# Double Angle Identities

Math Analysis - Trigonometry Identity - Double Angle - Sine

Tip: This isn't the place to ask a question because the teacher can't reply.

1 of 2 videos by AJ Speller

## Key Questions

• Double Angle Identities

$\sin 2 \theta = 2 \sin \theta \cos \theta$

$\cos 2 \theta = {\cos}^{2} \theta - {\sin}^{2} \theta = 2 {\cos}^{2} \theta - 1 = 1 - 2 {\sin}^{2} \theta$

$\tan 2 \theta = \frac{2 \tan \theta}{1 - {\tan}^{2} \theta}$

I hope that this was helpful.

• Substitute in the one angle you get into the identity.

For example, if you are asked to find $\sin 2 a$ and $a = 30$, then you substitute in $\sin 2 \left(30\right) = 2 \sin 30 \cos 30$ into your calculator, and that's how you solve equations with double angle identities.

• You can use the double angle ID to evaluate the exact value of a trigonometric expression when you can relate the expression of one of the double angle formulas to one of the special angles.

Ex.1 Evaluate $2 \sin \left(15\right) \cos \left(15\right)$

using $\sin \left(2 x\right) = 2 \sin x \cos x$, we observe that the left had side of this ID resembles the given question (with $x = 15$).
Thus we can convert the expression to
$2 \sin \left(15\right) \cos \left(15\right) = \sin \left(2 \cdot 15\right) = \sin \left(30\right) = \frac{1}{2}$

It is also possible to use known values of $\sin \left(x\right) , \cos \left(x\right) , \mathmr{and} \tan \left(x\right)$ to evaluate the exact value of a given trigonometric function of a double angle.

Ex.2 Given $\sin \left(x\right) = \frac{3}{5}$, find $\cos \left(2 x\right)$

To answer this question, we can then use the ID $\cos \left(2 x\right) = 1 - 2 {\sin}^{2} \left(x\right)$

Since $\sin \left(x\right) = \frac{3}{5}$, we can find ${\sin}^{2} \left(x\right) = \frac{9}{25}$

Therefore $\cos \left(2 x\right) = 1 - 2 {\sin}^{2} \left(x\right) = 1 - 2 \left(\frac{9}{25}\right) = 1 - \frac{18}{25} = \frac{7}{25}$

• You would need an expression to work with.

For example:
Given $\sin \alpha = \frac{3}{5}$ and $\cos \alpha = - \frac{4}{5}$, you could find $\sin 2 \alpha$ by using the double angle identity
$\sin 2 \alpha = 2 \sin \alpha \cos \alpha$.

$\sin 2 \alpha = 2 \left(\frac{3}{5}\right) \left(- \frac{4}{5}\right) = - \frac{24}{25}$.

You could find $\cos 2 \alpha$ by using any of:
$\cos 2 \alpha = {\cos}^{2} \alpha - {\sin}^{2} \alpha$
$\cos 2 \alpha = 1 - 2 {\sin}^{2} \alpha$
$\cos 2 \alpha = 2 {\cos}^{2} \alpha - 1$

In any case, you get $\cos \alpha = \frac{7}{25}$.

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