Question #168697

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


1
Expert's answer
2021-03-07T17:15:51-0500

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

Solution

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

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

3 sin\sin 4t + 8 cos\cos 4t = R[ sin\sin 4t cosα\cos \alpha + cos4t\cos 4t sinα\sin \alpha]

= (R cosα\cos \alpha) sin\sin 4t + (R sinα\sin \alpha) cos4t\cos 4t

Equating the coefficients of:

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

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

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

Hence: R = 32+82\sqrt{3^2 + 8^2} = 8.544003745

from trigonometric ratios: α=arctan83\alpha = \arctan \frac{8}{3} = 69.44395478 or 1.212025657 radians

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

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


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