Cos2theta

I'll state the question from my textbook here: If $\cos2\theta=0$, then $\begin^2$ $= (\cos^3 \theta + \sin^3 \theta)^2$ $= (\cos \theta + \sin \theta)^2(\cos^2 \theta - \cos \theta \sin \theta + \sin^2 \theta)^2$ $= (1+ \sin2\theta)(1-\sin2\theta + \sin^2 \theta \cos^2 \theta)$ $= (1+ \sin2\theta)(1-\sin2\theta) + (1 + \sin2\theta)\sin^2\theta \cos^2 \theta$ $= \cos^2 2\theta + \frac 14 (1 + \sin2\theta)\sin^2 2\theta$ $= \frac 14 (1 + \sin2\theta)\sin^2 2\theta$ Now since $\cos2\theta=0$, $\sin2\theta = \pm 1$. Therefore the above expression can take the values 0 and $\frac12$. My textbook gives the answer as $\frac12$. I don't see any grounds on rejecting the other answer of 0. Have I made a mistake somewhere? Or am I forgetting something? When in doubt, use the relations in the original problem. Let $\theta=\frac^2$$ Clearly, adding up all three rows of the matrix produces the zero vector, so the whole expression evaluates to zero. Your working is entirely correct: the book is wrong to omit 0 as an answer. The result of $\frac12$ is obtained with the other principal solution to $\cos2\theta=0$, $\theta=\frac\pi4$. You are right indeed by direct calculation we obtain $$\cos 2\theta=0\iff \theta=\frac4\implies \Delta^2=0$ If we compute the square of the matrix, we get $$ \begin $$ both possibilities being allowed by the hypothesis that $\cos2\theta=0$.

I'll state the question from my textbook here: If $\cos2\theta=0$, then $\begin^2$ $= (\cos^3 \theta + \sin^3 \theta)^2$ $= (\cos \theta + \sin \theta)^2(\cos^2 \theta - \cos \theta \sin \theta + \sin^2 \theta)^2$ $= (1+ \sin2\theta)(1-\sin2\theta + \sin^2 \theta \cos^2 \theta)$ $= (1+ \sin2\theta)(1-\sin2\theta) + (1 + \sin2\theta)\sin^2\theta \cos^2 \theta$ $= \cos^2 2\theta + \frac 14 (1 + \sin2\theta)\sin^2 2\theta$ $= \frac 14 (1 + \sin2\theta)\sin^2 2\theta$ Now since $\cos2\theta=0$, $\sin2\theta = \pm 1$. Therefore the above expression can take the values 0 and $\frac12$. My textbook gives the answer as $\frac12$. I don't see any grounds on rejecting the other answer of 0. Have I made a mistake somewhere? Or am I forgetting something? When in doubt, use the relations in the original problem. Let $\theta=\frac^2$$ Clearly, adding up all three rows of the matrix produces the zero vector, so the whole expression evaluates to zero. Your working is entirely correct: the book is wrong to omit 0 as an answer. The result of $\frac12$ is obtained with the other principal solution to $\cos2\theta=0$, $\theta=\frac\pi4$. You are right indeed by direct calculation we obtain $$\cos 2\theta=0\iff \theta=\frac4\implies \Delta^2=0$ If we compute the square of the matrix, we get $$ \begin $$ both possibilities being allowed by the hypothesis that $\cos2\theta=0$.