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I've been working on this problem on and off for a couple hours and have not been able to find out how to go about it even after looking through the other questions and some google searching.

Prove that the given equations are identities $$ \sin^4 \theta = \frac{3-4\cos2\theta+\cos4\theta}{8} $$

So I started off trying this

$$\sin^4 = (\sin^2\theta)^2=(1-\cos^2\theta)^2$$

$$(1-\cos^2\theta)^2 = \left[1-\left(\frac 12\cos2\theta + 1\right)\right]^2$$

But that hasn't led me to anywhere. Trying it from the other side hasn't yielded much results for me either.

$$\frac{(3-4\cos2\theta+\cos4\theta)}{8}= \frac18(3-4[2\cos^2\theta-1]+\cos4\theta)$$

$$\frac18(3-4[2\cos^2\theta-1]+\cos4\theta) = \frac18(3-4[2(1-\sin^2\theta)-1]+\cos4\theta)$$

From there I seem to have just hit a loop with $\sin^2$ and $\cos^2$ . I have A LOT of trouble with these types of problems in this section. I would greatly appreciate it if someone could help me out.

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4 Answers

up vote 3 down vote accepted

I appreciate that you've shown your work. I'm a bit low on time but let me give you a hint - if you are still stuck, comment and I'll help you out later.

Notice that in the intended answer, there are only cosines and there are no powers. So one strategy would be to do the following:

  1. First, express everything in terms of cosines. Your step where you wrote $\sin ^4 \theta = (1 - \cos^2\theta)^2$ looks great for this.
  2. Use a power reduction formula to write $\cos^2 \theta$ without powers. You wrote something that's very close to true. But instead it's $\cos^2 \theta = \dfrac{1 + \cos 2\theta}{2}$.
  3. Distribute the product.
  4. Oh no! We have another power! So do the power reduction again.
  5. Collect all the terms and check that your answer is right.
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Everyone has posted up great answers, but this one is what led me to finishing the problem. Thank you very much! –  Thumbtack Oct 23 '12 at 22:42
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You will want to use double-angle identities. For example, note that $$\cos 2\theta=2\cos^2\theta-1=1-2\sin^2\theta.$$ Maybe rewrite this as $$\sin^2\theta=\frac{1-\cos 2\theta}{2}.$$ Square both sides. We get $$\sin^4\theta=\frac{1}{4}(1-2\cos 2\theta+\cos^2 2\theta).$$ Getting warm! I will not finish things, but just remind you that one version of the double-angle identity above is $$\cos 4\theta=2\cos^2 2\theta -1.$$

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If you use $\cos(z) = \tfrac{1}{2}(e^{iz} + e^{-iz})$ and $\sin(z) = \tfrac{1}{2i}(e^{iz} - e^{-iz})$,

$$8 \sin^4 \theta = (3-4 \cos(2\theta)+\cos(4\theta))$$

becomes

$$8 (\tfrac{1}{2i}(e^{iz} - e^{-iz}))^4 = (3-4 \tfrac{1}{2}(e^{i2z} + e^{-i2z})+\tfrac{1}{2}(e^{i4z} + e^{-i4z}))$$

which you can prove by multiplying out the brackets

$$ 8 (-\tfrac{1}{4}e^{-2iz}-\tfrac{1}{4}e^{2iz}+\tfrac{1}{16}e^{-4iz}+\tfrac{1}{16}e^{4iz}+\tfrac{3}{8}) = (3-4 \tfrac{1}{2}(e^{i2z} + e^{-i2z})+\tfrac{1}{2}(e^{i4z} + e^{-i4z}))$$

and you can see this is equal by inspection so the trig identity is proven.

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You might have skipped a few steps by using the power reduction formula for $\sin^2\theta$.

$$\sin^4\theta=(\frac12-\frac12\cos2\theta)^2=\frac14(1-\cos2\theta)^2$$

You will notice this does not agree with what you have. You made a small mistake when using the power reduction formula

$$\cos^2x=\frac12+\frac12\cos2x$$

Anyway, from the form $\frac14(1-\cos2\theta)^2$, I would square it out, then use the power reduction formula above on the $\cos^22\theta$ term.

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