Question

Show that 12 sin⁡〖2θ-5 cos⁡〖2θ+4 〖cos θ-〗⁡〖5=cos⁡〖θ (24 sin⁡〖θ-10 cos⁡〖θ+4)〗 〗 〗 〗 〗 〗.

Answer 1

\[12 \sin 2\theta -5 \cos 2\theta+4\cos \theta-5=\cos \theta \left( 24 \sin \theta - 10 \cos \theta +4 \right)\]

Answer 2

writing C for cos(theta) and S for sin(theta).
You need to use the following
sin (2theta) = 2sin(theta).cos(theta) = 2SC
cos(2theta) = cos(theta).cos(theta) - sin(theta).sin(theta)
sin(theta) = 1-cos(theta).cos(theta)
LHS = 12sin(2theta) - 5 cos(2theta) + 4 cos(theta) - 5
= 12*2SC - 5(C^2 - S^2) + 4C - 5
= 24SC - 5C^2 + 5S^2 + 4C - 5
= 24SC - 5C^2 + 5(1-C^2) + 4C - 5
= 24SC - 5C^2 + 5 - 5C^2 + 4C - 5
= 24SC - 10C^2 + 4C....... the +5 & -5 wipe each other out
= C(24C - 10C + 4)
done.

Answer 3

thanks :)