Question

Trying to understand small calculation in problem...
Must express half angle trig function Sin^4(theta/2) in a form without powers. At beginning of process problem becomes sin^4(theta/2) = (sin^2(theta/2))^2 then (1-cos(theta)/2)^2 .... wondering why doesn't it become (1-cos^2(theta)/2)^2 ...? wouldn't it be squared? instead of just cos (theta)?

Answer 1

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

So...

\(\left(\sin ^2\left(\dfrac{\theta}{2}\right)\right)^2\implies\left(\dfrac{1-\cos ^2\left(\theta\right)}{2}\right)^2\) ?

Let me see...

Answer 2

Yah, it is what I thought. Half angle formula. The moving the half seems confusing but is because of that.

http://www.sosmath.com/trig/douangl/douangl.html

\(\sin^2\theta = \dfrac{1+\cos(2\theta)}{2}\)

Well, lets put the half inside there:

\(\sin^2\left(\dfrac{\theta}{2}\right) = \dfrac{1+\cos\left(\dfrac{2\theta}{2}\right)}{2}\)

Then the 2 cancels in the resultant fraction.

Answer 3

aaah... ok. thank you so much

Answer 4

Yah, you were thinking about the Pythagorean identity, but the /2 is moving from inside to outside the trug function, which is what caught you off guard. It is not really moving. just seems that way. =)

Answer 5

:) your right. Thank you Sorry about the late response for some reason my question kept timing out, and i couldn't bump it so I logged out and logged back in. Thank you for your help, your guys site is so great!...

Answer 6

No big deal. Glad I could point out the proper identity. I can understand why this one would be confusing!

If you plan on taking Calculus, get really good with trig identities. Work extra problems if you can. Really helps in the second half of single var. calc and everything after.