Double and triple angles

Put the two angles together

The sum formulae hold for any $x$ and $y$, so they hold when $y = x$. Then $x + y = 2x$:

• $\sin 2x = \sin x \cos x + \cos x \sin x = 2 \sin x \cos x$
• $\cos 2x = \cos x \cos x - \sin x \sin x = \cos^2 x - \sin^2 x$
• $\tan 2x = \dfrac{2 \tan x}{1 - \tan^2 x}$

Note that $\sin 2x$ is not $2 \sin x$. Doubling an angle does not double its sine. At $x = \dfrac{\pi}{2}$, for example, $\sin 2x = \sin \pi = 0$, while $2 \sin x = 2$.

Three forms of cos 2x

Use $\sin^2 x + \cos^2 x = 1$ to remove one of the two squares:

• $\cos 2x = \cos^2 x - \sin^2 x$
• $\cos 2x = 2\cos^2 x - 1$
• $\cos 2x = 1 - 2\sin^2 x$

Choose the form that fits the question. If you know only $\sin x$, use the last one.

Move the angle. The red point always sits on the blue curve. The grey point of the guess $2 \cos x$ is somewhere else.

Everything from tan x

Write $\sin 2x = \dfrac{2 \sin x \cos x}{\cos^2 x + \sin^2 x}$, since the bottom is $1$. Divide the top and the bottom by $\cos^2 x$:

• $\sin 2x = \dfrac{2 \tan x}{1 + \tan^2 x}$
• $\cos 2x = \dfrac{1 - \tan^2 x}{1 + \tan^2 x}$

So one number, $\tan x$, fixes $\sin 2x$, $\cos 2x$ and $\tan 2x$.

Half angles

Turn the last two forms of $\cos 2x$ round:

• $\sin^2 x = \dfrac{1 - \cos 2x}{2}$ and $\cos^2 x = \dfrac{1 + \cos 2x}{2}$

Now write $x$ for $2x$ and $\dfrac{x}{2}$ for $x$:

• $\sin \dfrac{x}{2} = \pm\sqrt{\dfrac{1 - \cos x}{2}}$ and $\cos \dfrac{x}{2} = \pm\sqrt{\dfrac{1 + \cos x}{2}}$
• $\tan \dfrac{x}{2} = \dfrac{\sin x}{1 + \cos x} = \dfrac{1 - \cos x}{\sin x}$

The square root gives the size. The quadrant of $\dfrac{x}{2}$ gives the sign. For example, $\tan 22.5^\circ = \dfrac{1 - \cos 45^\circ}{\sin 45^\circ} = \dfrac{1 - \frac{1}{\sqrt{2}}}{\frac{1}{\sqrt{2}}} = \sqrt{2} - 1$.

Triple angles

Write $3x = 2x + x$ and use the sum formula with the double angles. For the sine:

Replace $\cos^2 x$ by $1 - \sin^2 x$ and collect:

• $\sin 3x = 3\sin x - 4\sin^3 x$
• $\cos 3x = 4\cos^3 x - 3\cos x$
• $\tan 3x = \dfrac{3\tan x - \tan^3 x}{1 - 3\tan^2 x}$

In the sine formula the cube has a minus sign. In the cosine formula the cube comes first.

A worked example

$\tan x = \dfrac{3}{4}$, and $x$ lies in the third quadrant. Find $\sin 2x$, $\cos 2x$ and $\tan 2x$.

1. The ratios of $x$. A right triangle with sides $3$ and $4$ has hypotenuse $5$. In the third quadrant, sine and cosine are both negative: $\sin x = -\dfrac{3}{5}$ and $\cos x = -\dfrac{4}{5}$.
2. The double angles. $\sin 2x = 2\left(-\dfrac{3}{5}\right)\left(-\dfrac{4}{5}\right) = \dfrac{24}{25}$ and $\cos 2x = \dfrac{16}{25} - \dfrac{9}{25} = \dfrac{7}{25}$.
3. The tangent. $\tan 2x = \dfrac{24}{7}$. The formula in $\tan x$ agrees: $\dfrac{2 \cdot \frac{3}{4}}{1 - \frac{9}{16}} = \dfrac{24}{7}$.

Practise with new numbers

Try it in Python

The program works out $\sin 2x$ and $\cos 2x$ in two ways at every $30^\circ$ from $0^\circ$ to $180^\circ$.

Common mistakes

• $\sin 2x = 2\sin x$. It is false. The true formula is $2 \sin x \cos x$.
• The wrong form of $\cos 2x$. All three are right, but only one may fit the facts that you have.
• Dropping the $\pm$. In a half-angle formula, decide the sign from the quadrant of $\dfrac{x}{2}$, not of $x$.
• Mixing up the triple angles. $\sin 3x = 3\sin x - 4\sin^3 x$, but $\cos 3x = 4\cos^3 x - 3\cos x$.
• $\tan 2x$ when $\tan x = \pm 1$. The bottom $1 - \tan^2 x$ is $0$, and $\tan 2x$ is not defined: then $2x$ is an odd multiple of $\dfrac{\pi}{2}$.

Remember these

• $\sin 2x = 2\sin x \cos x = \dfrac{2\tan x}{1 + \tan^2 x}$
• $\cos 2x = \cos^2 x - \sin^2 x = 2\cos^2 x - 1 = 1 - 2\sin^2 x$
• $\tan 2x = \dfrac{2\tan x}{1 - \tan^2 x}$
• $\sin^2 x = \dfrac{1 - \cos 2x}{2}$ and $\cos^2 x = \dfrac{1 + \cos 2x}{2}$
• $\sin 3x = 3\sin x - 4\sin^3 x$ and $\cos 3x = 4\cos^3 x - 3\cos x$

Check yourself

Try each one on your own first. If you are stuck, open a hint. Then check with Show the answer.

1. $x$ is acute and $\sin x = \dfrac{3}{5}$. Find $\sin 2x$.

2. Write $\cos 4x$ in terms of $\cos x$ only.

3. Find the exact value of $\cos 22.5^\circ$.

4. Prove that $\cot x - \tan x = 2\cot 2x$.

5. Why can $\dfrac{1 - \cos 2x}{2}$ never be negative?