Question

Someone please help me!!!!​

Answer 1

Below.

Explanation:

Left side = 2 cos^2 ( π/4 - A/2) - 1

= 2 ( cos π/4 cos A/2 + sin π/4 sin A/2)^2 - 1

Now sin π/4 and cos π/4 = 1 /√2 so:

= 2 ( 1/√2 cos A/2 + 1/√2 sin A/2)^2 - 1

= 2 * 1/2( cos^2 A/2 + sin^2 A/2 + 2 sin A/2 cos A/2) - 1

But cos^2 a/2 + sin^2 A/2 = 1 so we have:

2 * 1/2( 1 + 2sin A/2 cos A/2) - 1

= 1 + 2 sin A/2 cos A/2 - 1

= 2 sin A/2 cos A/2

Using the identity 2 sin A cos A = sin 2A

2 sin A/2 cos A/2 = sin A = right side.

So left side = right side and the identity is proved.

Answer 2

Answer: see proof below

Explanation:

Use the Double Angle Identity: cos 2A = 2cos²A - 1

Use the Difference Identity: cos (A - B) = cosA · cosB + sinA · sinB

Use the Unit Circle to evaluate: cos (π/2) = 0 & sin (π/2) = 1

Proof LHS → RHS

`\text{Given:}\qquad \qquad \qquad \qquad 2\cos^2\bigg((\pi)/(4)-(A)/(2)\bigg)-1\\\\\\\text{Double Angle Identity:}\quad \cos2\bigg((\pi)/(4)-(A)/(2)\bigg)\\\\\\\text{Simplify:}\qquad \qquad \qquad \quad \cos\bigg((\pi)/(2)-A\bigg)\\\\\\\text{Difference Identity:}\qquad \cos(\pi)/(2)\cdot \cos A+\sin (\pi)/(2)\cdot \sin A\\\\\\\text{Unit Circle:}\qquad \qquad \qquad 0\cdot \cos A+1\cdot \sin A\\\\\\\text{Simplify:}\qquad \qquad \qquad \qquad \sin A`

LHS = RHS: sin A = sin A
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