Trigonometric Equations Question 18
Question: The number of pairs (x, y) satisfying the equations $ \sin x+\sin y=\sin (x+y) $ and $ |x|+|y|=1 $ is
Options:
A) 2
B) 4
C) 6
D) $ \infty $
Show Answer
Answer:
Correct Answer: C
Solution:
- The first equation can be written as $ 2\sin \frac{1}{2}(x+y)\cos \frac{1}{2}(x-y)=\sin(x+y) $ $ \therefore $ Either $ \sin \frac{1}{2}(x+y)=0 $ or $ \sin \frac{1}{2}x=0 $ or $ \sin \frac{1}{2}y=0 $ Thus $ x+y=-1,x-y=-1 $ . When $ x+y=0, $ we have to reject $ x+y=1 $ and check with the options or $ x+y=-1 $ and solve it with $ x-y=1 $ or $ x-y=-1 $ which gives $ ( \frac{1}{2},-\frac{1}{2} ) $ or $ ( -\frac{1}{2},,\frac{1}{2} ) $ as the possible solution. Again solving with $ x=0 $ , we get $ (0,\pm 1) $ and solving with $ y=0 $ , we get $ (\pm 1,0) $ as the other solution. Thus we have six pairs of solution for $ -4,,-3,,-2,,-1,,0,,1,,2,,3 $ and $ y $ .
BETA
