# Show (\sin(\theta)−\cos(2\theta))^2=1+\sin(\theta)−\sin(3\theta)−\frac12\cos(2\theta)+\frac12\cos(4\theta)

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

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kaluitagf
Method I. Write everything in terms of $$\displaystyle{\sin{{\left(\theta\right)}}}\ \text{ and }\ {\cos{{\left(\theta\right)}}}$$, 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 $$\displaystyle{\left({a}+{b}\right)}^{{2}}={a}^{{2}}+{2}{a}{b}+{b}^{{2}}$$
$$\displaystyle{\left({\sin{{x}}}−{\cos{{\left({2}{x}\right)}}}\right)}^{{2}}={{\sin}^{{2}}{x}}+{{\cos}^{{2}}{\left({2}{x}\right)}}−{2}{\sin{{x}}}{\cos{{\left({2}{x}\right)}}}$$
Then use the double angle formulas:
$$\displaystyle{{\sin}^{{2}}{x}}={\frac{{{1}−{{\cos}^{{2}}{x}}}}{{{2}}}},\ \ \ {{\cos}^{{2}}{\left({2}{x}\right)}}={\frac{{{1}+{\cos{{\left({4}{x}\right)}}}}}{{{2}}}}$$
and the general product to sum formula:
$$\displaystyle{\sin{{x}}}{\cos{{\left({2}{x}\right)}}}={\frac{{12}}{{−{\sin{{\left({x}\right)}}}+{\sin{{\left({3}{x}\right)}}}}}}$$
###### Not exactly what you’re looking for?
Louis Page

As we know the RHS, let's try to express $$\displaystyle{{\cos}^{{2}}{\left({2}\theta\right)}}$$ in terms of $$\displaystyle{\cos{{\left({4}\theta\right)}}}$$
$$\displaystyle{\left({\sin{\theta}}−{\cos{{2}}}\theta\right)}^{{2}}={{\sin}^{{2}}\theta}−{2}{\sin{\theta}}{\cos{{2}}}\theta+{{\cos}^{{2}}{2}}\theta$$
$$\displaystyle={{\sin}^{{2}}\theta}−{2}{\sin{\theta}}{\cos{{2}}}\theta+{\frac{{{1}+{\cos{{4}}}\theta}}{{{2}}}}$$
Now $$\displaystyle{\cos{{2}}}\theta={1}−{2}{{\sin}^{{2}}\theta}{w}{e}\ {u}{s}{e}\ {t}{h}{i}{s\ }toge{t}her\ {4}{{\sin}^{{3}}\theta}$$
$$\displaystyle{\left({\sin{\theta}}−{\cos{{2}}}\theta\right)}^{{2}}={{\sin}^{{2}}\theta}−{2}{\sin{\theta}}+{4}{{\sin}^{{3}}\theta}+{\frac{{12}}{+}}{\frac{{{\cos{{4}}}\theta}}{{{2}}}}$$
$$\displaystyle={{\sin}^{{2}}\theta}+{\sin{\theta}}-{\left({3}{\sin{\theta}}-{4}{{\sin}^{{3}}\theta}\right)}+{\frac{{12}}{+}}{\frac{{{\cos{{4}}}\theta}}{{{2}}}}$$
$$\displaystyle={\sin{\theta}}−{\sin{{3}}}\theta+{\frac{{{\cos{{4}}}\theta}}{{{2}}}}+{1}-{\frac{{12}}{+}}{{\sin}^{{2}}\theta}$$
$$\displaystyle={\sin{\theta}}−{\sin{{3}}}\theta+{\frac{{{\cos{{4}}}\theta}}{{{2}}}}+{1}−{\frac{{{1}−{2}{{\sin}^{{2}}\theta}}}{{{2}}}}$$
$$\displaystyle={1}+{\sin{\theta}}−{\sin{{3}}}\theta+{\frac{{{\cos{{4}}}\theta}}{{{2}}}}-{\frac{{{\cos{{2}}}\theta}}{{{2}}}}$$

Vasquez

Apply the identities
$$\cos 4\theta=2\cos^2 2\theta−1, \ \ \ \ \cos 2\theta=1−2\sin^2 \theta$$
$$\sin 3\theta=\sin \theta(2\cos 2\theta+1)$$
in simplifying the $$RHS$$
$$RHS=1+\sin\theta−\sin\theta(2\cos 2\theta+1)−\frac12(1−2\sin^2 \theta)+\frac12(2\cos^2 2\theta−1)$$
$$=\sin^2 \theta-2 \sin \theta \cos 2\theta+\cos^2 2\theta$$
$$=(\sin \theta-\cos 2\theta)^2=LHS$$