Question

Use the double angled formula to prove the following expressions.

(i) Given that cos(x+30)=3cos(x-30), prove that tanx= - root 3/ 2

(ii)prove that 1-cos2x/sin2x = tan x

(iii) verify that x= 180 is a solution of the equation 2x=2-2cos2x

(iv) Using the resulat from part (ii) or otherwise find the two other solutions 0<x<360 of the equation 2x= 2-2cos2x

Thanks

Answer 1

i) Given that $\cos(x + 30) = 3\cos(x - 30)$ , prove that $\tan x = -\text{root } 3/2$

$\cos (x + 30) = 3\cos (x - 30)\Leftrightarrow \cos x\cdot \cos 30 - \sin x\cdot \sin 30 = 3(\cos x\cdot \cos 30 + \sin x\cdot \sin 30)\Leftrightarrow$

Solution: $\frac{\sqrt{3}}{2} \cos x - \frac{1}{2} \sin x = \frac{3\sqrt{3}}{2} \cos x + \frac{3}{2} \sin x \Leftrightarrow \sqrt{3} \cos x = -2 \sin x \Leftrightarrow \tan x = -\frac{\sqrt{3}}{2}$ .

(ii) prove that $1 - \cos 2x / \sin 2x = \tan x$

Solution: $\frac{1 - \cos 2x}{\sin 2x} = \frac{1 - 1 - 2\sin^2 x}{2\sin x\cos x} = \frac{\sin x}{\cos x} = \tan x$

(iii) verify that $x = 180$ is a solution of the equation $2x = 2 - 2\cos 2x$

Solution: $x = 180^{\prime \prime} = \pi, 2x = 2 - 2\cos 2x \Leftrightarrow x = 1 - \cos 2x \Leftrightarrow \pi = 1 - \cos (2\pi) \Leftrightarrow \pi = 1 - 1 = 0 \Leftrightarrow \pi \neq 0$

(iv) Using the resultant from part (ii) or otherwise find the two other solutions $0 < x < 360$ of the equation $2x = 2 - 2\cos 2x$

Solution: $2x = 2 - 2\cos 2x \Leftrightarrow x = 1 - \cos 2x, \frac{1 - \cos 2x}{\sin 2x} = \tan x \Leftrightarrow \frac{x}{\sin 2x} = \tan x$

$g(x) = 1 - x, f(x) = \cos (2x), 0'' < x < 360''$ , that's why there are 2 solutions.