What Are the Trigonometric Identities?

The following identities are used extensively in trigonometry. You can use them when you solve trigonometric problems.

Formula

1.: $\cos^{2} α + \sin^{2} α = 1$
2.: $\sin (α + \frac{π}{2}) = \cos α$
3.: $\cos (α + \frac{π}{2}) = - \sin α$
4.: $\sin 2 α = 2 \sin α \cos α$
5.: $\begin{aligned} \cos 2 α & = \cos^{2} α - \sin^{2} α \\ = 2 \cos^{2} α - 1 \\ = 1 - 2 \sin^{2} α \end{aligned}$

$\cos 2 α = \cos^{2} α - \sin^{2} α = 2 \cos^{2} α - 1 = 1 - 2 \sin^{2} α$
6.: $\sin (α + β) = \sin α \cos β + \cos α \sin β$
7.: $\sin (α - β) = \sin α \cos β - \cos α \sin β$
8.: $\cos (α + β) = \cos α \cos β - \sin α \sin β$
9.: $\cos (α - β) = \cos α \cos β + \sin α \sin β$
10.: $\tan α = \frac{\sin α}{\cos α}$

Example 1

Show that $\cos (\frac{π}{4} + v) = \frac{\sqrt{2}}{2} (\cos v - \sin v)$

To show this, you use the formula

\cos (α + β) = \cos α \cos β - \sin α \sin β

That gives you

\begin{array}{l} \cos (\frac{π}{4} + v) & = \cos \frac{π}{4} \cos v - \sin \frac{π}{4} \sin v \\ = \frac{\sqrt{2}}{2} \cos v - \frac{\sqrt{2}}{2} \sin v \\ = \frac{\sqrt{2}}{2} (\cos v - \sin v) \end{array}

Example 2

Find the exact value of $\sin \frac{π}{12}$

To solve this problem you write that $\sin (α) = \sin (π - α)$ and use the trigonometric identity

\sin (α + β) = \sin α \cos β + \cos α \sin β

Then you get

\begin{array}{l} \sin (\frac{π}{12}) & = \sin (π - \frac{π}{12}) \\ = \sin (\frac{11 π}{12}) \\ = \sin (\frac{π}{6} + \frac{3 π}{4}) \\ = \sin (\frac{π}{6}) \cos (\frac{3 π}{4}) \\ + \cos (\frac{π}{6}) \sin (\frac{3 π}{4}) \\ = \frac{1}{2} \cdot (- \frac{\sqrt{2}}{2}) + \frac{\sqrt{3}}{2} \cdot \frac{\sqrt{2}}{2} \\ = \frac{\sqrt{6} - \sqrt{2}}{4} \end{array}

\begin{array}{l} \sin (\frac{π}{12}) & = \sin (π - \frac{π}{12}) \\ = \sin (\frac{11 π}{12}) \\ = \sin (\frac{π}{6} + \frac{3 π}{4}) \\ = \sin (\frac{π}{6}) \cos (\frac{3 π}{4}) + \cos (\frac{π}{6}) \sin (\frac{3 π}{4}) \\ = \frac{1}{2} \cdot (- \frac{\sqrt{2}}{2}) + \frac{\sqrt{3}}{2} \cdot \frac{\sqrt{2}}{2} \\ = \frac{\sqrt{6} - \sqrt{2}}{4} \end{array}

Example 3

Given $\sin v = \frac{\sqrt{3}}{2}$ , find $\cos v$

You use the formula

\cos^{2} α + \sin^{2} α = 1

which gives you

\begin{array}{l} \cos^{2} v + \sin^{2} v & = 1 \\ \cos^{2} v & = 1 - \sin^{2} v \end{array}

That means that

\begin{array}{l} \cos v & = \pm \sqrt{1 - \sin^{2} v} \\ = \pm \sqrt{1 - {(\frac{\sqrt{3}}{2})}^{2}} \\ = \pm \sqrt{1 - \frac{3}{4}} \\ = \pm \sqrt{\frac{1}{4}} \\ = \pm \frac{1}{2} \end{array}

Example 4

Given $\cos^{2} v + \sin^{2} v = \tan^{2} v$ , find $\sin v$

To solve this you have to use several of the trigonometric identities above, and then calculate $\sin v$ :

\begin{array}{l} 2 \cos^{2} v + \sin^{2} v & = \tan^{2} v \\ 1 - \sin^{2} v + \sin^{2} v & = \frac{\sin^{2} v}{\cos^{2} v} \\ 1 & = \frac{\sin^{2} v}{\cos^{2} v} | \cdot \cos^{2} v \\ \cos^{2} v & = \sin^{2} v \\ 1 - \sin^{2} v & = \sin^{2} v \\ 1 & = 2 \sin^{2} v | \div 2 \\ \frac{1}{2} & = \sin^{2} v \end{array}

\begin{array}{l} \cos^{2} v + \sin^{2} v & = \tan^{2} v \\ 1 - \sin^{2} v + \sin^{2} v & = \frac{\sin^{2} v}{\cos^{2} v} \\ 1 & = \frac{\sin^{2} v}{\cos^{2} v} & | \cdot \cos^{2} v \\ \cos^{2} v & = \sin^{2} v \\ 1 - \sin^{2} v & = \sin^{2} v \\ 1 & = 2 \sin^{2} v & | \div 2 \\ \frac{1}{2} & = \sin^{2} v \end{array}

That means

\sin v = \pm \sqrt{\frac{1}{2}} = \pm \frac{1}{\sqrt{2}} = \pm \frac{\sqrt{2}}{2}

Example 5

Show that $\cos (2 α) = \cos^{2} α - \sin^{2} α$

When you’re solving problems like this, you want to take logical steps to get to what you want to show:

\begin{array}{l} \cos (2 α) & = \cos (α + α) \\ = \cos α \cos α - \sin α \sin α \\ = \cos^{2} α - \sin^{2} α \end{array}

Q.E.D

Example 6

Show that $\sin (2 α) = 2 \sin α \cos α$

When you’re solving problems like this, you want to take logical steps to get to what you want to show:

\begin{array}{l} \sin (2 α) & = \sin (α + α) \\ = \sin α \cos α + \cos α \sin α \\ = 2 \sin α \sin α \end{array}

Q.E.D