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Question
If x = a sin θ and y = b cos θ, what is the value of b2x2 + a2y2?
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Solution
Given:
x = a sinθ and y = b cosθ
So, \[b^2 x^2 + a^2 y^2 = b^2 \left( asin\theta \right)^2 + a^2 \left( bcos\theta \right)^2 \]
\[ = a^2 b^2 \sin^2 \theta + a^2 b^2 \cos^2 \theta\]
\[ = a^2 b^2 \left( \sin^2 \theta + \cos^2 \theta \right)\]
We know that, `sin^2 θ+cos^2θ=1`
Therefore,
\[b^2 x^2 + a^2 y^2 = a^2 b^2\]
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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`
