Question

cosx=sin2x

Answer 1

What's the question Ma'am?

Answer 2

prove the identity

Answer 3

solve for x?

Answer 4

this isn't identity

Answer 5

try x=0
you see both sides are different

Answer 6

\[\sin(2x) = 2 \sin(x) \cos(x)\]

Answer 7

How can be this equal to only \(\cos(x)\) ??

Answer 8

I suggest, check your question again..

Answer 9

i'm pretty sure this was the question, but i don't have it with me

Answer 10

are you sure it wasn't to solve the equation?

Answer 11

yes it probably was

Answer 12

The identity is:
\(\normalsize\color{red}{ \displaystyle \sin(2x)=\color{black}{\sin(x+x)=\cos(x)\sin(x)+\cos(x)\sin(x)=} 2\sin(x)\cos(x)}\)
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
\(\normalsize\color{blue}{ \displaystyle \cos x(x)=\sin(2x)}\)
if you want to solve this one,

\(\normalsize\color{blue}{ \displaystyle \cos (x)=2\sin(x)\cos(x)}\)
\(\normalsize\color{blue}{ \displaystyle 0=2\sin(x)\cos(x)-\cos (x)}\)
\(\normalsize\color{blue}{ \displaystyle 0=\left(2\sin(x)-1\right)\cos (x)}\)
and on...

Answer 13

I have a typo in my first blue line it should say just cos(x)=sin(2x)

Answer 14

why is the 2 in front of the x and not in front of sin? all the ones i have trouble with are like that

Answer 15

also, cosx-cos2x=0 ?

Answer 16

new problem

Answer 17

The identity is:
\(\large\color{blue}{ \displaystyle \normalsize\color{red}{ \displaystyle \cos(2x)=\color{black}{\cos(x+x)=\cos(x)\cos(x)-\sin(x)\sin(x)=} \cos^2(x)-\sin^2(x)} }\)

Answer 18

and then, another identity is \(\large\color{blue}{ \displaystyle \normalsize\color{red}{ \displaystyle \sin^2(x)=1-\cos^2(x)} }\) .

Answer 19

using these 2 rules, you can re-write everything in terms of cos(x) and it will be sort of a quadratic equation

Answer 20

idk how to go on

Answer 21

well, please apply the first identity for cos(2x) to your prblem.

Answer 22

i've gotten to -2cosx+cosx-1=0