# Math Help - Express in terms of x:sin(...

1. ## Express in terms of x:sin(...

I have such a problem:

a) Find $\tan [sin^{-1} (\frac{2}{5}) + cos^{-1} (\frac{5}{10})] =$

(Make sure your answer is an algebraic expression with square roots but without trigonometric or inverse trigonometric functions.)

Well, I have got the answer (how could I not, with a calculator it wasn't a big deal)

but I'm not sure about this question - I don't really understand what I need to do?

b) Express in terms of $x:sin(2tan^{-1}(x))$

do I have to plug in the value obtained in a) or what

$8.88:sin(2tan^{-1}(8.88))$, doesn't make sense for me...

2. ## Re: Express in terms of x:sin(...

I'd consider (b) as separate problem. Here is the trick. Let $\theta = \tan^{-1}x \Longrightarrow x = \frac x1 = \tan \theta$

So now you can proceed algebraically using trig identities, or geometrically and make your own right triangle. Tangent = opposite over adjacent, and so you can make a right triangle where one of the acute angles is $\theta$, the opposite side is $x$ and the adjacent side is $1$. You can use Pythagoras to find the other side. With the fully populated triangle, you can find all trig functions of $\theta$. Would you know how to proceed?

3. ## Re: Express in terms of x:sin(...

BTW, for part (a), your calculator gave you an algebraic answer or a decimal expansion? Because you were asked for the former.

4. ## Re: Express in terms of x:sin(...

Originally Posted by Jhevon
I'd consider (b) as separate problem. Here is the trick. Let $\theta = \tan^{-1}x \Longrightarrow x = \frac x1 = \tan \theta$

So now you can proceed algebraically using trig identities, or geometrically and make your own right triangle. Tangent = opposite over adjacent, and so you can make a right triangle where one of the acute angles is $\theta$, the opposite side is $x$ and the adjacent side is $1$. You can use Pythagoras to find the other side. With the fully populated triangle, you can find all trig functions of $\theta$. Would you know how to proceed?
OK, so by Pthagorian's we have the hypotenuse = $\sqrt{1 + x^{2}}$, but further...

we would have $\sin2 \frac{x}{\sqrt{1 + x^{2}}}$ am I right... do I need to do something else?

5. ## Re: Express in terms of x:sin(...

Originally Posted by Jhevon
BTW, for part (a), your calculator gave you an algebraic answer or a decimal expansion? Because you were asked for the former.
dose that mean that I wasn't supposed to use the calculator, I'm working in webwork and, anyways, if I enter expressions in form of radicals and fractions it gives me the equivalent in decimal expansion - I obtained the decimal expansion,

was I supposed to apply the same method that you showed me for b) ?

6. ## Re: Express in terms of x:sin(...

Hello, dokrbb!

Here's the first one . . .

$\text{(a) Find }\:\tan \left[\sin^{-1}\!\left(\tfrac{2}{5}\right) + \cos^{-1}\!\left(\tfrac{5}{10}\right)\right]$

Let $\alpha = \sin^{-1}\!\left(\tfrac{2}{5}\right) \quad\Rightarrow\quad \sin\alpha = \tfrac{2}{5} \quad\Rightarrow\quad \tan\alpha = \tfrac{2}{\sqrt{21}}$ .[1]

Let $\beta = \cos^{-1}\!\left(\tfrac{1}{2}\right) \quad\Rightarrow\quad \cos\beta = \tfrac{1}{2} \quad\Rightarrow\quad \tan\beta = \sqrt{3}$ .[2]

We have: . $\tan(\alpha + \beta) \;=\;\frac{\tan\alpha + \tan\beta}{1 - \tan\alpha\tan\beta}$

Substitute [1] and [2]: . $\frac{\frac{2}{\sqrt{21}} + \sqrt{3}}{1-\left(\frac{2}{\sqrt{21}}\right)\left(\sqrt{3} \right)}$

Multiply by $\tfrac{\sqrt{21}}{\sqrt{21}}:\;\;\frac{2 + \sqrt{63}}{\sqrt{21} + 2\sqrt{3}} \;=\; \frac{2+3\sqrt{7}}{\sqrt{21} - 2\sqrt{3}}$

Multiply by $\tfrac{\sqrt{21}+2\sqrt{3}}{\sqrt{21}+2\sqrt{3}} :\;\;\frac{2+3\sqrt{7}}{\sqrt{21} - 2\sqrt{3}} \cdot \frac{\sqrt{21} + 2\sqrt{3}}{\sqrt{21} + 2\sqrt{3}}$

. . . . . . $=\;\frac{2\sqrt{21} + 4\sqrt{3} + 21\sqrt{3} + 6\sqrt{21}}{21 - 12}$

. . . . . . $=\;\frac{8\sqrt{21} + 25\sqrt{3}}{9}$

7. ## Re: Express in terms of x:sin(...

Originally Posted by Soroban
Hello, dokrbb!

Here's the first one . . .

Let $\alpha = \sin^{-1}\!\left(\tfrac{2}{5}\right) \quad\Rightarrow\quad \sin\alpha = \tfrac{2}{5} \quad\Rightarrow\quad \tan\alpha = \tfrac{2}{\sqrt{21}}$ .[1]

Let $\beta = \cos^{-1}\!\left(\tfrac{1}{2}\right) \quad\Rightarrow\quad \cos\beta = \tfrac{1}{2} \quad\Rightarrow\quad \tan\beta = \sqrt{3}$ .[2]

We have: . $\tan(\alpha + \beta) \;=\;\frac{\tan\alpha + \tan\beta}{1 - \tan\alpha\tan\beta}$

Substitute [1] and [2]: . $\frac{\frac{2}{\sqrt{21}} + \sqrt{3}}{1-\left(\frac{2}{\sqrt{21}}\right)\left(\sqrt{3} \right)}$

Multiply by $\tfrac{\sqrt{21}}{\sqrt{21}}:\;\;\frac{2 + \sqrt{63}}{\sqrt{21} + 2\sqrt{3}} \;=\; \frac{2+3\sqrt{7}}{\sqrt{21} - 2\sqrt{3}}$

Multiply by $\tfrac{\sqrt{21}+2\sqrt{3}}{\sqrt{21}+2\sqrt{3}} :\;\;\frac{2+3\sqrt{7}}{\sqrt{21} - 2\sqrt{3}} \cdot \frac{\sqrt{21} + 2\sqrt{3}}{\sqrt{21} + 2\sqrt{3}}$

. . . . . . $=\;\frac{2\sqrt{21} + 4\sqrt{3} + 21\sqrt{3} + 6\sqrt{21}}{21 - 12}$

. . . . . . $=\;\frac{8\sqrt{21} + 25\sqrt{3}}{9}$
that's perfect, thanks a lot - and I already figured how to do the second one: I used $sin(2x) = 2sin(x)cos(x)$