I have $\sin 2x=\frac 23$ , and I'm supposed to express $\sin^6 x+\cos^6 x$ as $\frac ab$ where $a, b$ are co-prime positive integers. This is what I did:
First, notice that $(\sin x +\cos x)^2=\sin^2 x+\cos^2 x+\sin 2x=1+ \frac 23=\frac53$ .
Now, from what was given we have $\sin x=\frac{1}{...
In this question, the only proof of the trigonometric identity:
$$\cos^6{\theta}+\sin^6{\theta}=\frac{1}{8}(5+3\cos{4\theta})$$
is via factoring the sum of cubes:
$$\cos^6{\theta}+\sin^6{\theta}=(\cos^2{\theta}+\sin^2{\theta})(\cos^4{\theta}-\cos^2{\theta}\sin^2{\theta}+\sin^4{\theta})$$
Is th...
@WeiZhong Do you mean why I posted the above question and search queries here? Well I was searching for other questions about $\cos^6x+\sin^6x=1−3\sin^2x\cos^2x$ (if they exists).
After all, purpose of this room is to help people who are looking for some question(s).
And, as and aside, since Approach0 is new, this room is also used for testing it.
Searching for questions is not easy. So this room was created for such situations - you can ask here for advice, maybe some other user will have better luck when searching the question.
No. This time are genuinely asking for help. Of course, it is possible that there is no duplicate and this trigonometric equality was posted on this site for the first time in the above question.
And a relevant answer can answer that post: http://math.stackexchange.com/questions/445549/find-int-frac-tan-2x-sqrt-cos6x-sin6x-dx
This is the top 10th search result from my query.
But I guess that is not considered a duplicate. Although this post contains an answer to that question, but the question in this post is not explicitly asking how to simplify \cos^6x+\sin^6x.