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)
They are both not right angles. You can easily tell because you can't draw a square in the corner.
