Answer to Question #168697 in Trigonometry for Richardson Ducote

Question

Express 3 sin 4t + 8 cos 4t in the form R sin ( ωt + α ), α    

Answer 1

Express 3 $\sin$ 4t + 8 $\cos$ 4t in the form R $\sin$ ($\omega t$ + $\alpha$), $\alpha$

Solution

Let 3 $\sin$ 4t + 8 $\cos$ 4t = R $\sin$ ($\omega t$ + $\alpha$)

By using compound angle formula $\sin (A + B) = \sin A \cos B + \cos A \sin B$ we get to solve the following:

3 $\sin$ 4t + 8 $\cos$ 4t = R[ $\sin$ 4t $\cos \alpha$ + $\cos 4t$ $\sin \alpha$]

= (R $\cos \alpha$) $\sin$ 4t + (R $\sin \alpha$) $\cos 4t$

Equating the coefficients of:

$\sin$ 4t gives: 3 = R $\cos \alpha$, from which, $\cos \alpha$ = $\frac{3}{R}$

And $\cos$ 4t gives: 8 = R $\sin \alpha$, from which, $\sin \alpha$ = $\frac{8}{R}$

There is only one quadrant where both $\sin \alpha$ and $\cos \alpha$ are positive and is the first.

Hence: R = $\sqrt{3^2 + 8^2}$ = 8.544003745

from trigonometric ratios: $\alpha = \arctan \frac{8}{3}$ = 69.44395478 or 1.212025657 radians

Hence 3 $\sin$ 4t + 8 $\cos$ 4t = 8.544003745 $\sin (4t + 1.212025657)$, 1.212025657

Answer: 8.544003745 $\sin$(4t + 1.212025657), 1.212025657