Question

Please someone help me!!!!!!​

Answer 1

see explanation

Explanation:

Using the identity

cos2Θ = 1 - 2sin²Θ, then

1 - 2sin²(
`(\pi )/(4)` -
`(0)/(2)` )

= cos [2(
`(\pi )/(4)` -
`(0)/(2)` )]

= sos(
`(\pi )/(2)` - Θ )

= cos
`(\pi )/(2)`cosΘ + sin

= 0 × cosΘ + 1 × sinΘ

= 0 + sinΘ

= sinΘ = right side

Answer 2

Answer: see proof below

Explanation:

Use the Difference Identity: sin (A + B) = sin A cos B - cos A sin B

Use the following Half-Angle Identities:

`\sin\bigg((A)/(2)\bigg)=\sqrt{(1-\cos A)/(2)}\\\\\cos\bigg((A)/(2)\bigg)=\sqrt{(1+\cos A)/(2)}`

Use the Pythagorean Identity: cos²A + sin²A = 1 --> sin²A = 1 - cos²A

Use the Unit Circle to evaluate:
`\cos(\pi)/(4)=\sin(\pi)/(4)=(1)/(\sqrt2)`

Proof LHS → RHS

`\text{Given:}\qquad \qquad \qquad 1-2\sin^2\bigg((\pi)/(4)-(\theta)/(2)\bigg)\\\\\text{Difference Identity:}\quad 1-2\bigg(\sin(\pi)/(4)\cdot \cos (\theta)/(2)-\cos (\pi)/(4)\cdot \sin(\theta)/(2)\bigg)^2\\\\\text{Unit Circle:}\qquad \qquad 1-2\bigg((1)/(\sqrt2)\cos (\theta)/(2)-(1)/(\sqrt2)\sin (\theta)/(2)\bigg)^2\\\\\\\text{Half-Angle Identity:}\quad 1-2\bigg((√(1+\cos A))/(2)-(√(1-\cos A))/(2)\bigg)^2`

`\text{Expand Binomial:}\quad 1-2\bigg((1+\cos A)/(4)-(2√(1-\cos^2 A))/(4)+(1-\cos A)/(4)\bigg)\\\\\text{Simplify:}\qquad \qquad \quad 1-2\bigg((2-2√(1-\cos^2 A))/(4)\bigg)\\\\\text{Pythagorean Identity:}\quad 1-(1)/(2)\bigg(2-2√(\sin^2 A)\bigg)\\\\\text{Simplify:}\qquad \qquad \qquad 1-(1)/(2)(2-2\sin A)\\\\\text{Distribute:}\qquad \qquad \qquad 1-(1-\sin A)\\\\.\qquad \qquad \qquad \qquad \quad =1-1+\sin A\\\\\text{Simplify:}\qquad \qquad \qquad \sin A`

RHS = LHS: sin A = sin A
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