Question 618404
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I presume you mean


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \sin^4(x)]


Which is to say


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \left(\sin(x)\right)^4]


As opposed to


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \sin(x^4)]


which is what you actually wrote.


Given that, write:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \sin^4(x)\ =\ \left(\sin^2(x)\right)\left(\sin^2(x)\right)]


Then use the Pythagorean identity:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \left(\sin^2(x)\right)\left(\sin^2(x)\right)\ =\ \left(1\ -\ \cos^2(x)\right)\left(1\ -\ \cos^2(x)\right)]


Then use the difference of two squares factorization:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \left(1\ -\ \cos^2(x)\right)\left(1\ -\ \cos^2(x)\right)\ =\ \left(1\ -\ \cos(x)\right)\left(1\ +\ \cos(x)\right)\left(1\ -\ \cos(x)\right)\left(1\ +\ \cos(x)\right)]


Which is the desired result.


John
*[tex \LARGE e^{i\pi}\ +\ 1\ =\ 0]
My calculator said it, I believe it, that settles it
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