hey its tri compund angle formulae proof.
Prove that 4sin(x+1/6Pi)sin(x-1/6pi) is identical to 3-4cos^2x
just tried to solve it but really confused.
thanks
have you seen these two idenities
$\displaystyle \cos(c-d)=\cos c \cos d+\sin c \sin d \ \ .............(1)$
$\displaystyle \cos(c+d)=\cos c \cos d-\sin c \sin d \ \ .............(2)$
on subtracting (2) from (1), you get
$\displaystyle \cos(c-d)-\cos(c+d)=2\sin c \sin d$
now LHS
$\displaystyle 4\sin(x+\frac{1}{6} \pi) \sin(x-\frac{1}{6} \pi) = 2\{2\sin(x+\frac{1}{6} \pi) \sin(x-\frac{1}{6} \pi)\}$
$\displaystyle = 2 \{ \cos \{(x+\frac{1}{6} \pi)-(x-\frac{1}{6} \pi)\} - \cos \{(x+\frac{1}{6} \pi)+(x-\frac{1}{6} \pi)\} \}$
$\displaystyle = 2 \{\cos \frac{\pi}{3} - \cos {2x}\}= 2 \{\frac{1}{2} - \cos {2x}\}$
$\displaystyle = 1- 2 \cos {2x} =1-2(2\cos^2x-1) =1-4\cos^2x+2$
$\displaystyle =3-4\cos^2x=RHS$