Write (Cos(5x))^2 using half angle formula

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November 1st 2010, 09:35 AM
CyanBC

Write (Cos(5x))^2 using half angle formula

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)

$<font color=$ \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.
November 1st 2010, 09:54 AM
Unknown008

$\sqrt{\dfrac{1+\cos\left(\frac{2\cdot 5x}{2}\right)}{2}}$

Why did you put $\cos(\frac{2\cdot 5x}{2})$

If you take

$\cos(x)= \sqrt{\dfrac{1+\cos(2x)}{2}}$

This becomes:

$\cos(5x)= \sqrt{\dfrac{1+\cos(10x)}{2}}$

But you don't need to take the square root...
November 1st 2010, 12:31 PM
CyanBC

I ended up with
(Cos(5x))^2 = 1/2+1/2(cos(10x))

but I must admit that I'm not sure how I came up with this answer I just made a few assumptions and thus guess works. Can someone help me step this out using trig identities?
November 1st 2010, 12:50 PM
e^(i*pi)

I suspect you're using the identity for $\cos(2 \theta)$ because it looks that way.

$\cos(2 \theta) = 2\cos^2 \theta - 1$ (If you don't know this then learn it, and learn it quickly! You'll need it)

Rearrange that using algera to get $\cos^2 \theta$ as the subject:

$\dfrac{1}{2}\,\left(\cos(2 \theta) +1\right) = \dfrac{1}{2} + \dfrac{1}{2}\cos(2 \theta)$

In your question use the above and simply subsitute $\theta = 5x$
November 1st 2010, 11:25 PM
Unknown008

Yep, that's it.

You can remember all the three forms of cos2theta in one go.

$\begin{array}{ccl}<br /> \cos2\theta &=& \cos^2\theta - \sin^2\theta \\<br /> &=& 2 \cos^2\theta - 1\\<br /> &=& 1 - 2\sin^2\theta \end{array}$

If you can remember at least the first one, you can use Pythagoras' Theorem to substitute the sin^2 and cos^2 ratios.
November 2nd 2010, 03:01 PM
CyanBC

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!