1. ## Trigo-system

Solve : $sin(x)sin(y)=\frac{\sqrt{3}}{4}$
$cos(x)cos(y)=\frac{\sqrt{3}}{4}$

2. Originally Posted by dhiab
Solve : $sin(x)sin(y)=\frac{\sqrt{3}}{4}$
$cos(x)cos(y)=\frac{\sqrt{3}}{4}$
Dear dhiab,

Use the trignometric identities given below. Then you would be able to solve your problem.

$cos(A+B)=cosAcosB-sinAsinB$

$cos(A-B)=cosAcosB+sinAsinB$

3. Originally Posted by dhiab
Solve : $sin(x)sin(y)=\frac{\sqrt{3}}{4}$
$cos(x)cos(y)=\frac{\sqrt{3}}{4}$
It follows that:
$\cos(x+y)=\cos(x)\cos(y)-\sin(x)\sin(y)=\frac{\sqrt{3}}{4}-\frac{\sqrt{3}}{4}=0$

Hence,

$x+y=\frac{\pi}{2}+n\pi, \qquad n\in\mathbb{Z}$

Now plug $y=\frac{\pi}{2}+n\pi$ into one of the above equations to learn what additional conditions $x$ and $y$ have to satisfy...

4. Originally Posted by dhiab
Solve : $sin(x)sin(y)=\frac{\sqrt{3}}{4}$
$cos(x)cos(y)=\frac{\sqrt{3}}{4}$
$cos(x)cos(y) + sin(x)sin(y) = \frac{\sqrt{3}}{2}]$

$cos(x-y) = \frac{\sqrt{3}}{2}$

x - y = π/6 .......(1)

If you subtract the above two equation you will get

cos(x+y) = 0 of x+y = π/2.....(2)

From eq.1 and 2, find x and y.

5. Hello dhiab
Originally Posted by dhiab
Solve : $sin(x)sin(y)=\frac{\sqrt{3}}{4}$
$cos(x)cos(y)=\frac{\sqrt{3}}{4}$
Using $\sin x \sin y = \tfrac12\big(\cos(x-y) -\cos(x+y)\big)$:
$\sin x\sin y=\frac{\sqrt{3}}{4}$

$\Rightarrow \cos(x-y)-\cos(x+y) = \frac{\sqrt3}{2}$ ... (1)
Similarly:

$\cos x \cos y =\frac{\sqrt{3}}{4}$

$\Rightarrow \cos(x-y)+\cos(x+y) = \frac{\sqrt3}{2}$ ... (2)

$\cos(x-y) =\frac{\sqrt3}{2}$

$\Rightarrow x-y = 2n\pi\pm\frac{\pi}{6}$ ... (3)

Subtract (1) and (2):
$\cos(x+y)=0$

$\Rightarrow x+y = 2n\pi\pm\frac{\pi}{2}$ ...(4)

$x=2n\pi \pm \frac{\pi}{3}$
$\Rightarrow y = 2n\pi\pm\frac{\pi}{6}$