Why did you put
If you take
This becomes:
But you don't need to take the square root...
I need to know how to write some thing like (Cos(5x))^2 using a half angle formula, Such that I have something that looks like
(Cos(5x))^2=_______+_______cos(___x)
I know with a half angle formula that
Cos(x/2)=+/- sqrt((1+cos(x))/2)
but that ^2 is throwing me off, I don't know what to do with it per se.
Would something like = +/- sqrt((1+cos(5x)^2)/2)
Then using a power reducing formula work?
cos(x)^2=(1+cos2x)/(2)
\sqrt{\frac{1+\cos\!\left(\frac{2\cdot 5x}{2}\right)}{2}}" alt="\sqrt{\frac{1+\cos\!\left(\frac{2\cdot 5x}{2}\right)}{2}}" />
(i do not know why font is coming up)
I'm not sure what to do with 5x if I'm supposed to end up with 2x
would it be cos(2(5x)) or am I going down the wrong path?
Any suggestions?
Thank you for your time.
I suspect you're using the identity for because it looks that way.
(If you don't know this then learn it, and learn it quickly! You'll need it)
Rearrange that using algera to get as the subject:
In your question use the above and simply subsitute
Ok, I think I get it. If I take 5x out of the equation and plug it back in later.
The power reduction formula
cos^2 A = (1 + cos2A)/2
seems a bit of a better shortcut to cos2x=2cosx^2-1
*sigh*
Anyway. Thanks for all the help! I'm getting much better at throwing this stuff around!