Answer to Question #251971 in Trigonometry for aisme

Question

(a) Show that sin(3x) + sin(x) = 4sin(x)cos^2(x)

(b) Find all the angles between 0 and $\pi$ which satisfy the equation sin(3x) + sin(x) = 2(cos^2)(x).

Answer 1

(a) Let us show that $\sin(3x) + \sin(x) = 4\sin(x)\cos^2(x).$ Indeed,

$\sin(3x) + \sin(x)=2\sin (\frac{3x+x}2)\cos (\frac{3x-x}2)=2\sin (2x)\cos (x)\\=2\cdot2\sin(x)\cos(x)\cos(x)=4\sin(x)\cos^2(x).$

(b) Let us find all the angles between 0 and $\pi$ which satisfy the equation $\sin(3x) + \sin(x) = 2\cos^2(x),$ which is equivalent to $4\sin(x)\cos^2(x)=2\cos^2(x),$ and hence to $2\cos^2(x)(2\sin(x)-1)=0.$ It follows that $\cos(x)=0$ or $\sin(x)=\frac{1}2,$ and hence $x=\frac{\pi}2+\pi n$ or $x=\frac{\pi}6+2\pi n$ or $x=\frac{5\pi}6+2\pi n, \ n\in\Z.$ Therefore, the angles between 0 and $\pi$ which satisfy the equation are $\frac{\pi}2,\ \frac{\pi}6,\ \frac{5\pi}6.$