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

pierdoodsu 2022-01-02 Answered
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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Expert Answer

kaluitagf
Answered 2022-01-03 Author has 5757 answers
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:
\(\displaystyle{\sin{{x}}}{\cos{{\left({2}{x}\right)}}}={\frac{{12}}{{−{\sin{{\left({x}\right)}}}+{\sin{{\left({3}{x}\right)}}}}}}\)
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Louis Page
Answered 2022-01-04 Author has 3991 answers

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 \(\displaystyle{\cos{{2}}}\theta={1}−{2}{{\sin}^{{2}}\theta}{w}{e}\ {u}{s}{e}\ {t}{h}{i}{s\ }toge{t}her\ {4}{{\sin}^{{3}}\theta}\)
\(\displaystyle{\left({\sin{\theta}}−{\cos{{2}}}\theta\right)}^{{2}}={{\sin}^{{2}}\theta}−{2}{\sin{\theta}}+{4}{{\sin}^{{3}}\theta}+{\frac{{12}}{+}}{\frac{{{\cos{{4}}}\theta}}{{{2}}}}\)
\(\displaystyle={{\sin}^{{2}}\theta}+{\sin{\theta}}-{\left({3}{\sin{\theta}}-{4}{{\sin}^{{3}}\theta}\right)}+{\frac{{12}}{+}}{\frac{{{\cos{{4}}}\theta}}{{{2}}}}\)
\(\displaystyle={\sin{\theta}}−{\sin{{3}}}\theta+{\frac{{{\cos{{4}}}\theta}}{{{2}}}}+{1}-{\frac{{12}}{+}}{{\sin}^{{2}}\theta}\)
\(\displaystyle={\sin{\theta}}−{\sin{{3}}}\theta+{\frac{{{\cos{{4}}}\theta}}{{{2}}}}+{1}−{\frac{{{1}−{2}{{\sin}^{{2}}\theta}}}{{{2}}}}\)
\(\displaystyle={1}+{\sin{\theta}}−{\sin{{3}}}\theta+{\frac{{{\cos{{4}}}\theta}}{{{2}}}}-{\frac{{{\cos{{2}}}\theta}}{{{2}}}}\)

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Vasquez
Answered 2022-01-08 Author has 9499 answers

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 \(RHS\)
\(RHS=1+\sin\theta−\sin\theta(2\cos 2\theta+1)−\frac12(1−2\sin^2 \theta)+\frac12(2\cos^2 2\theta−1)\)
\(=\sin^2 \theta-2 \sin \theta \cos 2\theta+\cos^2 2\theta\)
\(=(\sin \theta-\cos 2\theta)^2=LHS\)

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Find the value of \(\displaystyle{{\cos}^{{-{1}}}{\left(\sqrt{{{\frac{{{2}+\sqrt{{3}}}}{{{4}}}}}}\right)}}\)
I am trying to solve:
\(\displaystyle{{\sin}^{{-{1}}}{\cot{{\left({{\cos}^{{-{1}}}{\left(\sqrt{{{\frac{{{2}+\sqrt{{3}}}}{{{4}}}}}}\right)}}+{{\cos}^{{-{1}}}{\left({\frac{{\sqrt{{{12}}}}}{{{4}}}}\right)}}+{{\csc}^{{-{1}}}{\left(\sqrt{{2}}\right)}}\right)}}}}\)
My solution is as follow:
\(\displaystyle{T}={{\sin}^{{-{1}}}{\cot{{\left({{\cos}^{{-{1}}}{\left(\sqrt{{{\frac{{{2}+\sqrt{{3}}}}{{{4}}}}}}\right)}}+{{\cos}^{{-{1}}}{\left({\frac{{\sqrt{{{12}}}}}{{{4}}}}\right)}}+{{\csc}^{{-{1}}}{\left(\sqrt{{2}}\right)}}\right)}}}}\)
Since:
\(\displaystyle{{\csc}^{{-{1}}}\sqrt{{2}}}={{\sin}^{{-{1}}}{\left({\frac{{{1}}}{{\sqrt{{2}}}}}\right)}}={\frac{{\pi}}{{{4}}}};\ {{\cos}^{{-{1}}}{\left({\frac{{\sqrt{{{12}}}}}{{{4}}}}\right)}}={{\cos}^{{-{1}}}{\left({\frac{{\sqrt{{3}}}}{{{2}}}}\right)}}={\frac{{\pi}}{{{6}}}}\)
Then:
\(\displaystyle{T}={{\sin}^{{-{1}}}{\cot{{\left({{\cos}^{{-{1}}}{\left(\sqrt{{{\frac{{{2}+\sqrt{{3}}}}{{{4}}}}}}\right)}}+{\frac{{\pi}}{{{4}}}}+{\frac{{\pi}}{{{6}}}}\right)}}}}\)
I am not able to proceed further.
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