Question

If cosθ = x, then = ± √1+x/2.

A. sin θ/2
B. cos2θ
C. sin2θ
D. cos θ/2

Answer 1

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`\cos( \alpha ) = x`

`{sin}^(2)( \alpha ) = 1 - {cos}^(2)(x)`

`{sin}^(2)( \alpha ) = 1 - ({x})^(2)`

`\sin( \alpha ) = ± \sqrt{1 - {x}^(2) }`

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Look :

`\sin(2 \alpha ) = 2. \sin( \alpha ) . \cos( \alpha )`

`\sin(2 \alpha ) = 2 * ( ± \sqrt{1 - {x}^(2) })(x) \\`

`\sin(2 \alpha ) = ± \: 2 \: x \: \sqrt{1 - {x}^(2) }`

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`\cos(2 \alpha ) = {cos}^(2)(x) - {sin}^(2)(x)`

`\cos(2 \alpha ) = {x}^(2) - ({± \: \sqrt{1 - {x}^(2) } })^(2) \\`

`\cos(2 \alpha ) = {x}^(2) - |1 - {x}^(2) |`

If | 1 - x² | ≥ 0 :

`\cos(2 \alpha ) = {x}^(2) - (1 - {x}^(2))`

`\cos(2 \alpha ) = {x}^(2) - 1 + {x}^(2)`

`\cos(2 \alpha ) = 2 {x}^(2) - 1`

If | 1 - x² | < 0 :

`\cos(2 \alpha ) = {x}^(2) - ( - 1 )(1 - {x}^(2) ) \\`

`\cos(2 \alpha ) = {x}^(2) + 1 - {x}^(2)`

`\cos(2 \alpha ) = 1`

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`{sin}^(2)( ( \alpha )/(2) ) = (1 - \cos( \alpha ) )/(2) \\`

`{sin}^(2)( ( \alpha )/(2) ) = (1 - x)/(2) \\`

`\sin( ( \alpha )/(2) ) = ± \sqrt{ (1 - x)/(2) } \\`

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`{cos}^(2)( ( \alpha )/(2) ) = (1 + \cos( \alpha ) )/(2) \\`

`{cos}^(2)( ( \alpha )/(2) ) = (1 + x)/(2) \\`

`\cos( ( \alpha )/(2) ) = ± \sqrt{ (1 + x)/(2) } \\`

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Thus the correct answer is (( D )) .

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