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प्रश्न
If sin x = `15/17` and cos y = `12/13, 0 < x < pi/2, 0 < y < pi/2` find the value of sin(x + y)
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उत्तर
Given sin x = `15/17, 0 < x < pi/2`
We have cos2x + sin2x = 1
∴ cos2x = 1 – sin2x
= `1 - (15/17)^2`
= `1 - 225/289`
cos2x = `(289 - 225)/289 = 64/289`
cos x = `+- sqrt(64/289)`
= `+- 8/17`
Given that `0 < x < pi/2`, that is x lies in the first quadrant
∴ cos x is positive.
cos x = `8/17`
Also given cos y = `12/13, 0 < x < pi/2`
We have cos2y + sin2y = 1
sin2y = 1 – cos2y
sin2y = `1 - (12/13)^2 = 1 - 14/169`
sin2y = `(169 - 144)/169 = 25/169`
sin y = `+- sqrt(25/169) = +- 5/13`
Since `0 < y < pi/2, y lies in the first quadrant sin y is positive.
∴ sin y = `5/13`
sin x = `15/17`
sin y = `5/13`
cos x = `8/17`
cos y = `12/13`
sin(x + y) = sin x cos y + cos x sin y
= `15/17 * 12/13 + 8/17 * 5/13`
sin(x + y) = `180/221 + 40/221`
= `220/221`
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