Using Power reduction to solve cos^4t dt?
Invalid question.
?
invalid?
@Isaiah.Feynman its cos^4 (t)
So what do you want to solve there?
well we have to use power reduction. wouldnt you reduce it to cos^2(t) cos^2(t)?
Yeah.
then(1+cos2x/2)(1+cos2x/2) ?
Where did you get that from?
cos^2x dx= integral of 1+cos2x/2
No. What you mean is the trig identity \[\cos^{2}x = 1-\sin^{2}x\]
its a power reduction formula?
No.
You could use the ummmm Half-Angle Formula's a bunch of times. Or you could use the Cosine Redux Formula:\[\Large I_n\quad=\quad \frac{1}{n}\cos^{n-1}x \sin x +\frac{n-1}{n}I_{n-2}\]
Where I_n is,\[\Large I_n\quad=\quad \int\limits \cos^n x\;dx\]
Although, if you haven't learned about that yet, then maybe half-angles are easier :)
\[\Large \cos^4x \quad=\quad [\cos^2x]^2 \quad=\quad \left[\frac{1}{2}(1+\cos2x)\right]^2\]
Gotta expand that out and do some more half angle after that ^^
1/4(x+1/2sin2x)?
Hmm I'm confused.. did you ignore the square on the outside? D:
wouldnt it just be another 1/2?
No :(
hahaha. sorry. I dont really understand half angles
This is what expanding the brackets gives us,\[\Large \frac{1}{4}(1+2\cos2x+\cos^22x)\]That part make sense? :o
I multiplied out the outer square
ohhh i know this. duh
What is that reindeer doing on that lady's face? +_+ That's kinda rude.
You'll want to apply the Half-Angle Formula again to this orange term before you can integrate. \[\Large \frac{1}{4}(1+2\cos2x+\color{orangered}{\cos^22x})\]
Doing reindeer things :D
pshhh
hmm so youd plug in the 1+cos2x/2 into it?
No the angle `doubles` each time we apply the rule. So we multiply another 2 inside of the cosine.
\[\Large \cos^22x\quad=\quad \frac{1}{2}(1+\cos(2\cdot2x)) \quad=\quad\frac{1}{2}(1+\cos4x)\] Understand how that works? :) Dubs. The. Angle.
yup! thanks<3
@skay your smartscore graph is a straight line. Can you get it's equation? :P
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