What Are Half Angle Identities?

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Just like in algebra, when you start learning trigonometry, you'll accumulate sets of formulas that are useful for problem-solving. One such set is the half-angle identities, which you can use for two purposes. One is to convert trigonometric functions of (​θ​/2) into functions in terms of the more familiar (and more easily manipulated) ​θ​. The other is to find the actual value of trigonometric functions of ​θ​, when ​θ​ can be expressed as half of a more familiar angle.

Reviewing the Half-Angle Identities

Many math textbooks will list four primary half-angle identities. But by applying a mix of algebra and trigonometry, these equations can be massaged into a number of useful forms. You don't necessarily have to memorize all of these (unless your teacher insists), but you should, at least, understand how to use them:

Half-Angle Identity for Sine

\sin\bigg(\frac{θ}{2}\bigg) = ±\sqrt{\frac{1 - \cosθ}{2}}

Half-Angle Identity for Cosine

\cos\bigg(\frac{θ}{2}\bigg) = ±\sqrt{\frac{1 + \cosθ}{2}}

Half-Angle Identities for Tangent

\tan\bigg(\frac{θ}{2}\bigg) = ±\sqrt{\frac{1 -\cosθ}{1 + \cosθ}} \\ \,\\ \tan\bigg(\frac{θ}{2}\bigg) = \frac{\sinθ}{1 + \cosθ} \\ \,\\ \tan\bigg(\frac{θ}{2}\bigg) = \frac{1 - \cosθ}{\sinθ} \\ \,\\ \tan\bigg(\frac{θ}{2}\bigg) = \cscθ - \cotθ

Half-Angle Identities for Cotangent

\cot\bigg(\frac{θ}{2}\bigg) = ±\sqrt{\frac{1 + \cosθ}{1 - \cosθ}} \\ \,\\ \cot\bigg(\frac{θ}{2}\bigg) = \frac{\sinθ}{1 - \cosθ} \\ \,\\ \cot\bigg(\frac{θ}{2}\bigg) = \frac{1 + \cosθ}{\sinθ} \\ \,\\ \cot\bigg(\frac{θ}{2}\bigg) = \cscθ + \cotθ

An Example of Using Half-Angle Identities

So how do you use half-angle identities? The first step is recognizing that you're dealing with an angle that's half of a more familiar angle.

1. Find θ

2. imagine that you're asked to find the sine of the angle 15 degrees. This isn't one of the angles most students will memorize the values of trig functions for. But if you let 15 degrees be equal to θ/2 and then solve for θ, you'll find that:

\frac{θ}{2} = 15 \\ θ = 30

Because the resulting θ, 30 degrees, is a more familiar angle, using the half-angle formula here will be helpful.

3. Choose a Half-Angle Formula

4. Because you've been asked to find the sine, there's really just one half-angle formula to choose from:

\sin\bigg(\frac{θ}{2}\bigg) = ±\sqrt{\frac{1 - \cosθ}{2}}

Substituting in ​θ​/2 = 15 degrees and ​θ​ = 30 degrees gives you:

\sin(15) = ±\sqrt{\frac{1 - \cos(30)}{2}}

If you'd been asked to find the tangent or cotangent, both of which half multiply ways of expressing their half-angle identity, you'd simply choose the version that looked easiest to work.

5. Resolve the ± Sign

6. The ± sign at the beginning of some half-angle identities means that the root in question could be positive or negative. You can resolve this ambiguity by using your knowledge of trigonometric functions in quadrants. Here's a quick recap of which trig functions return ​positive​ values in which quadrants:

• Quadrant I: all trig functions
• Quadrant II: only sine and cosecant
• Quadrant III: only tangent and cotangent
• Quadrant IV: only cosine and secant

Because in this case your angle θ represents 30 degrees, which falls in Quadrant I, you know that the sine value it returns will be positive. So you can drop the ± sign and simply evaluate:

\sin(15) = \sqrt{\frac{1 - \cos(30)}{2}}
7. Substitute the Familiar Values

8. Substitute in the familiar, known value of cos(30). In this case, use the exact values (as opposed to decimal approximations from a chart):

\sin(15) = \sqrt{\frac{1 - \sqrt{3/2}}{2}}

10. Next, simplify the right side of your equation to find a value for sin(15). Start by multiplying the expression under the radical by 2/2, which gives you:

\sin(15) = \sqrt{\frac{2(1 - \sqrt{3/2})}{4}}

This simplifies to:

\sin(15) = \sqrt{\frac{2 - \sqrt{3}}{4}}

You can then factor out the square root of 4:

\sin(15) = \frac{1}{2}\sqrt{2 - \sqrt{3}}

In most cases, this is about as far as you'd simplify. While the result may not be terribly pretty, you've translated the sine of an unfamiliar angle into an exact quantity.