# Hard Questions on Inverse Trigonometry

• Nov 15th 2009, 10:09 AM
Sunyata
Hard Questions on Inverse Trigonometry
Could someone please demonstrate to me how to solve these equations:

a) arccos(x) + arccos(x*sqrt(3)) = pi/2

b) 2arcsin(x*sqrt(6)) + arcsin(4x) = pi/2</SPAN>
• Nov 15th 2009, 10:16 AM
skeeter
Quote:

Originally Posted by Sunyata
Could someone please demonstrate to me how to solve these equations:

a) arccos(x) + arccos(x*sqrt(3)) = pi/2

b) 2arcsin(x*sqrt(6)) + arcsin(4x) = pi/2</SPAN>

see this thread ... solution process is the same.

http://www.mathhelpforum.com/math-he...ease-help.html
• Nov 15th 2009, 12:24 PM
Soroban
Hello, Sunyata!

Quote:

$\displaystyle (a)\;\;\arccos(x) + \arccos(x\sqrt{3}) \:=\:\frac{\pi}{2}$
Let: .$\displaystyle \begin{array}{ccccccccccc}\alpha \;=\; \arccos(x) &\Longrightarrow & \cos\alpha \;=\; x & \Longrightarrow & \sin\alpha \;=\; \sqrt{1-x^2} \\ \beta \;=\; \arccos(x\sqrt{3}) & \Longrightarrow & \cos\beta \;=\; x\sqrt{3} & \Longrightarrow & \sin\beta \;=\; \sqrt{1-3x^2} \end{array}$ .[1]

The equation becomes: .$\displaystyle \alpha + \beta \;=\;\frac{\pi}{2}$

Take the cosine: .$\displaystyle \cos(\alpha + \beta) \;=\;\cos\left(\tfrac{\pi}{2}\right)$

. . which becomes: .$\displaystyle \cos\alpha\cos\beta - \sin\alpha\sin\beta \;=\;0$

Substitute [1]: .$\displaystyle (x)(x\sqrt{3}) - (\sqrt{1-x^2})(\sqrt{1-3x^2}) \:=\:0$

. . . . . . $\displaystyle x^2\sqrt{3} - \sqrt{1-4x^2 + 3x^4} \:=\:0 \quad\Rightarrow\quad x^2\sqrt{3} \:=\:\sqrt{1 - 4x^2 + 3x^4}$

Square both sides: .$\displaystyle 3x^4 \:=\:1 - 4x^2 + 3x^4 \quad\Rightarrow\quad 4x^2 \:=\:1 \quad\Rightarrow\quad x^2 \:=\:\frac{1}{4}$

. . Therefore: .$\displaystyle x \;=\;\pm\frac{1}{2}$

• Nov 15th 2009, 12:53 PM
skeeter
$\displaystyle x = -\frac{1}{2}$ is an extraneous solution
• Nov 15th 2009, 01:55 PM
Soroban
Absolutely right, skeeter . . . *blush*

Thanks for the correction.

• Nov 15th 2009, 04:21 PM
xwrathbringerx
Hmmm how exactly do u find the domain for x, skeeter and Soroban?
• Nov 15th 2009, 04:25 PM
skeeter
Quote:

Originally Posted by xwrathbringerx
Hmmm how exactly do u find the limit, skeeter and Soroban?

this problem does not involve a limit ... what "limit" are you talking about?
• Nov 15th 2009, 04:28 PM
xwrathbringerx
Oops i mean domain for x. (Doh)

Sori
• Nov 15th 2009, 05:01 PM
skeeter