More Trigonometric Identities

I already proved the angle difference theorem for the cosine yesterday. Starting from this theorem, I prove a few more trigonometric identities.

Complementary angles

The complementary angle identities are pretty obvious if you look at the unit circle definition of the trigonometric functions. $$\cos (\pi/2 - \alpha) = \sin \alpha$$

Apply the cosine difference identity. \begin{align*} \cos (\pi/2 - \alpha) &= \cos (\pi/2) \cos \alpha + \sin (\pi/2) \sin \alpha\\ &= 0 \cos \alpha + 1 \sin \alpha \\ &= \sin \alpha \end{align*}
$$\sin (\pi/2 - \alpha) = \cos \alpha$$
Apply the complementary angles identity for cosine backwards. \begin{align*} \sin (\pi/2 - \alpha) &= \cos (\pi/2 - \pi/2 + \alpha) \\ &= \cos \alpha \end{align*}
$$\tan (\pi/2 - \alpha) = \cot \alpha$$
Follows from the definitions of tangent and cotangent in terms of sine and cosine, $$\tan \alpha = \frac{\sin \alpha}{\cos \alpha}$$ and $$\cot \alpha = \tan^{-1} \alpha$$, and the previous two identities. \begin{align*} \tan (\pi/2 - \alpha) &= \frac{\sin (\pi/2 - \alpha)}{\cos (\pi/2 - \alpha)} = \frac{\cos \alpha}{\sin \alpha}\\ &= \cot \alpha \end{align*}

Negative angles

The negative angle identities are again pretty obvious if you look at the symmetries of the trigonometric functions at the unit circle. $$\cos(-\alpha) = \cos \alpha$$

Use $$-\alpha = 0 - \alpha$$ and apply the cosine difference identity. \begin{align*} \cos (-\alpha) &= \cos (0 - \alpha)\\ &= \cos 0 \cos \alpha + \sin 0 \sin \alpha\\ &= \cos \alpha \end{align*}
$$\sin(-\alpha) = -\sin \alpha$$
Use the observation that the angles $$0$$ and $$2\pi$$ are represented by the same point on the unit circle, apply the complementary angle identity for sine/cosine and the cosine difference identity. \begin{align*} \sin (-\alpha) &= \sin (2\pi - \alpha)\\ &= \cos (\pi/2 - 2\pi + \alpha)\\ &= \cos (\alpha - 3\pi / 2)\\ &= \cos \alpha \cos (3\pi / 2) + \sin \alpha \sin (3\pi / 2)\\ &= (\cos \alpha) 0 + \sin \alpha (-1)\\ &= -\sin \alpha \end{align*}
$$\tan(-\alpha) = -\tan \alpha$$
Use the definition of the tangent in terms of sine and cosine ($$\tan \alpha = \frac{\sin \alpha}{\cos \alpha}$$) and the previous two identities. \begin{align*} \tan (-\alpha) &= \frac{\sin (-\alpha)}{\cos (-\alpha)} \\ &= \frac{- \sin \alpha}{\cos \alpha} \\ &= - \tan \alpha \end{align*}

Angle sums and differences

The angle sum and difference identities are less obvious. \begin{align*} \cos (\alpha - \beta) &= \cos \alpha \cos \beta + \sin \alpha \sin \beta \\ \cos (\alpha + \beta) &= \cos \alpha \cos \beta - \sin \alpha \sin \beta \\ \end{align*}

We already know that the first one holds and we'll use it to prove the cosine sum identity with the help of the negative angle identity. \begin{align*} \cos (\alpha + \beta) &= \cos (\alpha - (- \beta)) \\ &= \cos \alpha \cos (-\beta) + \sin \alpha \sin (- \beta) \\ &= \cos \alpha \cos \beta + \sin \alpha (- \sin \beta) \\ &= \cos \alpha \cos \beta - \sin \alpha \sin \beta \\ \end{align*}
$$\sin (\alpha + \beta) = \sin \alpha \cos \beta + \cos \alpha \sin \beta$$
Convert the sine to cosine via the complementary angle, apply the cosine difference identity and finally the complementary angle again. \begin{align*} \sin (\alpha + \beta) &= \cos (\pi/2 - \alpha - \beta) \\ &= \cos (\pi/2 - \alpha) \cos \beta + \sin (\pi/2 - \alpha) \sin \beta \\ &= \sin \alpha \cos \beta + \cos \alpha \sin \beta \end{align*}
$$\sin (\alpha - \beta) = \sin \alpha \cos \beta - \cos \alpha \sin \beta$$
Use the sine sum identity and negative angles. \begin{align*} \sin (\alpha - \beta) &= \sin (\alpha + (-\beta)) \\ &= \sin \alpha \cos (- \beta) + \cos \alpha \sin (-\beta) \\ &= \sin \alpha \cos (\beta) - \cos \alpha \sin \beta \end{align*}