Trigonometric Formula

Crelicx

I have a question about one of the Trig formulas.

I know $$\displaystyle sin^2(x)=\frac{1}{2}(1-\cos{(2x)})$$

I’m wondering if this formula works for any real number for the coefficient in the argument for sin(x)?

If so would it look like this? $$\displaystyle sin^2{(\alpha{x})}=\frac{1}{2}(1-=cos^2{({2}\alpha{x})})$$

Or like this? $$\displaystyle sin^2{(\alpha{x})}=\frac{1}{2\alpha}(1-=cos^2{({2}\alpha{x})})$$

Or...

Also if the coefficient can be any real number I'm assuming it works for $$\displaystyle cos^2(x)=\frac{1}{2}(1+\cos{(2x)})$$ as well?

Crelicx

Sorry if there's any confusion.

When I reviewed my post I did not notice that I put an equal sign between the negative and the $$\displaystyle cos^2{({2}\alpha{x})})$$

I meant to ask if the formula would look like this $$\displaystyle sin^2{(\alpha{x})}=\frac{1}{2}(1-cos^2{({2}\alpha{x})})$$

Or like this $$\displaystyle sin^2{(\alpha{x})}=\frac{1}{2\alpha}(1-cos^2{({2}\alpha{x})})$$

I also didn't see an option to edit my post, and I'm sorry if this counts as an unnecessary bump.

Debsta

MHF Helper
I know $$\displaystyle sin^2(x)=\frac{1}{2}(1-\cos{(2x)})$$
Correct.

If so would it look like this? $$\displaystyle sin^2{(\alpha{x})}=\frac{1}{2}(1-cos^2{({2}\alpha{x})})$$
The cos is not squared.
It should be:

$$\displaystyle sin^2{(\alpha{x})}=\frac{1}{2}(1-cos{({2}\alpha{x})})$$

Or like this $$\displaystyle sin^2{(\alpha{x})}=\frac{1}{2\alpha}(1-cos^2{({2}\alpha{x})})$$
No!

Last edited:

skeeter

MHF Helper
The power reduction identity is $\sin^2{u}=\dfrac{1-\cos(2u)}{2}$

let $u=\alpha x$ ...

$\sin^2(\alpha x)=\dfrac{1-\cos(2\alpha x)}{2}$