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प्रश्न
If θ = 15°, find the value of: cos3θ - sin6θ + 3sin(5θ + 15°) - 2 tan23θ
उत्तर
θ = 15°
`(3)/(2)cos3θ - sin6θ + 3sin(5θ + 15°) - 2tan^2 3θ`
= `(3)/(2)cos 3 xx 15° - sin6 xx 15° + 3sin(5 xx 15° + 15°) -2tan^2 3 xx 15°`
= `(3)/(2)cos45° - sin90° + 3sin90° - 2tan^2 45°`
= `(3)/(2) xx (1)/sqrt(2) - 1 + 3 xx 1 - 2 xx (1)^2`
= `(3)/(2sqrt(2)) - 1 + 3 - 2`
= `(3)/(2sqrt(2))`
= `(3)/(2sqrt(2)) xx sqrt(2)/sqrt(2)`
= `(3sqrt(2))/(4)`.
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