Question

Use the squared identities to simplify 2sin^2xsin^2x.

Answer 1

\[2 \sin ^2x \sin ^2x=\frac{ \left( 2\sin ^2x \right)\left( 2 \sin ^2x \right) }{ 2 }=\frac{ \left( 1-\cos 2x \right) \left( 1-\cos 2x \right)}{ 2}\]

Answer 2

A.3+4cos(2x)+cos(4x)/4
B.3-4cos(2x)+cos(4x)/4
C.3-4cos(2x)-cos(4x)/4
D.3+4cos(2x)-cos(4x)/4

Answer 3

@Surjithayer

Answer 4

\[=\frac{ 1-2\cos 2 x+\cos ^22x }{ 2 }=\frac{ 2-4 \cos 2x+2\cos ^2 2 x }{ 4 }\]
\[=\frac{ 2-\cos 2x+1+\cos 4x }{ 4 }=?\]

Answer 5

correction -4 cos 2x not -cos 2x

Answer 6

3-4cos2x+cos4x/4? so its B? right? @surjithayer

Answer 7

correct.

Answer 8

@surjithayer Thank you so much sir :D can you show me the steps on how you got the final answer? :)

Answer 9

\[\cos 2x=\cos \left( x+x \right)=\cos x \cos x-\sin x \sin x=\cos ^2x-\sin ^2x\]
\[\cos 2x=\cos ^2x-(1-\cos ^2x)=2\cos ^2x-1,1+\cos 2x==2\cos ^2x\]
\[\cos 2x=1-\sin ^2x-\sin ^2x=1-2\sin ^2x,2\sin ^2x=1-\cos 2x\]

Answer 10

i multiplied and divided the numerator and denominator by 2