#1**+2 **

This one is a little tricky, Cupcake !!!!

Not really kosher, but since we assume that 1 - tan^2x is not equal to 0, we can multiply both sides by 1 - tan^2x and get that

2tanx = 1 - tan^2 x rearrange as

tan^2 x + 2 tanx - 1 = 0 let tan x = m and we have that

m^2 + 2m - 1 = 0

m^2 + 2m = 1 complete the square on m

m^2 + 2m + 1 = 1 + 1

(m + 1)^2 = 2 take both roots

m + 1 = ±√2

m = ±√2 - 1

So

tan x = √2 - 1.....take the arctan (√2 - 1) = m = 22.5° = pi/8 + n*pi

And

tan x = - √2 - 1.....take the arctan (-√2 - 1 ) = x = -67.5°.....add 180 to this = 112.5° = [225/2]° = (5/4)pi / 2 =

(5/8)pi + n * pi

CPhill Apr 26, 2019

#3**+3 **

** Trigonometric Identities:**

\(\large\dfrac{2\tan(x)}{1-\tan^2(x)} = 1\)

\(\begin{array}{|rcll|} \hline && \mathbf{ \dfrac{2\tan(x)}{1-\tan^2(x)} } \\\\ &=& \dfrac{ \tan(x)+\tan(x)}{1-\tan(x)\tan(x)} \quad | \quad \boxed{ \text{Formula: } \tan(x+y) = \dfrac{\tan(x)+\tan(y)}{1-\tan(x)\tan(y)} } \\\\ &\mathbf{=}& \mathbf{\tan(2x)} \\ \hline \end{array}\)

\(\begin{array}{|rcll|} \hline \tan(2x) &=& 1 \\ 2x &=& \arctan(1) + n\pi,\ n \in \mathbb{Z} \quad | \quad \arctan(1) = \dfrac{\pi}{4} \\ 2x &=& \dfrac{\pi}{4} + n\pi \\ \mathbf{x} &\mathbf{=}& \mathbf{\dfrac{\pi}{8} + n\dfrac{\pi}{2}} \\ \hline \end{array}\)

\(\text{Split } \mathbf{\dfrac{\pi}{8} + n\dfrac{\pi}{2}}\text{ in two solutions:}\)

\(\begin{array}{|rcll|} \hline x & = & \dfrac{\pi}{8} + n\dfrac{\pi}{2} \\ x_1 &=& \dfrac{\pi}{8} + n\left(\dfrac{\pi}{2}+\dfrac{\pi}{2} \right) \\ \mathbf{x_1} &\mathbf{=}& \mathbf{\dfrac{\pi}{8} + n\pi} \\\\ x_2 &=& \dfrac{\pi}{8}+\dfrac{\pi}{2} + n\left(\dfrac{\pi}{2}+\dfrac{\pi}{2} \right) \\ \mathbf{x_2} &\mathbf{=}& \mathbf{\dfrac{5\pi}{8} + n\pi} \\ \hline \end{array}\)

heureka Apr 26, 2019