Question

solve sin-1x + sin-1y = 2pi/3 and cos-1x - cos-1y = pi/3

Answer 1

Answer:

The solution is

x= 1/2 and y= 1

Step-by-step explanation:

Formula used;

$sin^{-1}x+cos^{-1}x=\frac{\pi}{2}$

$Now,\\\\sin^{-1}x+sin^{-1}y=\frac{2\pi}{3}\\\\\frac{\pi}{2}-cos^{-1}x+\frac{\pi}{2}-cos^{-1}y=\frac{2\pi}{3}\\\\\pi-(cos^{-1}x+cos^{-1}y)=\frac{2\pi}{3}\\\\cos^{-1}x+cos^{-1}y=\frac{3\pi-2\pi}{3}\\\\cos^{-1}x+cos^{-1}y=\frac{\pi}{3}......(1)\\\\cos^{-1}x-cos^{-1}y=\frac{\pi}{3}......(2)$

Adding (1) and (2) we get

$2cos^{-1}x=2\frac{\pi}{3}\\\\cos^{-1}x=\frac{\pi}{3}\\\\x=cos\frac{\pi}{3}=\frac{1}{2}$

put $cos^{-1}x=\frac{\pi}{3}$ in (1) we get

$\frac{\pi}{3}+cos^{-1}y=\frac{\pi}{3}\\\\cos^{-1}y=0\\\\y = cos0=1$

Answer 2

Answer: The solution is,

x = √3/2 and y = 1

Step-by-step explanation:

Here, the given equations,

$sin^{-1} x + sin^{-1} y = \frac{2\pi}{3}$ -------(1)

$cos^{-1}-cos^{-1}=\frac{\pi}{3}$ -------(2)

Since,

$sin^{1}x+cos^{-1}=\frac{\pi}{2}\implies cos^{-1}=\frac{\pi}{2}-sin^{-1}x$

Similarly,

$cos^{-1}y=\frac{\pi}{2}-sin^{-1}y$,

Thus, from equation (2),

$\frac{\pi}{2}-sin^{-1}x-\frac{\pi}{2}+sin^{-1}y=\frac{\pi}{3}$

$\implies -sin^{-1}x+sin^{-1}y=\frac{\pi}{3}$  ---------(3),

Equation (1) + equation (3),

We get,

$2sin^{-1}y=\frac{2\pi}{3}+\frac{\pi}{3}=\pi$

$\implies sin^{-1}y = \frac{\pi}{2}$

$\implies y = 1$

Now, Equation (1) - Equation (3),

We get,

$2sin^{-1}x=\frac{2\pi}{3}-\frac{\pi}{3}=\frac{\pi}{3}$

$\implies sin^{-1}x = \frac{\pi}{6}$

$\implies x = \frac{\sqrt{3}}{2}$