# Thread: Hard Questions on Inverse Trigonometry

1. ## 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>

2. 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

3. Hello, Sunyata!

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

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

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

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

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

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

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

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

4. $x = -\frac{1}{2}$ is an extraneous solution

5. Absolutely right, skeeter . . . *blush*

Thanks for the correction.

6. Hmmm how exactly do u find the domain for x, skeeter and Soroban?

7. 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?

8. Oops i mean domain for x.

Sori