Double-Angle and Half-Angle Identities

The trigonometric identities are our best means to simplify expressions involving trig functions, so the more we have in our arsenal the better. In this article, we'll be looking specifically at the double-angle and half-angle identities. Let's get started!

The double-angle trigonometric identities

The double-angle trigonometric identities are special cases of Bhaskaracharya's formulas where u = v, but they come up frequently enough to be worth studying independently. There are a total of five of them:

1. $\sin \left(2u\right)=2\sin \left(u\right)\cos \left(u\right)$

2. $\cos \left(2u\right)={\cos }^2\left(u\right)-{\sin }^2\left(u\right)$

3. $\cos \left(2u\right)=2{\cos }^2\left(u\right)-1$

4. $\cos \left(2u\right)=1-2{\sin }^2\left(u\right)$

5. $\tan \left(2u\right)=\frac{2\tan \left(u\right)}{1-{\tan }^2\left(u\right)}$

You might be surprised that 3 of the 5 are dedicated to cos (2u), but it just means that we have more options to choose from. For example, the fourth one above is particularly useful if the rest of the expression is in terms of sine.

Let's put these into practice by rewriting the following expression in a simpler form: $2\sin \left(5p\right)\cos \left(5p\right)$ .

In this case, u is 5p. The format of our expression lends itself to the double-angle formula for sine, $\sin \left(2u\right)=2\sin \left(u\right)\cos \left(u\right)$ . When we apply the formula, we'll get:

$2\sin \left(5p\right)\cos \left(5p\right)=\sin (2\cdot 5p)$

$\sin \left(10p\right)$

That's much simpler than what we started with! There's another group of trigonometric identities that are derived from the formulas above by solving for ${\sin }^2\left(u\right)$ , ${\cos }^2\left(u\right)$ or ${\tan }^2\left(u\right)$ . They are called power-reducing identities and look like this:

1. ${\sin }^2\left(u\right)=\frac{1-\cos \left(2u\right)}{2}$

2. ${\cos }^2\left(u\right)=\frac{1+\cos \left(2u\right)}{2}$

3. ${\tan }^2\left(u\right)=\frac{1-\cos \left(2u\right)}{1+\cos \left(2u\right)}$

We never know what might help us solve a particularly challenging problem, so it's worth committing these identities to memory as well.

The half-angle trigonometric identities

The half-angle trigonometric identities are derived from those above as well, except we replace the 2u with θ and take the square root of both sides. There are five of them in all:

Again, we have three different identities for $\tan \left(\frac{\theta }{2}\right)$ , giving us lots of options to help us simplify expressions involving it.

Practice Questions

a. Use trigonometric identities to determine the exact value of $\cos \left(15°\right)$ .

None of the trigonometric identities can be applied to $\cos \left(15°\right)$ as it sits, but we can rewrite it as $\cos \left(\frac{30°}{2}\right)$ . In that format, this problem lends itself well to the half-angle identity of cosine, $\cos \frac{\theta }{2}=\pm \sqrt{\frac{1+\cos \theta }{2}}$ . In this case, $\theta =30$ . When we apply the formula, the math will look like this:

$\cos \frac{30}{2}=\pm \sqrt{\frac{1+\cos 30}{2}}$

$=\pm \sqrt{\frac{1+\frac{\sqrt{3}}{2}}{2}}$

$=\pm \sqrt{\frac{2+\sqrt{3}}{4}}$

$=\pm \frac{\sqrt{2+\sqrt{3}}}{2}$

All square roots give us positive and negative numbers, but we know that a 15-degree angle is in quadrant I where the cosine is always positive. Therefore, our final answer is $\frac{\sqrt{2+\sqrt{3}}}{2}$

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Applying trigonometric identities is equal parts rote memorization and quantitative problem-solving, meaning that there are two distinct areas where students could struggle. If your student needs help with the math involved in applying the formulas above or remembering what the double-angle and half-angle identities are and when to apply them, a private tutor can provide additional practice problems until their pupil feels ready to move on. A tutor can also tailor lessons to a student's unique needs to improve their study efficiency. Contact the Educational Directors at Varsity Tutors today to learn more.