Question

If cos 2θ = 0, determine  `[(theta, costheta, sintheta),(costheta, sintheta, 0),(sintheta, 0, costheta)]^2`

Answer 1

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`