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who can simplify the following term in a simplest way?

$$ 2\cos^2(x)\sin^2(x) + \cos^4(x) + \sin^4(x)$$

(The answer is 1). Thanks for any suggestions.

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    $\begingroup$ The general approach would be to search for a suitable transform for this function. In our case the best one probably is the Fourier transform. The trigonometric "laws" are then automatically applied when calculating the function in the Fourier domain. Now you did not ask for this, so I write it in a comment instead of an answer. $\endgroup$ Commented Jul 21, 2015 at 21:24

4 Answers 4

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$$ 2\cos^2(x)\sin^2(x) + \cos^4(x) + \sin^4(x)$$

$$(a+b)^2=2ab+a^2+b^2$$

$$=(\cos^2x+\sin^2x)^2$$

$$=1^2$$

$$=\boxed{\color{red}1}$$

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HINT :

$$2ab+a^2+b^2=(a+b)^2$$ with $$\cos^2 (x)+\sin^2(x)=1.$$

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    $\begingroup$ Ok, sorry. How could I have overlooked this ... :-) $\endgroup$
    – Breaking M
    Commented Jul 21, 2015 at 21:12
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Although the answer is straight forward but here are some steps if you find helpful
Notice, $$2\cos^2 x\sin^2 x+\cos^4 x+\sin^4 x$$ $$=(\cos^2 x)^2+(\sin^2 x)^2+2(\cos^2 x)(\sin^2 x)$$ Now, assume $\alpha =\cos^2 x$ & $\beta=\sin^2 x$ & apply $\alpha^2+\beta^2+2\alpha \beta=(\alpha+\beta)^2$$$=(\cos^2x+\sin^2 x)^2$$$$=(1)^2=1 \quad (\text{since,}\ \cos^2 x+\sin^2 x=1)$$

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$$2\cos^2x\sin^2x + \cos^4x + \sin^4x = (\cos^2x + \sin^2x)^2$$

I'm sure you can work out what $n$ should be. Mathlove already gave you a big clue there!

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    $\begingroup$ you have put $n$ - but it should be $2$... $\endgroup$ Commented Jul 21, 2015 at 21:14
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    $\begingroup$ @johannesvalks, I've left that deliberately for the OP to work out You've spoilt it now! The downvote was so unnecessary! $\endgroup$
    – John_dydx
    Commented Jul 21, 2015 at 21:15
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    $\begingroup$ I did not downvoted your post. It appears there are others doing that... I did actually an upvote... $\endgroup$ Commented Jul 21, 2015 at 21:17
  • $\begingroup$ @johannesvalks, Thanks! $\endgroup$
    – John_dydx
    Commented Jul 21, 2015 at 21:51
  • $\begingroup$ You are welcome! $\endgroup$ Commented Jul 21, 2015 at 21:53

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