Let x and y be real numbers such that sin⁡xsin⁡y=3 and cos⁡xcos⁡y=12. The value of sin⁡2xsin⁡2y+cos⁡2xcos⁡2y can be expressed in the form pq, where p and q are relatively ' positive integers. Find p+q.

sin⁡2xsin⁡2y+cos⁡2xcos⁡2y=2sin⁡xcos⁡x2sin⁡ycos⁡y+cos2x−sin2xcos2y−sin2y=3sin⁡ycos⁡y2sin⁡ycos⁡y+(14)cos2y−9sin2ycos2y−sin2y=32+(14)cos2y−9sin2ycos2y−sin2yOn the other hand:1=cos2x+sin2x=14cos2y+9(1−cos2y),&nbsp;whence:&nbsp;cos2y=3235,&nbsp;sin2y=335We finally havesin2xsin2y+cos2xcos2y=32+143235-93353235-335=4958p+q=49+58=107

We have an answer to "Let x and y be real numbers such..."

Kendall Daniels

Answered question

2022-03-24

Let x and y be real numbers such that

\frac{\sin x}{\sin y} = 3

and

\frac{\cos x}{\cos y} = \frac{1}{2}

. The value of

\frac{\sin 2 x}{\sin 2 y} + \frac{\cos 2 x}{\cos 2 y}

can be expressed in the form

\frac{p}{q}

, where p and q are relatively ' positive integers. Find p+q.

Answer & Explanation

Roy Brady

Beginner2022-03-25Added 19 answers

$\frac{\sin 2 x}{\sin 2 y} + \frac{\cos 2 x}{\cos 2 y} = \frac{2 \sin x \cos x}{2 \sin y \cos y} + \frac{\cos^{2} x - \sin^{2} x}{\cos^{2} y - \sin^{2} y}$
$= \frac{3 \sin y \cos y}{2 \sin y \cos y} + \frac{(\frac{1}{4}) \cos^{2} y - 9 \sin^{2} y}{\cos^{2} y - \sin^{2} y}$
$= \frac{3}{2} + \frac{(\frac{1}{4}) \cos^{2} y - 9 \sin^{2} y}{\cos^{2} y - \sin^{2} y}$
On the other hand:
$1 = \cos^{2} x + \sin^{2} x = \frac{1}{4} \cos^{2} y + 9 (1 - \cos^{2} y), whence: \cos^{2} y = \frac{32}{35}, \sin^{2} y = \frac{3}{35}$
We finally have
$\frac{\sin 2 x}{\sin 2 y} + \frac{\cos 2 x}{\cos 2 y} = \frac{3}{2} + \frac{\frac{1}{4} \frac{32}{35} - 9 \frac{3}{35}}{\frac{32}{35} - \frac{3}{35}} = \frac{49}{58}$
$p + q = 49 + 58 = 107$

Do you have a similar question?

Recalculate according to your conditions!

New Questions in Precalculus

Didn't find what you were looking for?