Sum and Difference Identities

Working with algebraic expressions involving trigonometric functions such as sine and cosine can feel challenging because they require different techniques compared to basic operations like addition and division. However, this doesn't mean that we're powerless to manipulate such expressions. There are multiple trigonometric identities that we can use to rewrite these expressions into a format that might be easier to work with.

In this article, we'll be exploring the Bhaskaracharya sum and difference identities, named after the mathematician who discovered them. First, we'll look at what these identities are. After that, we'll study how to apply them. Let's get started!

The sum identities

There are three sum identities to remember:

That might look confusing, but there's one for each of the trigonometric functions (sine, cosine, and tangent). The tangent identity only involves tangents, while sine mixes sines and cosines while cosine is $\cos \cos -\sin \sin$ . The rote memorization isn't that bad.

Let's try a practice problem by expressing $\sin \left(2x\right)cos\left(5x\right)+\cos \left(2x\right)sin\left(5x\right)$ in a simpler form with a trigonometric identity. Our $u$ value is $2x$ and our $v$ value is $5x$ . Plugging these values into the sum identity for sine, we get:

$\sin \left(2x+5x\right)=\sin \left(2x\right)cos\left(5x\right)+\cos \left(2x\right)sin\left(5x\right)$

$\sin \left(2x+5x\right)=\sin \left(7x\right)$

That makes it much easier to work with, right?

The difference identities

Just as there are three sum identities (one for each trigonometric function), there are also three difference identities:

They look scary, but a closer examination reveals that they're the same as the sum identities except all of the signs are reversed. If it's a + sign in the sum identity, it's a minus sign in the difference identity and vice versa. As such, the memory tricks you use for the sum identities can be applied to the difference identities too.

Let's try using one of these identities to write the following expression in a simpler form:

$\sin \left(\frac{\pi }{6}\right)cos\left(\frac{\pi }{4}\right)-\cos \left(\frac{\pi }{6}\right)sin\left(\frac{\pi }{4}\right)$

The format of the expression lends itself to the sine difference identity:

$\sin \left(u-v\right)=\sin \left(u\right)cos\left(v\right)-\cos \left(u\right)sin\left(v\right)$

Our $u$ value is $\frac{\pi }{6}$ and our $v$ value is $\frac{\pi }{4}$ , so let's plug them in:

$\sin \left(\frac{\pi }{6}-\frac{\pi }{4}\right)=\sin \left(\frac{\pi }{6}\right)cos\left(\frac{\pi }{4}\right)-\cos \left(\frac{\pi }{6}\right)sin\left(\frac{\pi }{4}\right)$

$\sin \left(\frac{\pi }{6}-\frac{\pi }{4}\right)=\sin \left(\frac{2\pi -3\pi }{12}\right)$

$\sin \left(-\frac{\pi }{12}\right)$

We should always go slow and check our work, as a mistake during an early step will lead to further mistakes later on. Likewise, we don't want to let the presence of the pi symbol trip us up.

Practice Questions

a. Evaluate $\tan \left(15°\right)$ .

We can rewrite $\tan \left(15°\right)$ as:

$\tan \left(45°-30°\right)$

Not only does this give us easier values to work with, but it also allows us to apply the tangent difference identity:

$\frac{\tan \left(u\right)-\tan \left(v\right)}{1+\tan \left(u\right)\tan \left(v\right)}$

$\frac{\tan \left(45°\right)-\tan \left(30°\right)}{1+\tan \left(45°\right)\tan \left(30°\right)}$

$\frac{1+\frac{\sqrt{3}}{3}}{1+1\times \frac{\sqrt{3}}{3}}$

$3-\frac{\sqrt{3}}{3}+\sqrt{3}$

$[3-\frac{\sqrt{3}}{3}+\sqrt{3}]\times [3-\frac{\sqrt{3}}{3}-\sqrt{3}]$

$12-6\frac{\sqrt{3}}{6}$

$2-\sqrt{3}$

b. Evaluate $\cos \left(255°\right)$

We can change this expression to:

$\cos \left(300-45\right)$

That makes our u value 300 and our v value 45 when we apply the following difference identity:

$\cos \left(u+v\right)=\cos \left(u\right)cos\left(v\right)+\sin \left(u\right)sin\left(v\right)$

Now, all we have to do is work out the math:

$\cos \left(300\right)\cos \left(45\right)+\sin \left(300\right)\sin \left(45\right)\to \frac{1}{2}\times \frac{\sqrt{2}}{2}+\left(\frac{-\sqrt{3}}{2}\times \frac{\sqrt{2}}{2}\right)$

$\frac{\sqrt{2}-\sqrt{6}}{4}$ is our final answer.

c. Evaluate $\sin \left(165°\right)$

We can rewrite $\sin \left(165\right)$ as $\sin \left(135+30\right)$ , allowing us to apply the trigonometric sum identity for sine:

$\sin \left(u+v\right)=\sin \left(u\right)cos\left(v\right)+\cos \left(u\right)sin\left(v\right)$

Our u value is 135 and our v value is 30, so we can sub the variables in and work out the math:

$\sin \left(135\right)\cos \left(30\right)+\cos \left(135\right)\sin \left(30\right)\to \frac{\sqrt{2}}{2}\times \frac{\sqrt{3}}{2}+-\frac{\sqrt{2}}{2}\times \frac{1}{2}$

$\frac{\sqrt{6}-\sqrt{2}}{4}$ is our final answer

Topics related to the Sum and Difference Identities

Cosine Function

Inverse Trigonometric Functions

Finding the Area of a Triangle Using Sine

Flashcards covering the Sum and Difference Identities

Trigonometry Flashcards

CLEP Precalculus Flashcards

Practice tests covering the Sum and Difference Identities

Trigonometry Diagnostic Tests

Precalculus Diagnostic Tests

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