How to derive cos(2θ) = 1 - 2sin²(θ)

[Moonspex]

Hello everyone! This is my first post here so pardon me if it's a little too simple... I just can't figure out where this equation came from (or rather how it got to that point):

cos(2 theta) = 1 - 2sin^2(theta)

Does it have something to do with the identity cos^2 + sin^2 = 1, and if so, how does it apply? (I just started my first year at college and I find myself wondering how I got here!)

 

[gts87]

I would say first consider $$cos(2\theta) = cos(\theta + \theta)$$, then use the sum to product formula $$cos(\alpha + \beta) = (cos\alpha)(cos\beta) - (sin\alpha)(sin\beta)$$. See if you can go from there.

 

[NoMoreExams]

Remember that

$$e^{i \theta} = cos(\theta) + i \cdot sin(\theta)$$

So let's square both sides to get

$$e^{2i \theta} = cos^{2}(\theta) + 2i \cdot cos(\theta)sin(\theta) - sin^{2}(\theta)$$

But note also that

$$e^{2i \theta} = cos(2 \theta) + i \cdot sin(\theta)$$

So matching up the real parts we get:

$$cos(2 \theta) = cos^{2}(\theta) - sin^{2}(\theta)$$

and matching up the imaginary parts we get:

$$sin(2 \theta) = 2 sin(\theta)cos(\theta)$$

 

[Moonspex]

Ok, so the question is actually

$$\int sin^{2}(x)dx$$

Thanks to your help I got to

$$\int sin^{2}(x)dx = \int 1/2 - cos(2\theta)/2 dx$$

Now, the answer says this integration equals

$$1/2x - sin(2\theta)/4 + C$$

My question now is how does the cos bit integrate to the sin bit?

 

[HallsofIvy]

?? you should have learned that d(sin x)/dx= cos(x) and so $\int cos(x) dx= sin(x)+ C$ before you start trying to integrate sin(2x)!

 

[Moonspex]

So the dividend is just integrated normally; how does the divisor go from 2 to 4?

 

[Thaakisfox]

because $$\frac{d}{dx}\sin(2x)=2*\cos(2x)$$ and so $$\int \cos (2x) dx = \frac{\sin(2x)}{2}+C$$

So you have an extra 2 in the denominator, hence it becomes 2x2 which is 4...

 

[Moonspex]

Just figured it out a minute ago! It's frustrating when I know it's something so simple... thanks everyone!

 

[MrSparky]

lol guys if his asking a basic question like this i don't think it requires integration
cos 2x = cos (x+x)
cos (x+x) = cos x . cos x - sin x . sin x
= cos^2 x - sin^2 x
using the identity cos^2 x + sin^2 x = 1
sin^2 x= 1-cos^2 x
cos^2 x +cos^2 x - 1 = cos 2x
2 cos^2 x -1=cos 2x
There no integration or w/e these guys used xD

 

[Gib Z]

Please, a) Don't revive old threads as you did with your other post, and b), read the entire thread before commenting.

 

[optics.tech]

$$cos(A+B)=cosA \ cosB - sinA \ sinB$$

so

$$cos2\theta=cos(\theta+\theta)$$

$$=cos\theta cos\theta-sin\theta sin\theta$$

$$=cos^2\theta-sin^2\theta$$

remember that:

$$sin^2\theta+cos^2\theta=1$$


for $$sin^2\theta=1-cos^2\theta$$:


$$cos2\theta=cos^2\theta-sin^2\theta$$

$$cos2\theta=cos^2\theta-(1-cos^2\theta)$$

$$cos2\theta=cos^2\theta-1+cos^2\theta$$

$$cos2\theta=2cos^2\theta-1$$

$$cos2\theta+1=2cos^2\theta$$

$$cos^2\theta=\frac{cos2\theta+1}{2}$$


for $$cos^2\theta=1-sin^2\theta$$:


$$cos2\theta=cos^2\theta-sin^2\theta$$

$$cos2\theta=(1-sin^2\theta)-sin^2\theta$$

$$cos2\theta=1-2sin^2\theta$$

$$2sin^2\theta=1-cos2\theta$$

$$sin^2\theta=\frac{1-cos2\theta}{2}$$

So

$$\int \ sin^2x \ dx$$

$$=\int \ \frac{1-cos2x}{2} \ dx$$

$$=\frac{1}{2}\int \ 1-cos2x \ dx$$

let $$u=2x$$, then $$\frac{du}{dx}=2$$, $$du=2 \ dx$$

So

$$\frac{1}{2} \int \ 1-cos2x \ dx$$

$$=\frac{1}{2} \ \frac{1}{2} \int \ (1-cos2x)(2 \ dx)$$

$$=\frac{1}{2} \ \frac{1}{2} \int \ (1-cos \ u) \ du$$

$$=\frac{1}{4} \ \int \ (1-cos \ u) \ du$$

$$=\frac{1}{4} (u-sin \ u) + C$$

$$=\frac{1}{4} (2x-sin \ 2x) + C$$

 

[Gib Z]

That's correct, I don't see the problem.

 

[NoMoreExams]

If you are unsure, differentiate your answer...