hello there, I have a question about identifying the two strengths of the Philippines that may contribute to a greater integration among countries in the Asian region. can you help me out in this one?
Answers
Explanation:
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▪Given :-
\begin{gathered} A = \begin{bmatrix} \sf cos \theta& \sf sin \theta \\ \sf - sin \theta& \sf cos \theta \end{bmatrix}\end{gathered}A=[cosθ−sinθsinθcosθ]
And
B=A+A^4B=A+A4
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▪To Calculate :-
det(B)
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▪Solution :-
\begin{gathered} A = \begin{bmatrix} \sf cos \theta& \sf sin \theta \\ \sf - sin \theta& \sf cos \theta \end{bmatrix}\end{gathered}A=[cosθ−sinθsinθcosθ]
So,
\begin{gathered} \sf A {}^{2} = \begin{bmatrix} \sf cos \theta& \sf sin \theta \\ \sf - sin \theta& \sf cos \theta \end{bmatrix} \\ \\ = \begin{bmatrix} \sf cos \theta& \sf sin \theta \\ \sf - sin \theta& \sf cos \theta \end{bmatrix} \\ \\ \begin{bmatrix} \sf cos \theta& \sf sin \theta \\ \sf - sin \theta& \sf cos \theta \end{bmatrix} \begin{bmatrix} \sf cos \theta& \sf sin \theta \\ \sf - sin \theta& \sf cos \theta \end{bmatrix} \\ \\ = \small \begin{bmatrix} \sf cos {}^{2} \theta - {sin}^{2} \theta& \sf sin \theta cos \theta + sin \theta cos \theta \\ \sf - sin \theta cos \theta - sin \theta cos \theta& \sf - {sin}^{2} \theta + cos {}^{2} \theta \end{bmatrix} \\ \\ = \begin{bmatrix} \sf cos 2\theta& \sf sin 2\theta \\ \sf - sin2 \theta& \sf cos2 \theta \end{bmatrix} \end{gathered}A2=[cosθ−sinθsinθcosθ]=[cosθ−sinθsinθcosθ][cosθ−sinθsinθcosθ][cosθ−sinθsinθcosθ]=[cos2θ−sin2θ−sinθcosθ−sinθcosθsinθcosθ+sinθcosθ−sin2θ+cos2θ]=[cos2θ−sin2θsin2θcos2θ]
Similarly,
\begin{gathered}A {}^{4} = \begin{bmatrix} \sf cos 4\theta& \sf sin 4\theta \\ \sf - sin 4\theta& \sf cos4 \theta \end{bmatrix}\end{gathered}A4=[cos4θ−sin4θsin4θcos4θ]
As,
Given Matrix
B = A + A {}^{4}B=A+A4
So,
\begin{gathered} \sf B= \begin{bmatrix} \sf cos \theta& \sf sin \theta \\ \sf - sin \theta& \sf cos \theta \end{bmatrix}+ \begin{bmatrix} \sf cos 4\theta& \sf sin 4\theta \\ \sf - sin 4\theta& \sf cos4 \theta \end{bmatrix} \\ \\ = \begin{bmatrix} \sf cos \theta + cos 4\theta& \sf sin \theta + sin 4\theta \\ \sf -( sin \theta + sin 4\theta)& \sf cos \theta + cos4 \theta \end{bmatrix} \end{gathered}B=[cosθ−sinθsinθcosθ]+[cos4θ−sin4θsin4θcos4θ]=[cosθ+cos4θ−(sinθ+sin4θ)sinθ+sin4θcosθ+cos4θ]
\begin{gathered} \bf \small\therefore det(B) = {(cos \theta + cos4 \theta)}^{2} + {(sin \theta + sin4 \theta)}^{2} \\ \\ = \sf {cos}^{2} \theta + {cos}^{2} 4\theta + 2 cos\theta cos4 \theta \\ + {sin}^{2} \theta \sf+ {sin}^{2} 4\theta + 2 sin\theta sin4 \theta \\ \\ = \sf 2 + 2cos(3 \theta)\end{gathered}∴det(B)=(cosθ+cos4θ)2+(sinθ+sin4θ)2=cos2θ+cos24θ+2cosθcos4θ+sin2θ+sin24θ+2sinθsin4θ=2+2cos(3θ)
\begin{gathered} \sf So, at \: \theta = \frac{\pi}{5} \\ \\ \sf det(B) = 2 + 2cos \frac{3\pi}{5} \\ \\ = \sf4 {cos}^{2} ( \frac{3\pi}{10} ) \\ \\ = \sf4(\frac{ \sqrt{10 - 2 \sqrt{5} } }{4} \: {)}^{2} \\ \\\large \colorbox{lime}{ \underline{\boxed{ \color{magenta}\bf det(B)= \frac{1}{4} (10 - 2 \sqrt{5} \: )}}}\end{gathered}So,atθ=5πdet(B)=2+2cos53π=4cos2(103π)=4(410−25)2 det(B)=4