Question

Prove that
\[4\sin(x+{1\over6}\pi)\sin(x-{1\over6}\pi)=3-4\cos^2x\]

Answer 1

sin(a+b) = sina*cosb + sinb*cosa
sin(a-b) = sina*cosb - sinb*cosa

a = x
b = pi/6
sin b = 1/2
cos b = sqrt3/2

Answer 2

Yes, I know that. But how do I work with the 4sin?

Answer 3

4sin(x+1/6π)sin(x−1/6π) = -2{ cos[(x+1/6π)+(x-1/6π)] - cos[(x+1/6π)-(x-1/6π)] }
= 2cos 1/3π -2 cos 2x = 1 - 2[ 2(cosx)^2 -1] = 1- 4(cosx)^2 +2 = 3-4(cosx)^2

Answer 4

\[\rightarrow 4(\frac{\sqrt{3}}{2}\sin x+\frac{1}{2} \cos x)(\frac{\sqrt{3}}{2}\sin x-\frac{1}{2} \cos x)\]
\[=4(\frac{3}{4}\sin^{2} x - \frac{1}{4}\cos^{2} x)\]
\[=3 \sin^{2} x - \cos^{2} x\]

Answer 5

first step : product to sum formula
second step: double angle formula

Answer 6

@Callisto How? I don't understand...

Answer 7

sinAsinB = -1/2 [cos(A+B)-cos(A-B) ]
put A= x+1/6π and B = x-1/6π
then you'll get 2cos 1/3π -2 cos 2x

Answer 8

then use double angle formula, cos2x = 2(cosx)^2 -1 to reduce 2cos 1/3π -2 cos 2x
ps : 2cos 1/3π = 2 (0.5) =1
-2 cos 2x = -2[2(cosx)^2 -1] = -4(cosx)^2 +2
combine the two, which gives 1-4(cosx)^2 +2 that is 3-4(cosx)^2

Answer 9

I've never learned the product formula...? Is there another way to do it without the product formula?

Answer 10

hmm dumbcow has shown you lol only left one step for you that is 3(sinx)^2 - (cosx)^2 =
3 [ 1-(cosx)^2 ] - (cosx)^2 = 3 - 4 (cosx)^2

Answer 11

how does 3(sinx)^2 become 3[ 1- (cosx)^2]?

Answer 12

\[\sin^2x+\cos^2x=1\]
so
\[\sin^2x=1-\cos^2x\]

Answer 13

Oh Ok. Thanks.