# Thread: hey one more trig question im stuck on

1. ## hey one more trig question im stuck on

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

2. Haven't tried it myself but try using the identity

$\sin(u)\sin(v) = \frac{1}{2}[\cos(u-v) - \cos(u+v)]$

EDIT - Having re-read the question I dont think it'll work.

Haven't tried it myself but try using the identity

$\sin(u)\sin(v) = \frac{1}{2}[\cos(u-v) - \cos(u+v)]$

EDIT - Having re-read the question I dont think it'll work.
yep lol havent seen that idenity in my life

4. Originally Posted by elliegurl
yep lol havent seen that idenity in my life
have you seen these two idenities
$\cos(c-d)=\cos c \cos d+\sin c \sin d \ \ .............(1)$
$\cos(c+d)=\cos c \cos d-\sin c \sin d \ \ .............(2)$
on subtracting (2) from (1), you get
$\cos(c-d)-\cos(c+d)=2\sin c \sin d$

now LHS
$4\sin(x+\frac{1}{6} \pi) \sin(x-\frac{1}{6} \pi) = 2\{2\sin(x+\frac{1}{6} \pi) \sin(x-\frac{1}{6} \pi)\}$
$= 2 \{ \cos \{(x+\frac{1}{6} \pi)-(x-\frac{1}{6} \pi)\} - \cos \{(x+\frac{1}{6} \pi)+(x-\frac{1}{6} \pi)\} \}$
$= 2 \{\cos \frac{\pi}{3} - \cos {2x}\}= 2 \{\frac{1}{2} - \cos {2x}\}$
$= 1- 2 \cos {2x} =1-2(2\cos^2x-1) =1-4\cos^2x+2$
$=3-4\cos^2x=RHS$