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प्रश्न
Find the value of cos 2A, A lies in the first quadrant, when sin A = `4/5`
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उत्तर
we know sin2A + cos2A = 1
cos2 A = 1 – sin2A
= `1 - (4/5)^2`
= `1 - 16/25`
= `(25 - 16)/25`
= `9/25`
cos A = `+- sqrt(9/25)`
= `+- 3/5`
Since A lies in the first quadrant, cos A is positive
∴ cos A = `3/5`
cos 2A = cos2A – sin2A
= `(3/5)^2 - (4/5)^2`
= `9/25 - 16/25`
= `(9 - 16)/25`
= `(-7)/25`
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