If cos−1x−cos−12y =α where −1−1≤x≤1,−2≤y≤2,x≤2y then for all 4x2−4xycosα +y2 is equal to

cos(cos−1x−cos−12y ) =cosα cosα ⇒x×2y+1−x21−4y2 ⇒(cosα −2xy ) 2=(1−x2)(1−4y2 )x2+4y2−xycosα =1−cos2α =sin2α

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If ^{ -1 }{ x } - ^{ -1 }{ { y }{ 2 } } = where -1-1 x 1,-2

Question

If $cos^{- 1} x - cos^{- 1} \frac{y}{2} = α$ where $- 1 - 1 \leq x \leq 1, - 2 \leq y \leq 2, x \leq \frac{y}{2}$ then for all $4 x^{2} - 4 x y cos α + y^{2}$ is equal to

4 sin^{2} α - 2 x^{2} y^{2}

4 cos^{2} α + 2 x^{2} y^{2}

4 sin^{2} α

2 sin^{2} α

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Mila Discussed

cos^{- 1} x - cos^{- 1} \frac{y}{2} = α

where

- 1 - 1 \leq x \leq 1, - 2 \leq y \leq 2, x \leq \frac{y}{2}

then for all

4 x^{2} - 4 x y cos α + y^{2}

is equal to

15 mins ago

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15 mins ago

cos (cos^{- 1} x - cos^{- 1} \frac{y}{2}) = cos α

cos α \Rightarrow x \times \frac{y}{2} + 1 - x^{2} 1 - \frac{y ^{2}}{4} \Rightarrow (cos α - \frac{x y}{2})^{2} = (1 - x^{2}) (1 - \frac{y ^{2}}{4})

x^{2} + \frac{y ^{2}}{4} - x y cos α = 1 - cos^{2} α = sin^{2} α

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Question Text	If $cos^{- 1} x - cos^{- 1} \frac{y}{2} = α$ where $- 1 - 1 \leq x \leq 1, - 2 \leq y \leq 2, x \leq \frac{y}{2}$ then for all $4 x^{2} - 4 x y cos α + y^{2}$ is equal to
Answer Type	Text solution:1
Upvotes	152