4u trig proof (1 Viewer)

pikachu975 · Feb 12, 2017

http://prntscr.com/e7naqo

Thanks for any help

InteGrand · Feb 12, 2017

pikachu975 said:
http://prntscr.com/e7naqo

Thanks for any help

$\noindent One method is as follows. Note that $\cos^{2}x = \frac{1}{2}\left(1+c\right)$ and $\sin^{2}x = \frac{1}{2}\left(1-c\right)$, where $c \equiv \cos 2x$. Thus$

$\begin{align*}\cos^{6} x + \sin^{6}x &= \left(\cos^{2} x\right)^{3} + \left(\sin^{2} x\right)^{3}\\ &= \frac{1}{8}\left((1+c)^{3} + (1-c)^{3}\right) \\ &= \frac{1}{8}\left(1 + 3c^{2} + 1 + 3c^{2}\right) \\ &= \frac{1}{4}+\frac{3}{4}c^{2},\end{align*}$

$\noindent as required.$

pikachu975 · Feb 12, 2017

InteGrand said:
$\noindent One method is as follows. Note that $\cos^{2}x = \frac{1}{2}\left(1+c\right)$ and $\sin^{2}x = \frac{1}{2}\left(1-c\right)$, where $c \equiv \cos 2x$. Thus$

$\begin{align*}\cos^{6} x + \sin^{6}x &= \left(\cos^{2} x\right)^{3} + \left(\sin^{2} x\right)^{3}\\ &= \frac{1}{8}\left((1+c)^{3} + (1-c)^{3}\right) \\ &= \frac{1}{8}\left(1 + 3c^{2} + 1 + 3c^{2}\right) \\ &= \frac{1}{4}+\frac{3}{4}c^{2},\end{align*}$

$\noindent as required.$

Thanks for this, is there a way to go from RHS to LHS too?

InteGrand · Feb 12, 2017

pikachu975 said:
Thanks for this, is there a way to go from RHS to LHS too?

Yeah, e.g. reverse the above steps.

Mahan1 · Feb 13, 2017

pikachu975 said:
http://prntscr.com/e7naqo

Thanks for any help

I thought to share another method to tackle the question:
$\frac{1}{4}+ \frac{3}{4}cos^{2}2x = \frac{1}{4}+ \frac{3}{4}(1-sin^{2}2x)$
$= 1 -3sin^{2}x \ cos^{2}x$
$= 1-3sin^{2}x \ cos^{2}x \ (sin^{2}x \ + cos^{2}x) = 1-3sin^{4}x\ cos^{2}x -3sin^{2}x \ cos^{4}x$
$= sin^{6}x \ +cos^{6}x$
The last line is from the fact that $1 = (sin^{2}x \ + cos^{2}x)^{3}$

juantheron · Feb 14, 2017

$Using $\sin^2 x+\cos^2 x = 1\Rightarrow \sin^2 x+\cos^2 x+(-1) = 0$

$Now if $a+b+c=0\;,$ Then $a^3+b^3+c^3=3abc$

$So $(\sin^2 x)^3+(\cos^2 x)^3+(-1)^3=-3\sin^2 x\cdot \cos^2 x$

$So $\sin^6 x+\cos^6 x= \frac{1}{4}\left[4-3(\sin 2x)^2\right] = \frac{1}{4}\left[1+3\cos^2 (2x)\right].$

pikachu975 · Feb 14, 2017

juantheron said:
$Using $\sin^2 x+\cos^2 x = 1\Rightarrow \sin^2 x+\cos^2 x+(-1) = 0$

$Now if $a+b+c=0\;,$ Then $a^3+b^3+c^3=3abc$

$So $(\sin^2 x)^3+(\cos^2 x)^3+(-1)^3=-3\sin^2 x\cdot \cos^2 x$

$So $\sin^6 x+\cos^6 x= \frac{1}{4}\left[4-3(\sin 2x)^2\right] = \frac{1}{4}\left[1+3\cos^2 (2x)\right].$

Where is the 3abc from? Thanks btw

InteGrand · Feb 14, 2017

pikachu975 said:
Where is the 3abc from? Thanks btw

$\noindent That fact follows from the factorisation$

$a^{3} + b^{3} + c^{3} -3abc = (a+b+c)\left(a^{2} + b^{2} + c^{2} - ab -bc -ca\right).$

4u trig proof (1 Viewer)

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