Question

How show this identity?

(sinx)(sin 2x)+(cosx)(Cos 2x) = cosx

Answer 1

think a difference/sum identity for cosine

Answer 2

Still not understanding your reply? Can you expand on that please.

Answer 3

do you know the difference/sum identity for cosine?

Answer 4

yes.

Answer 5

that is the identity I'm asking you to use

Answer 6

if you don't know how to use it
then please state the identity I'm asking you to use

Answer 7

Cos (x+y) = Cos(x) Cos(y) - Sin (x) Sin(y)
Cos(x-y) Cos (x) Cos(y)+ Sin(X) Sin(y)

Answer 8

do you notice your last one is the same as the above but instead of y you have 2x?

Answer 9

I see...

Answer 10

hint: as freckles indicated \(\bf cos({\color{brown}{ \alpha}} - {\color{blue}{ \beta}})= cos({\color{brown}{ \alpha}})cos({\color{blue}{ \beta}}) + sin({\color{brown}{ \alpha}})sin({\color{blue}{ \beta}})\)

Answer 11

Let me see if I can figure this one out. Will let you know if I run into a problem in a few minutes.

Answer 12

What do I do with 2x?

Answer 13

\[\cos(x)\cos(y)+\sin(x)\cos(y) \\ \text{ is the same as what you wrote } \\ \cos(x)\cos(2x)+\sin(x)\sin(2x) \\ \text{ except } y \text{ in place of } 2x \\ \text{ and you just said that you know the following equality } \\ \cos(x) \cos(y)+\sin(x) \cos(y)=\cos(x-y) \\ \text{ why not just replace } y \text{ with } 2x ?\]

Answer 14

so you should have cos(x-?)

Answer 15

what is the question mark

Answer 16

2x

Answer 17

yes

Answer 18

can you simplify cos(x-2x)?

Answer 19

Yes it would be: Cosx

Answer 20

right cos(x-2x)=cos(-x)
but cos is an even function so
cos(-x)=cos(x)

Answer 21

Wow! You are awesome! Can I use this trick for problems like this without going the long way. Just substitute (x+y) or (x-Y)?

Answer 22

maybe I'm not sure
this is the only way I would have done the problem :p

I guess you were thinking about using double angle identities?

Answer 23

Yes I was trying to use double angle identity. How would I use double angle identity to solve this problem because I think this is what my professor is looking for?

Answer 24

\[\sin (x) [2 \sin(x) \cos(x)]+\cos(x)[ \cos^2(x)-\sin^2(x)]= \cos(x) \\ 2 \sin^2(x) \cos(x)+\cos^3(x)-\cos(x) \sin^2(x) \\ \sin^2(x) \cos(x)+\cos^3(x) \text{ combined like terms } \\ (1-\cos^2(x))\cos(x)+\cos^3(x) \text{ by pythagorean identity }\]
one step left
see if you can see it

Answer 25

(sin x)squared (cos x) +(Cos x)squared
1 (cos x) = Cos X

Answer 26

\((\sin x)(\color{red}{\sin 2x})+(\cos x)(\color{green}{\cos 2x}) = \cos x\)

\((\sin x)(\color{red}{2\sin x \cos x})+(\cos x)(\color{green}{\cos^2 x - \sin^2 x}) = \cos x\)

\(2 \sin^2 x \cos x + \cos ^3 x - \sin^2 x \cos x = \cos x\)

\(\sin^2 x \cos x + \cos ^3 x = \cos x\)

\(\cos x(\color{blue}{\sin^2 x + \cos^2 x}) = \cos x\)

\(\cos x \times \color{blue}{1}= \cos x\)

\(\cos x = \cos x\)

Answer 27

Ok, I see math student 55 and Freckles. Thank you sooo much!

Answer 28

You're welcome.