Question

Please someone help me...​

Answer 1

Answer:

they take the same form

Step-by-step 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, $\sin^2 (2A)=4\sin^2A\cos^2A$