Question

Please someone help me...​

Answer 1

they take the same form

Explanation:

factor (1 - 1/4 sin ^2 (2A) ) (cos ^ 2 (A) -sin ^2(A))

= ( -sin 2A/ 2) + 1 ) (sin (2A)/ 2) -1

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

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

Answer 2

use
`a^2-b^2=(a+b)(a-b)`

to get
`(\cos^3A-\sin^3A)(\cos^3A+\sin^3A)`

then use
`a^3+b^3=(a+b)(a^2+b^2-ab)`

and
`a^3-b^3=(a-b)(a^2+b^2+ab)`

also,
`\sin^2\theta+\cos^2\theta=1`

to get
`(\cos A-\sin A)(1+\sin A\cos A)(\cos A+ \sin A)(1-\sin A\cos A)`

then again use the first identity In both pairs, i.e.

`(\cos A-\sin A)(\cos A+ \sin A) \cdot (1+\sin A\cos A)(1-\sin A\cos A)`

to get
`\cos 2A (1-\sin^2A\cos^2A)`

multiply and divide by 4 to get the RHS.

because,
`\sin(2A)= 2\sin A \cos A`

squaring both sides,
`\sin^2 (2A)=4\sin^2A\cos^2A`