Since sin(2x)=2sinxcosx, we can plug that in to get sin(4x)=2sin(2x)cos(2x)=2*2sinxcosxcos(2x)=4sinxcosxcos(2x). Since cos(2x) = cos^2x-sin^2x, we plug that in. In addition, cos4x=cos^2(2x)-sin^2(2x). Next, since cos^2x=(1+cos(2x))/2 and sin^2x= (1-cos(2x))/2, we plug those in to end up with 4sinxcosxcos(2x)-((1+cos(2x))/2-(1-cos(2x))/2)
=4sinxcosxcos(2x)-(2cos(2x)/2)=4sinxcosxcos(2x)-cos(2x)
=cos(2x)*(4sinxcosx-1). Since sinxcosx=sin(2x), we plug that back in to end up with cos(2x)*(4sin(2x)-1)
I don’t know tbh cause I don’t know what that is
Answer:distributive
Step-by-step explanation:
Answer:
Yes, they are equivalent
Step-by-step explanation:
Divide the second numerator by the first numerator
Divide the second denominator by the first denominator
Compare the quotients and if they are the same then the fractions are equivalent
Answer:
no you cant
Step-by-step explanation:
