Simplify-write as single function?

Pingu

Junior Member
Joined
Dec 22, 2005
Messages
56
The question tells me to simplify the expression Cos2X*CosX+Sin2X*SinX
It says I have to write it as a single function.
What is a single function and how do I simplify Cos2X*CosX+Sin2X*SinX into a single function?
 
G'day, Pingu.

f(x)=cos(2x) cosx+sin(2x)sinx\displaystyle f(x) = \cos{(2x)} \cdot\ cos{x} \, + \, \sin{(2x)} \cdot \sin{x} passes as a function, so it may better to ask to write the expression as a single trigonometric function.

By that I mean to simplify it down to just cos(x)\displaystyle cos(x) or just sin2(2x)\displaystyle \sin^2{(2x)}, for examples, as opposed to having both sin and cos in there.

A lot of this work is a bit of trial and error, playing with identities and hoping you get a nice result out. It can be quite difficult to see the end of the path from the beginning; you can just make your way until you can see it. With practise, your foresight improves.

One option to start with would be to put cos(2x)\displaystyle \cos{(2x)} and sin(2x)\displaystyle \sin{(2x)} in terms of cosx\displaystyle \cos{x} and sinx\displaystyle \sin{x}, respectively:

You should be familiar with the following identities
. . \(\displaystyle \L \cos{2x} = 2\cos^2{x} - 1\)

. . \(\displaystyle \L \sin{2x} = 2\sin{x}\cos{x}\)

So
. . \(\displaystyle \L \cos{2x} \cdot \cos{x} + \sin{2x} \cdot \sin{x}\)
becomes
. . \(\displaystyle \L (2\cos^2{x} - 1)\cos{x} \, + \, (2\sin{x}\cos{x}) \cdot \sin{x}\)
Simplify:
. . \(\displaystyle \L 2\cos^3{x} \, - \, \cos{x} \, + \, 2\sin^2{x}\cos{x}\)

See if you click onto the next step. (Give it a few minutes.)
 
There are several identities you can use.

cos2xcosx+sin2xsinx\displaystyle cos2xcosx+sin2xsinx

cos2x=cos2xsin2x\displaystyle cos2x=cos^{2}x-sin^{2}x

(cos2xsin2x)cosx+2sinxcosxsinx\displaystyle (cos^{2}x-sin^{2}x)cosx+2sinxcosxsinx

cos3xsin2xcosx+2sin2xcosx\displaystyle cos^{3}x-sin^{2}xcosx+2sin^{2}xcosx

cosx(cos2xsin2x+2sin2x)\displaystyle cosx(cos^{2}x-sin^{2}x+2sin^{2}x)

cosx(cos2x+sin2x)\displaystyle cosx(cos^{2}x+sin^{2}x)

Now you should see something familiar in the parentheses.
 
Or an alternative
cos(a-b)=cos(a)cos(b)+sin(a)sin(b)
where a=2x and b=x
 
Thnak you for helping me

I was able to solve it by using the

cos(a-b)=cos(a)cos(b)+sin(a)sin(b)
where a=2x and b=x
method.
 
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