Question

Please someone help me!!!!!!​

Answer 1

Answer:  see proof below

Step-by-step explanation:

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

Use the following Half-Angle Identities:  

$\sin\bigg(\dfrac{A}{2}\bigg)=\sqrt{\dfrac{1-\cos A}{2}}\\\\\cos\bigg(\dfrac{A}{2}\bigg)=\sqrt{\dfrac{1+\cos A}{2}}$

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

Use the Unit Circle to evaluate: $\cos\dfrac{\pi}{4}=\sin\dfrac{\pi}{4}=\dfrac{1}{\sqrt2}$

Proof LHS → RHS

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

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

RHS = LHS:   sin A = sin A  $\checkmark$

Answer 2

Answer:

see explanation

Step-by-step explanation:

Using the identity

cos2Θ = 1 - 2sin²Θ, then

1 - 2sin²($\frac{\pi }{4}$ - $\frac{0}{2}$ )

= cos [2($\frac{\pi }{4}$ - $\frac{0}{2}$ )]

= sos($\frac{\pi }{2}$ - Θ )

= cos$\frac{\pi }{2}$cosΘ + sin

= 0 × cosΘ + 1 × sinΘ

= 0 + sinΘ

= sinΘ = right side