trigo inverse

**prasum** · March 28th 2011, 01:11 AM

solve x 2tan(inverse)(sin x)=tan(inverse)(2sec x) 0<x<90

**Prove It** · March 28th 2011, 01:13 AM

Is this your equation?

$\displaystyle 2\tan^{-1}{(\sin{x})} = \tan^{-1}{(2\sec{x})}$ for $\displaystyle 0^{\circ} < x < 90^{\circ}$ ...

**Soroban** · March 28th 2011, 06:50 AM

Hello, prasum!

$\text{Solve for }x\!:\;2\tan^{\text{-}1}(\sin x) \;=\; \tan^{\text{-}1}(2\sec x),\;\text{ for }0^o < x < 90^o$

Let: . $\begin{Bmatrix}\alpha &=& \tan^{\text{-}1}(\sin x) & \Rightarrow& \tan\alpha &=& \sin x \\ \beta &=& \tan^{\text{-}1}(2\sec x) & \Rightarrow & \tan\beta &=& 2\sec x \end{Bmatrix}$ .[1]

And the equation becomes: . $2\alpha \:=\:\beta$

Take the tangent of both sides: . $\tan2\alpha \;=\;\tan\beta$

. . And we have: . $\displaystyle \frac{2\tan\alpha}{1 - \tan^2\alpha} \;=\;\tan\beta$

Substitute [1]: . $\dfrac{\sin x}{1-\sin^2\!x} \;=\;2\sec x$

. . And we have: . $\dfrac{\sin x}{\cos^2\!x} \;=\;\dfrac{2}{\cos x}$

Since $x \ne 90^o,\;\cos x \ne 0$

$\text{Multiply by }\cos x\!:\;\;\dfrac{\sin x}{\cos x} \;=\;2 \quad\Rightarrow\quad \tan x \:=\:2$

Therefore: . $x \;=\;\tan^{\text{-}1}2 \;\approx\;63.4^o$

**prasum** · March 28th 2011, 07:03 AM

Originally Posted by Soroban

Hello, prasum!

Let: . $\begin{Bmatrix}\alpha &=& \tan^{\text{-}1}(\sin x) & \Rightarrow& \tan\alpha &=& \sin x \\ \beta &=& \tan^{\text{-}1}(2\sec x) & \Rightarrow & \tan\beta &=& 2\sec x \end{Bmatrix}$ .[1]

And the equation becomes: . $2\alpha \:=\:\beta$

Take the tangent of both sides: . $\tan2\alpha \;=\;\tan\beta$

. . And we have: . $\displaystyle \frac{2\tan\alpha}{1 - \tan^2\alpha} \;=\;\tan\beta$

Substitute [1]: . $\dfrac{\sin x}{1-\sin^2\!x} \;=\;2\sec x$

. . And we have: . $\dfrac{\sin x}{\cos^2\!x} \;=\;\dfrac{2}{\cos x}$

Since $x \ne 90^o,\;\cos x \ne 0$

$\text{Multiply by }\cos x\!:\;\;\dfrac{\sin x}{\cos x} \;=\;2 \quad\Rightarrow\quad \tan x \:=\:2$

Therefore: . $x \;=\;\tan^{\text{-}1}2 \;\approx\;63.4^o$

thanks

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