+3 votes
in JEE by kratos

If sinx=3÷5, cosy=-12÷13, where x and y both lie in second quadrant find the value of sin(x+y)

1 Answer

+4 votes
by kratos
 
Best answer

We know that sin(x)2 + cos(x)2 = 1

(3/5)2 + cos(x)2 = 1
9/25 + cos(x)2 = 25/25
cos(x)2 = 16/25
cos(x) = +/- (4/5)
Since it is in second quadrant cos(x) = -4/5
sin(y)2 + (-12/13)2 = 1
sin(y)2 + 144/169 = 169/169
sin(y)2 = 25/169
sin(y) = +/- 5/13
Since it is in second quadrant sin(y) = 5/13
sin(x) = 3/5
cos(x) = -4/5
sin(y) = 5/13
cos(y) = -12/13
sin(x + y) => sin(x)cos(y) + sin(y)cos(x)

=> (3/5) (-12/13) + (5/13) (-4/5)

=> -36 / 65 + -20 / 65

=> -56/65

So, sin(x+y) = -56/65

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