Prove the identity: cos(3t)=cos^3(t)-3sin^2(t)cos(t)

1 answer:

3 0

Solutions

To solve the given equation lets use double-angle theorem. Double-angle formulas allow the expression of trigonometric functions of angles equal to 2α in terms of α, which can simplify the functions and make it easier to perform more complex calculations, such as integration, on them.

cos(3t)=cos^3(t)-3sin^2(t)cos(t)

cos(A+B) = cosAcosB - sinAsinB

Now cos(3t) = cos(2t+t) 

⇒cos(2t)cos(t) - sin(2t)sin(t) 

⇒(2cos^2(t)-1)cos(t) - 2sin(t)cos(t)sin(t) 

⇒ 2cos^(3)t - cos(t) - 2sin^2(t)cos(t) 

⇒ 2cos^3(t) - cos(t) - 2(1-cos^2(t))cos(t) 

⇒2cos^3(t) - cos(t) - 2cos(t) + 2cos^3(t) 

⇒4cos^3(t) - 3cos(t)

You might be interested in

Find the value of tan(sin^-1(1/2))

suter [353]

If you know that $\sin\dfrac\pi3=\dfrac12$ , then you know right away

$\tan\left(\sin^{-1}\dfrac12\right)=\tan\dfrac\pi3=\dfrac1{\sqrt}3=\dfrac{\sqrt3}3$

###

Otherwise, you can derive the same result. Let $\theta=\sin^{-1}\dfrac12$ , so that $\sin\theta=\dfrac12$ . $\sin^{-1}$ is bounded, so we know $-\dfrac\pi2\le\theta\le\dfrac\pi2$ . For these values of $\theta$ , we always have $\cos\theta\ge0$ .