C3 Trigonometry

I'm stuck on this question.

I began like this:

Let 'X' = x
And 'Y' = (3x-pi/4)
Therefore
X + Y = 4x - pi/4
X - Y = pi/4 - 2x

So the equation can be rewritten as:

(sqrt2)2cos(2x - pi/8)cos(pi/8 - x) = sinx

I don't know what to do next though, nor am I sure my sqrt two is in the right place.

start by expanding the lump with root2 in front using the compound angle
put the numbers in
simplify fully
then you will see where the hence comes from

Alrighty, I did what you said and got:

cos x + cos 3x + sin 3x sqrt 2 = sin x

I can't see where the hence comes from, though.

I'm stuck on this question.

I began like this:

Let 'X' = x
And 'Y' = (3x-pi/4)
Therefore
X + Y = 4x - pi/4
X - Y = pi/4 - 2x

So the equation can be rewritten as:

(sqrt2)2cos(2x - pi/8)cos(pi/8 - x) = sinx

I don't know what to do next though, nor am I sure my sqrt two is in the right place.

you have a square root which should not be there

cosx + cos3x in the LHS

and

sinx - sin3x in RHS

any better?

Ah, alright, I looked through it again and saw where the square root should have disappeared.

I ended up getting:
2cosx - 2sin^2xcosx = 2sinx - 2sin x cos^2 - 2sin^3x
through using the identity in the question and the compound formula to change everything into terms of the angle x.
I'll continue with it tomorrow, but does that look alright so far?
Apologies, I'm just finding this question tricky.

Hi,

I cheated and used the 'or otherwise' method, I will try and do the other method later as well if I have time.

http://postimg.org/image/czv09jooh/

Kind regards

JB

How did you go from 4cos^3x-2cosx-4sin^3x-2sinx=0

to cosx(2cosx^2-1) + sinx(1-2sin^2x) = 0 ???

Should be 2cosx(2cosx^2-1) + 2sinx(1-2sin^2x) no?

Hi Hassan,

You are correct in saying it should be multiplied by 2, but remember it is being set equal to 0 so you can divide both sides by 2. Basically 2cos2x=0 has the same roots as cos2x=0. http://postimg.org/image/rjpwtok9l/

I have uploaded some more working, and a graph to help you visualize why you can divide by 2.

http://postimg.org/image/4lthtbue1/

I hope that helps

Kind regards

JB