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प्रश्न
Factorize: sin3θ + cos3θ
Hence, prove the following identity:
`(sin^3θ + cos^3θ)/(sin θ + cos θ) + sin θ cos θ = 1`
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उत्तर
sin3θ + cos3θ
= (sin θ + cos θ)(sin2θ + cos2 – sin θ cos θ)
= (sin θ + cos θ)(1 – sin θ cos θ). ...(i)
L.H.S = `(sin^3θ + cos^3θ)/(sinθ + cosθ) + sinθcosθ`
= `((sinθ + cosθ)(1 - sinθcosθ))/((sinθ + cosθ)) + sinθcosθ` ...(From(i))
= 1 – sin θ cos θ + sin θ.cos θ
Simplify by cancelling – sin θ cos θ and + sin θ.cos θ
= 1
= R.H.S.
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Find the value of sin2θ + cos2θ
Solution:
In Δ ABC, ∠ABC = 90°, ∠C = θ°
AB2 + BC2 = `square` .....(Pythagoras theorem)
Divide both sides by AC2
`"AB"^2/"AC"^2 + "BC"^2/"AC"^2 = "AC"^2/"AC"^2`
∴ `("AB"^2/"AC"^2) + ("BC"^2/"AC"^2) = 1`
But `"AB"/"AC" = square and "BC"/"AC" = square`
∴ `sin^2 theta + cos^2 theta = square`
