# Math Help - Inverse Trig Functions and Advance Trig

1. ## Inverse Trig Functions and Advance Trig

You are given 5 cos wt = A cos (wt - pi / 6) + 5 cos (wt + Z)

for all t, where w is a fixed constant

Find, exactly, all possible values for A > 0 and Z

Im really having trouble with this question.

2. $5\cos(\omega t) = A\cos\left(\omega t - \frac{\pi}{6}\right) + 5\cos(\omega t + Z)$

$5\cos(\omega t) = A\left[\cos(\omega t)\cos\left(\frac{\pi}{6}\right)+\sin(\omega t)\sin\left(\frac{\pi}{6}\right)\right] + 5[\cos(\omega t)\cos(Z) - sin(\omega t)\sin(Z)]$

$5\cos(\omega t) = A\left[\cos(\omega t)\left(\frac{\sqrt{3}}{2}\right)+\sin(\omega t)\left(\frac{1}{2}\right)\right] + 5[\cos(\omega t)\cos(Z) - sin(\omega t)\sin(Z)]$

$5\cos(\omega t) = \left[\frac{A\sqrt{3}}{2} + 5\cos(z)\right]\cos(\omega t)+ \left[\frac{A}{2} - 5\sin(Z)\right]\sin(\omega t)$

equating coefficients on the left and right side ...

(1) $\frac{A}{2} - 5\sin(Z) = 0$

(2) $\frac{A\sqrt{3}}{2} + 5\cos(z) = 5$

from equation (1) ...

$A = 10\sin(Z)$

substitute into equation (2) ...

$5\sqrt{3}\sin(Z) + 5\cos(Z) = 5$

$\sqrt{3}\sin(Z) + \cos(Z) = 1$

square both sides ...

$3\sin^2(Z) + 2\sqrt{3}\sin(Z)\cos(Z) + \cos^2(Z) = 1$

$2\sin^2(Z) + 2\sqrt{3}\sin(Z)\cos(Z) = 0$

$2\sin(Z)[\sin(Z) + \sqrt{3} \cos(Z)] = 0$

o.k. ... all the "grunt" work is done. Take the last equation, solve for Z, then go back and solve for A. Buena suerte.