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# Trigonometric Identities

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4 Trig

Apr 25, 2019

#1
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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   Apr 26, 2019
edited by CPhill  Apr 26, 2019
#2
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Thank you!! Cupcake  Apr 26, 2019
#3
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Trigonometric Identities:

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

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

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

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

$$\begin{array}{|rcll|} \hline x & = & \dfrac{\pi}{8} + n\dfrac{\pi}{2} \\ x_1 &=& \dfrac{\pi}{8} + n\left(\dfrac{\pi}{2}+\dfrac{\pi}{2} \right) \\ \mathbf{x_1} &\mathbf{=}& \mathbf{\dfrac{\pi}{8} + n\pi} \\\\ x_2 &=& \dfrac{\pi}{8}+\dfrac{\pi}{2} + n\left(\dfrac{\pi}{2}+\dfrac{\pi}{2} \right) \\ \mathbf{x_2} &\mathbf{=}& \mathbf{\dfrac{5\pi}{8} + n\pi} \\ \hline \end{array}$$ Apr 26, 2019
#4
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Thank you!

Cupcake  Apr 26, 2019