□If cos=tan, then the value of 2sin+sin2+sin3+sin4 is (A) 2 (B) 3 (C) 1 (D) 0 (E) none of the

Question

above

Answer 1

From the given information, we have $\cos \theta = \tan \theta = \frac{\sin \theta}{\cos \theta}.$ Since $\cos \theta \neq 0$ (otherwise $\cos \theta = \tan \theta$ would be undefined), we can multiply both sides by $\cos \theta$ to get $\cos^2 \theta = \sin \theta.$

Using the identity $\sin^2 \theta + \cos^2 \theta = 1,$ we have $\sin^2 \theta + \sin \theta = 1.$ Quadratic formula gives $\sin \theta = \frac{-1 \pm \sqrt{5}}{2},$ so $\sin \theta = \frac{-1 - \sqrt{5}}{2}$ because $\sin \theta$ is negative when $180^\circ < \theta < 270^\circ.$

Therefore,
\begin{align*}
2 \sin \theta + \sin 2 \theta + \sin 3 \theta + \sin 4 \theta &= 2 \cdot \left( \frac{-1 - \sqrt{5}}{2} \right) + \sin \left( 2 \cdot \frac{-1 - \sqrt{5}}{2} \right) + \sin \left( 3 \cdot \frac{-1 - \sqrt{5}}{2} \right) + \sin \left( 4 \cdot \frac{-1 - \sqrt{5}}{2} \right) \\
&= - (1 + \sqrt{5} ) + \left( - \sqrt{1 - \left( \frac{-1 - \sqrt{5}}{2} \right)^2 } \right) + \left( - \sqrt{1 - \left( \frac{-1 - \sqrt{5}}{2} \right)^2 } \right) + \sqrt{1 - \left( \frac{-1 - \sqrt{5}}{2} \right)^2} .
\end{align*}Note that the last two terms cancel with the first two terms, so the answer is $\boxed{0}.$