Precalculus Review

Trigonometric formulas

Double-angle and half-angle formulas. By letting

in Trigonometric additions and subtractions we can derive the double-angle formulas:

$\displaystyle \cos 2u = \cos^2 u - \sin^2 u = 1 - 2 \sin^2 u = 2 \cos^2 u - 1$ and $\displaystyle \sin 2u = 2 \sin u \cos u.$

By letting $u = \dfrac{v}{2}$ in the double-angle formulas, we can verify the following half-angle formulas:

$\displaystyle \sin\dfrac{v}{2} = \pm \sqrt{\dfrac{1 - \cos v}{2}}; \quad \cos\dfrac{v}{2} = \pm \sqrt{\dfrac{1 + \cos v}{2}},$

and

$\displaystyle \tan\dfrac{v}{2} = \pm \sqrt{\dfrac{1 - \cos v}{1 + \cos v}} = \dfrac{1 - \cos v}{\sin v} = \dfrac{\sin v}{1 + \cos v}.$

Sum-to-product formulas. By letting $u = \dfrac{\alpha + \beta}{2}$ and $v = \dfrac{\alpha - \beta}{2}$ in Trigonometric additions and subtractions, we can verify the following formulas:

$\begin{displaymath}\begin{array}{ll} \sin\alpha + \sin\beta = 2 \sin\dfrac{\al... ...{\alpha + \beta}{2} \sin\dfrac{\alpha - \beta}{2}. \end{array}\end{displaymath}$

Sum of trigonometric functions having common period. Let $\theta = \tan^{-1}(b/a)$ if $a \ge 0$ (we set $\theta = \pi + \tan^{-1}(b/a)$ if ). Then we have $\cos\theta = \dfrac{a}{\sqrt{a^2 + b^2}}$ and $\sin\theta = \dfrac{b}{\sqrt{a^2 + b^2}}$ , and obtain

$\displaystyle a \cos\omega x + b \sin\omega x = \sqrt{a^2 + b^2}[\cos\omega x ... ...theta + \sin\omega x \sin\theta] = \sqrt{a^2 + b^2}\cos(\omega x - \theta).$