# tangent of inverse trignometric function

• Feb 4th 2007, 11:54 AM
viet
tangent of inverse trignometric function
a) express in term of X :
$\displaystyle \sin\left( 2 \tan^{-1}(x)\right)$

b) Find: $\displaystyle \tan\left( \sin^{-1}(\frac {5}{8}) + \cos^{-1}(\frac {1}{6})\right)$
• Feb 4th 2007, 12:48 PM
Soroban
Hello, viet!

Quote:

a) Express in term of $\displaystyle x$: .$\displaystyle \sin\left(2\tan^{-1}(x)\right)$

We will need the identity: .$\displaystyle \sin2\theta \:=\:2\sin\theta\cos\theta$

Recall that an inverse trig value is an angle.

Let $\displaystyle \theta = \tan^{-1}(x)\quad\Rightarrow\quad\tan\theta = x$

Since $\displaystyle \tan\theta = \frac{x}{1} = \frac{opp}{adj}$, we have: $\displaystyle opp = x,\:adj = 1$
. . and Pythagorus gives us: $\displaystyle hyp = \sqrt{x^2+1}$
Hence: .$\displaystyle \sin\theta = \frac{x}{\sqrt{x^2+1}},\;\cos\theta = \frac{1}{\sqrt{x^2+1}}$

The problem becomes: .$\displaystyle \sin(2\theta) \:=\:2\sin\theta\cos\theta$

. . . . . . $\displaystyle =\:2\left(\frac{x}{\sqrt{x^2+1}}\right)\left(\frac {1}{\sqrt{x^2+1}}\right) \;=\;\boxed{\frac{2x}{x^2+1}}$

Quote:

b) Find: .$\displaystyle \tan\left[\sin^{-1}\left(\frac {5}{8}\right) + \cos^{-1}\left(\frac {1}{6}\right)\right]$

We have: .$\displaystyle \tan\left[\underbrace{\sin^{-1}\left(\frac{5}{8}\right)}_{\alpha} + \underbrace{\cos^{-1}\left(\frac{1}{6}\right)}_{\beta}\right]$

Then: .$\displaystyle \sin\alpha = \frac{5}{8}\quad\Rightarrow\quad opp = 5,\:hyp = 8\quad\Rightarrow\quad adj = \sqrt{39}$
. . Hence: .$\displaystyle \tan\alpha = \frac{5}{\sqrt{39}}$

And: .$\displaystyle \cos\beta = \frac{1}{6}\quad\Rightarrow\quad adj = 1,\:hyp = 6\quad\Rightarrow\quad opp = \sqrt{35}$
. . Hence: .$\displaystyle \tan\beta = \frac{\sqrt{35}}{1} = \sqrt{35}$

The problem becomes: .$\displaystyle \tan(\alpha + \beta)\;=\;\frac{\tan\alpha + \tan\beta}{1 - \tan\alpha\tan\beta} \;=\;$ $\displaystyle \frac{\frac{5}{\sqrt{39}} + \sqrt{35}}{1 - \frac{5}{\sqrt{39}}\cdot\sqrt{35}}$

Multiply top and bottom by $\displaystyle \boxed{\sqrt{39}\!:\;\;\frac{5 + \sqrt{1365}}{\sqrt{39} - 5\sqrt{35}}}$

I hope I didn't make any stupid blunders . . .
• Feb 4th 2007, 06:23 PM
thedoge
I used your method for part B on a similar problem and it is not working.

numbers were 5/6 for the sin, 5/9 for the cos.

ended with (((5/sqrt(11))+(sqrt(56)/5))/(1-(5/sqrt(11))*(sqrt(56)/5)))*sqrt(11)
but it is wrong
• Feb 4th 2007, 07:41 PM
Soroban
Hello, thedoge!

Quote:

I used your method for part B on a similar problem and it is not working.

Numbers were 5/6 for the sin, 5/9 for the cos.

Ended with: (((5/sqrt(11))+(sqrt(56)/5))/(1-(5/sqrt(11))*(sqrt(56)/5)))*sqrt(11) .??

but it is wrong

Evidently, your preliminary work is correct.
. . $\displaystyle \tan\alpha = \frac{5}{\sqrt{11}},\;\;\tan\beta = \frac{\sqrt{56}}{5}$

Identity: .$\displaystyle \tan(\alpha + \beta) \;=\:\frac{\tan\alpha + \tan\beta}{1 - \tan\alpha\cdot\tan\beta}$

We have: .$\displaystyle \tan(\alpha + \beta) \;=\;\frac{\frac{5}{\sqrt{11}} + \frac{\sqrt{56}}{5}} {1 - \left(\frac{5}{\sqrt{11}}\right)\!\left(\frac{\sqr t{56}}{5}\right)}$

Multiply top and bottom by $\displaystyle 5\sqrt{11}:\;\;\frac{25 + \sqrt{56}\sqrt{11}}{5\sqrt{11} - 5\sqrt{56}}$

• Feb 4th 2007, 07:46 PM
thedoge
Ah ok. Thank you:)