# solving trigo equation

• Aug 28th 2009, 09:14 AM
wintersoltice
solving trigo equation
$\displaystyle 5cos^2 \theta + 12 cos \theta sin \theta = a + b cos (2\theta + y)$
Find a , b and y.

what i tried to do is:
$\displaystyle 5cos^2 \theta + 12 cos \theta sin \theta$
$\displaystyle =5(cos 2\theta + sin^2 \theta)+ 6(2 cos \theta sin \theta)$
$\displaystyle =5sin^2 \theta + 5 cos 2\theta + 6sin2\theta$
$\displaystyle =5sin^2 \theta + \sqrt{61} cos (2\theta-50.2^\circ)$

then....stuck....is $\displaystyle a= 5sin^2 \theta$ ?
or it can be further simplified/manipulated??
$\displaystyle a= 5sin^2 \theta$ doesn't seem to be the final answer.....
• Aug 28th 2009, 11:02 AM
Opalg
The question asks you to write $\displaystyle 5\cos^2\theta+12\cos\theta\sin\theta$ in terms of $\displaystyle 2\theta$. So I would start by using the formulas $\displaystyle \cos^2\theta = \tfrac12(1+\cos(2\theta))$ and $\displaystyle 2\cos\theta\sin\theta = \sin(2\theta)$. That gives

$\displaystyle 5\cos^2\theta+12\cos\theta\sin\theta = \tfrac52(1+\cos(2\theta)) + 6\sin(2\theta)$. Next step would be to write this as $\displaystyle \tfrac52 + \tfrac{13}2\bigl(\tfrac5{13}\cos(2\theta) + \tfrac{12}{13}\sin(2\theta)\bigr).$ Can you see why, and how to take it from there?
• Aug 28th 2009, 11:11 AM
running-gag
Quote:

Originally Posted by wintersoltice
$\displaystyle 5cos^2 \theta + 12 cos \theta sin \theta = a + b cos (2\theta + y)$
Find a and b.
Hence, solve $\displaystyle 5cos^2 \theta + 12 cos \theta sin \theta = 0$.

what i tried to do is:
$\displaystyle 5cos^2 \theta + 12 cos \theta sin \theta$
$\displaystyle =5(cos 2\theta + sin^2 \theta)+ 6(2 cos \theta sin \theta)$
$\displaystyle =5sin^2 \theta + 5 cos 2\theta + 6sin2\theta$
$\displaystyle =5sin^2 \theta + \sqrt{61} cos (2\theta-50.2^\circ)$

Hi

$\displaystyle 5 \cos^2 \theta + 12 \cos \theta sin \theta = a + b \cos (2\theta + y)$

$\displaystyle \int_{0}^{\pi}(5 \cos^2 \theta + 12 \cos \theta \sin \theta) d \theta = \int_{0}^{\pi}(a + b \cos (2\theta + y)) d \theta$

$\displaystyle \int_{0}^{\pi}(5 \frac{\cos 2\theta +1}{2} + 6 \sin 2\theta) d \theta = \int_{0}^{\pi}(a + b \cos (2\theta + y)) d \theta$

$\displaystyle \frac{5}{2} \pi = a \pi$

$\displaystyle a = \frac{5}{2}$

$\displaystyle 5 \cos^2 \theta -\frac{5}{2} + 12 \cos \theta \sin \theta = b \cos (2\theta + y)$

To find b substitute $\displaystyle 5 \cos^2 \theta -\frac{5}{2} = \frac{5}{2} \cos 2\theta$ and $\displaystyle 12 \cos \theta \sin \theta = 6 \sin 2\theta$
• Aug 28th 2009, 10:45 PM
wintersoltice
i see .....

so....

$\displaystyle 5cos^2 \theta + 12 cos \theta sin\theta$

$\displaystyle =\frac{5}{2}(1+cos2\theta) + 6 sin 2\theta$

$\displaystyle =\frac{5}{2} + \frac{5}{2}cos2\theta + 6 sin2\theta$

$\displaystyle =\frac{5}{2} + \frac{13}{2} cos (2\theta - 67.4degree)$

so.....$\displaystyle a=\frac{5}{2} , b=\frac{13}{2} , y=-67.4degree$

thanks:)