# Trigonometric Identities

## Identity and equation

The difference between these two equalities $$x(x-1)=0$$ and $$x(x-1)=x^2-x$$, is that the first one is true for some values of $$x$$, and the second is true for all $$x$$. An equality that is true for all values is an identity. In some books the symbol $$\equiv$$ is used to denote identities, as in $$x(x-1)\equiv x^2-x$$.

## Identities from the unit circle

Which of following identities are true?

 $$\sin \alpha = \sin (-\alpha)$$ $$\sin \alpha = \cos (90^\circ -\alpha)$$ $$\sin \alpha = \sin (180^\circ -\alpha)$$ $$\sin \alpha = \cos (270^\circ -\alpha)$$ $$\sin \alpha = -\sin (-\alpha)$$ $$\cos \alpha = \sin (90^\circ -\alpha)$$ $$\sin \alpha = -\sin (180^\circ -\alpha)$$ $$\cos \alpha = \sin (270^\circ -\alpha)$$ $$\cos \alpha = \cos (-\alpha)$$ $$\sin \alpha = -\cos (90^\circ -\alpha)$$ $$\cos \alpha = \cos (180^\circ -\alpha)$$ $$\sin \alpha = -\cos (270^\circ -\alpha)$$ $$\cos \alpha = -\cos (-\alpha)$$ $$\cos \alpha = -\sin (90^\circ -\alpha)$$ $$\cos \alpha = -\cos (180^\circ -\alpha)$$ $$\cos \alpha = -\sin (270^\circ -\alpha)$$

In the applet above, an angle in each quadrant is deduced from the explanation below.

Furthermore, following identities are used:

$\sin(\beta -90^\circ)=-\sin(90^\circ -\beta)=-\cos \beta$

$\cos(\beta-180^\circ)=\cos(180^\circ-\beta)=-\cos\beta$

$\sin(\beta -270^\circ)=-\sin(270^\circ -\beta)=\cos \beta$

In this demonstration, $$\alpha$$ is restricted to $$0^\circ \lt \alpha \lt 90^\circ$$ since it is an angle in a right-angled triangle. There is no restriction on $$\beta$$, it can be in any quadrant. The restriction seen in the applet, $$0 \le \beta \lt 360^\circ$$, stems from the fact that GeoGebra can only show those angles.

Using the identities $$\sin(-\beta)=-\sin(\beta)$$ and $$\cos(-\beta)=\cos(\beta)$$ we get that:

$\sin(\alpha-\beta)=\sin(\alpha + (-\beta))=\sin \alpha \cos (-\beta) + \cos \alpha \sin (-\beta)=\sin \alpha \cos \beta - \cos \alpha \sin \beta$

$\sin(\alpha+\beta)=\sin \alpha \cos \beta + \cos \alpha \sin \beta$

$\sin(\alpha-\beta)=\sin \alpha \cos \beta - \cos \alpha \sin \beta$

Apart from the identities that are used for explaining the addition formula for sine, following identities are also used in the applet above:

$\cos(\beta -90^\circ)=\cos(90^\circ -\beta)=\sin \beta$

$\sin(\beta-180^\circ)=-\sin(180^\circ-\beta)=-\sin\beta$

$\cos(\beta -270^\circ)=\cos(270^\circ -\beta)=-\sin \beta$

Using the identities $$\sin(-\beta)=-\sin(\beta)$$ and $$\cos(-\beta)=\cos(\beta)$$ we get that:

$\cos(\alpha-\beta)=\cos(\alpha + (-\beta))=\cos \alpha \cos (-\beta) - \sin \alpha \sin (-\beta)=\cos \alpha \cos \beta + \sin \alpha \sin \beta$

$\cos(\alpha+\beta)=\cos \alpha \cos \beta - \sin \alpha \sin \beta$

$\cos(\alpha-\beta)=\cos \alpha \cos \beta + \sin \alpha \sin \beta$

## When both angles are larger than 90°

The addition formulas are true even when both angles are larger than $$90^\circ$$. Assume that $$90^\circ\lt \alpha \lt 180^\circ$$. Let $$\alpha ' =\alpha -90^\circ$$. From the symmetry of the unit circle we get that $$\sin \alpha =\sin(90^\circ +\alpha ')=-\cos \alpha '$$ and $$\cos \alpha=\cos(90^\circ+\alpha ')=-\sin\alpha '$$. Using these identities we get:

$\sin(\alpha + \beta)=\sin(90^\circ + \alpha ' + \beta)=-\cos(\alpha ' +\beta )$

We can now use the addition formula since $$\alpha ' \lt 90^\circ$$.

$-\cos(\alpha ' +\beta )=-\cos \alpha ' \cos \beta +\sin \alpha ' \sin\beta = \sin\alpha\cos\beta+\sin\beta\cos\alpha$

In a similar way we can prove that all addition formulas are true for all angles.