Trigonometric EquationMedium
Question
The number of pairs (x, y) satisfying the equations sin x + sin y = sin (x + y) and |y| = 1 is-
Options
A.2
B.4
C.6
D.None of these
Solution
sin x + sin y = sin (x + y) we can write
2sin
(x + y) cos
(x - y) = 2sin
(x + y) cos
(x + y)
or 2sin
(x +y) {cos 
(x - y) - cos 
(x + y) = 0
or 2sin
(x + y). 2sin 
x sin 
y = 0
Either 2sin
= 0 or sin 
= 0 or sin
= 0
⇒ x + y = 0, x = 0, y = 0 and |x| + |y| = 1
⇒ x + y = 1, x - y = 1
x + y = -1, x - y = 1-
When x + y = 0, we have to reject x + y = 1 or -1 and solve it with x - y = 1 or x - y = -1
which gives
or 
as the possible solution. Again solving with x = 0, we get (0 ± 1) and by solving with y = 0, we get (±1, 0) as the other solution. Thus we have 6 pairs of solutions for x and y
2sin
(x + y) cos (x - y) = 2sin(x + y) cos(x + y)or 2sin
(x +y) {cos (x - y) - cos (x + y) = 0or 2sin
(x + y). 2sin x sin y = 0Either 2sin
= 0 or sin = 0 or sin = 0⇒ x + y = 0, x = 0, y = 0 and |x| + |y| = 1
⇒ x + y = 1, x - y = 1
x + y = -1, x - y = 1-
When x + y = 0, we have to reject x + y = 1 or -1 and solve it with x - y = 1 or x - y = -1
which gives
or as the possible solution. Again solving with x = 0, we get (0 ± 1) and by solving with y = 0, we get (±1, 0) as the other solution. Thus we have 6 pairs of solutions for x and yCreate a free account to view solution
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