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Question
If cos 2θ = 0, determine `[(theta, costheta, sintheta),(costheta, sintheta, 0),(sintheta, 0, costheta)]^2`
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Solution
Given cos 2θ = 0
Let A = `[(theta, costheta, sintheta),(costheta, sintheta, 0),(sintheta, 0, costheta)]^2`
A = `[|(sintheta, 0),(0, costheta)| - cos|(costheta, 0),(sintheta, costheta)| + sintheta|(costheta, sintheta),(sintheta, 0)|]^2`
A = `[0 - cos theta(costheta - 0) + sintheta(0 - sin^2theta)]^2`
A = `[- cos3theta - sintheta]2`
A = `[cos^3theta + sin^3theta]^2` .......(1)
cos 2θ = 0
⇒ 2θ = `pi/2`
⇒ θ = `pi/4`
Substituting θ = `pi/4` in equation (1) we get
A = `[cos^3pi/4 + sin^3 pi/4]^2`
= `[(1/sqrt(2))^3 + (1/sqrt(2))^3]^2`
= `[2 xx (1/sqrt(2))^3]^2`
= `[2 xx 1/(2sqrt(2))]^2`
=`[1/sqrt(2)]^2`
A = `1/2`
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