023 Q6)

Your turn— tap to reveal the worked answer (9709/33 Oct/Nov 2023 Q6)

Write $\cot^2\theta=\dfrac{\cos^2\theta}{\sin^2\theta}=\dfrac{1-\sin^2\theta}{\sin^2\theta}$ and use the sine form $\cos2\theta=1-2\sin^2\theta$ , then multiply through by $\sin^2\theta$ :

$(1-\sin^2\theta)+\sin^2\theta(2-4\sin^2\theta)=4\sin^2\theta$

Answer

4\sin^4\theta+3\sin^2\theta-1=0\ \checkmark