Trig identity (1 Viewer)

[Joshmosh2]

Which of the following is an expression for sin8x/sinx
a)cos4xcos2xcosx
b)2cos4xcos2xcosx
c)4cos4xcos2xcosx
d)8cos4xcos2xcosx

I have no idea how to deal with the sin8x
Any help or suggestions are greatly appreciated

Update: Found a cheap method to find the answer, but I want to know what to do with the sin8x

 

[bakstar]

Use Double Angle Formula

sin2x = 2sinxcosx

 

[Joshmosh2]

Use Double Angle Formula

sin2x = 2sinxcosx

Yes, I definately took that into account, but how would that be expressed?

sin8x = 2(sin4xcos4x)
= 2((2sin2xcos2x)(2cos^2x-2sin^2x))...

 

[Thank_You]

I got it!

Answer is D

HOW?

OK,

= (2sin4xcos4x)/sinx

Since cos4x is in the all the answers, I'm going to leave it alone.

= 2sinx4x/sinx x cos4x (Double angled sin4x)
= 2(2sin2xcos2x)/sinx x cos4x (Double angled sin2x)
= 4(2sinxcosx)(cos^2(x)-sin^2(x))/sinx x cos4x
= 8cos4xcosx[(cos^2(x)-sin^2(x)] (Crossed out sinx)
= 8cos4xcosx (2cos^2(x)-1) (Using trig identity: sin^2(x) + cos^2(x) =1)
= 8cos4xcosx (1+cos2x -1 ) (Combining the sin and cos identity, I get: 1+cos2x =2cos^2(x), thus substituting..)
= 8cos4xcos2xcosx

Hence answer is D