Question #0a796

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Answer 1 · 2018-02-17T01:59:17

$3.7 cos(pi t-5.953)$ i.e. the constants are $A = 3.7$ , $omega = pi$ and $phi =2pi-tan^{-1}(12/35)=5.953$

To express the given function in the form $A cos (omega t + phi)$ , notice that according to the trigonometric sum rule

$A cos (omega t + phi) = A [cos (omega t )cos (phi)-sin(omega t)sin(phi)] = -A sin(phi) sin(omega t) + A cos(phi)cos(omega t)$

Comparing with the expression given, we have $omega = pi$ nd

$A sin(phi) = -1.2, qquad A cos(phi) = 3.5$

Thus

$A^2 = = (A sin(phi))^2 + (A cos(phi))^2= (-1.2)^2+(3.5)^2 = 13.69 = 3.7^2$

So, $A = 3.7$

Now,
$tan(phi) = {A sin(phi)}/{A cos(phi)} = -{1.2}/{3.5} = -0.3429$

Since $sin(phi)$ is negative while $cos(phi)$ is positive, the angle $phi$ must be in the fourth quadrant. So

$phi =2pi-tan^{-1}(12/35)=5.953$