# [SOLVED] Identity and general sloution problem

#### dojo

I'm stuck on this revision.

Using identities for cos(C+D) and cos(C-D) prove that
cos A + cos B = 2cos(1/2(A+B))cos(1/2(A-B))

Hence, find in terms of pi the eneral sloution of the equation:
cos5x + cosx = cos3x

I know the general solution for cos will be -+x + 2nPi

#### Sudharaka

I'm stuck on this revision.

Using identities for cos(C+D) and cos(C-D) prove that
cos A + cos B = 2cos(1/2(A+B))cos(1/2(A-B))

Hence, find in terms of pi the eneral sloution of the equation:
cos5x + cosx = cos3x

I know the general solution for cos will be -+x + 2nPi
Dear dojo,

First write down the identities for cos(C+D) and cos(C-D) and add these two equations.

i.e: $$\displaystyle cos(C+D)=cosC~cosD-sinC~sinD$$

$$\displaystyle cos(C-D)=cosC~cosD+sinC~sinD$$

$$\displaystyle cos(C+D)+cos(C-D)=2cosC~cosD$$

Now substitute $$\displaystyle C=\frac{A+B}{2}~and~D=\frac{A-B}{2}$$

#### dojo

I see. Out of interest why are they added?

With regards to the general angle not sure where to go on this one. Apply the formaul you derived and that give 2cos3xcos2x = cos3x but where now?

#### Sudharaka

I see. Out of interest why are they added?

With regards to the general angle not sure where to go on this one. Apply the formaul you derived and that give 2cos3xcos2x = cos3x but where now?
Dear dojo,

Note that in the equation, $$\displaystyle cos A + cos B = 2cos\frac{A+B}{2}cos\frac{A-B}{2}$$ there are two cosine terms added together in the left hand side and two cosine terms multiplied in the RHS side. By adding the two equations;cos(C+D) and cos(C-D) we can get the equation, $$\displaystyle cos(C+D)+cos(C-D)=2cosC~cosD$$ which is of the same form.

For the second part,

$$\displaystyle 2cos3xcos2x = cos3x$$

$$\displaystyle cos3x(2cos2x-1)=0$$

Hope you can continue from here.

#### dojo

Yes i think this helps

So either:
(1) cos 3x = 0
(2) 2cos 2x = 0

So general solutionswill be:
(1):
x = +/- pi/8 + nPi

and (2)
x = +/- pi/12 + 2/3*nPi which simplifies 2/3(+/- pi/8 + nPi)

#### Sudharaka

Yes i think this helps

So either:
(1) cos 3x = 0
(2) 2cos 2x = 0

So general solutionswill be:
(1):
x = +/- pi/8 + nPi

and (2)
x = +/- pi/12 + 2/3*nPi which simplifies 2/3(+/- pi/8 + nPi)

Dear dojo,

$$\displaystyle cos3x(2cos2x-1)=0\Rightarrow{cos3x=0~or~cos2x=\frac{1}{2}}$$

Hence your second answer is incorrect. Also note that if $$\displaystyle cos\theta=\alpha\Rightarrow{\theta=2n\pi\pm{cos^{-1}\alpha}~\forall{n\in{Z}}}$$ Therefore, for the first one,

$$\displaystyle cos3x=0$$

$$\displaystyle 3x=2n\pi\pm{cos^{-1}0}~\forall{n\in{Z}}$$

$$\displaystyle x=\frac{2n}{3}\pi\pm{\frac{\pi}{6}}~\forall{n\in{Z}}$$

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#### dojo

Schoolboy error (Wink) - it should be +/- pi/6 + nPi

#### Sudharaka

Schoolboy error (Wink) - it should be +/- pi/6 + nPi
Dear dojo,

When, $$\displaystyle cos3x=0\Rightarrow{x=\frac{2n\pi}{3}\pm{\frac{\pi}{6}}}$$

When, $$\displaystyle cos2x=\frac{1}{2}\Rightarrow{x=n\pi\pm{\frac{\pi}{6}}}$$