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प्रश्न
If 2 cos 2A = `sqrt3` and A is acute,
find:
(i) A
(ii) sin 3A
(iii) sin2 (75° - A) + cos2 (45° +A)
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उत्तर
(i) 2 cos 2A = `sqrt3`
cos 2A = `(sqrt3)/(2)`
cos 2A = cos 30°
2A = 30°
A = 15°
(ii) sin 3A = sin 3(15°)
= sin 45°
= `(1)/(sqrt2)`
(iii) sin2(75° – A ) + cos2 (45 + A) = sin2 ( 75° –15°) + (cos2 ( 45° + 15°)
= sin2 60° + cos2 60°
= `(sqrt3/2)^2 + (1/2)^2`
= `(3)/(4) + (1)/(4)`
= 1
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