Trigonometric Formula

Jun 2016
4
0
Phoenix
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?

Thanks in advance.
 
Jun 2016
4
0
Phoenix
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
Oct 2009
1,346
623
Brisbane
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
Jun 2008
16,217
6,765
North Texas
The power reduction identity is $\sin^2{u}=\dfrac{1-\cos(2u)}{2}$

let $u=\alpha x$ ...

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