pierdoodsu

2022-01-02

Show ${\left(\mathrm{sin}\left(\theta \right)-\mathrm{cos}\left(2\theta \right)\right)}^{2}=1+\mathrm{sin}\left(\theta \right)-\mathrm{sin}\left(3\theta \right)-\frac{12}{\mathrm{cos}\left(2\theta \right)}+\frac{12}{\mathrm{cos}\left(4\theta \right)}$

kaluitagf

Method I. Write everything in terms of , and then compare. Tedious, but doable.
Method II. If you know what a Fourier series is, you may determine the coefficients on the right-hand side by integration.
Method III. First expand the left by ${\left(a+b\right)}^{2}={a}^{2}+2ab+{b}^{2}$
${\left(\mathrm{sin}x-\mathrm{cos}\left(2x\right)\right)}^{2}={\mathrm{sin}}^{2}x+{\mathrm{cos}}^{2}\left(2x\right)-2\mathrm{sin}x\mathrm{cos}\left(2x\right)$
Then use the double angle formulas:

and the general product to sum formula:
$\mathrm{sin}x\mathrm{cos}\left(2x\right)=\frac{12}{-\mathrm{sin}\left(x\right)+\mathrm{sin}\left(3x\right)}$

Louis Page

As we know the RHS, let's try to express ${\mathrm{cos}}^{2}\left(2\theta \right)$ in terms of $\mathrm{cos}\left(4\theta \right)$
${\left(\mathrm{sin}\theta -\mathrm{cos}2\theta \right)}^{2}={\mathrm{sin}}^{2}\theta -2\mathrm{sin}\theta \mathrm{cos}2\theta +{\mathrm{cos}}^{2}2\theta$
$={\mathrm{sin}}^{2}\theta -2\mathrm{sin}\theta \mathrm{cos}2\theta +\frac{1+\mathrm{cos}4\theta }{2}$
Now
${\left(\mathrm{sin}\theta -\mathrm{cos}2\theta \right)}^{2}={\mathrm{sin}}^{2}\theta -2\mathrm{sin}\theta +4{\mathrm{sin}}^{3}\theta +\frac{12}{+}\frac{\mathrm{cos}4\theta }{2}$
$={\mathrm{sin}}^{2}\theta +\mathrm{sin}\theta -\left(3\mathrm{sin}\theta -4{\mathrm{sin}}^{3}\theta \right)+\frac{12}{+}\frac{\mathrm{cos}4\theta }{2}$
$=\mathrm{sin}\theta -\mathrm{sin}3\theta +\frac{\mathrm{cos}4\theta }{2}+1-\frac{12}{+}{\mathrm{sin}}^{2}\theta$
$=\mathrm{sin}\theta -\mathrm{sin}3\theta +\frac{\mathrm{cos}4\theta }{2}+1-\frac{1-2{\mathrm{sin}}^{2}\theta }{2}$
$=1+\mathrm{sin}\theta -\mathrm{sin}3\theta +\frac{\mathrm{cos}4\theta }{2}-\frac{\mathrm{cos}2\theta }{2}$

Vasquez

Apply the identities

$\mathrm{sin}3\theta =\mathrm{sin}\theta \left(2\mathrm{cos}2\theta +1\right)$
in simplifying the $RHS$
$RHS=1+\mathrm{sin}\theta -\mathrm{sin}\theta \left(2\mathrm{cos}2\theta +1\right)-\frac{1}{2}\left(1-2{\mathrm{sin}}^{2}\theta \right)+\frac{1}{2}\left(2{\mathrm{cos}}^{2}2\theta -1\right)$
$={\mathrm{sin}}^{2}\theta -2\mathrm{sin}\theta \mathrm{cos}2\theta +{\mathrm{cos}}^{2}2\theta$
$=\left(\mathrm{sin}\theta -\mathrm{cos}2\theta {\right)}^{2}=LHS$

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