Hard Questions on Inverse Trigonometry

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November 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>
November 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
November 15th 2009, 12:24 PM
Soroban

Hello, Sunyata!

Quote:

$(a)\;\;\arccos(x) + \arccos(x\sqrt{3}) \:=\:\frac{\pi}{2}$

Let: . $\begin{array}{ccccccccccc}\alpha \;=\; \arccos(x) &\Longrightarrow & \cos\alpha \;=\; x & \Longrightarrow & \sin\alpha \;=\; \sqrt{1-x^2} \\<br /> \beta \;=\; \arccos(x\sqrt{3}) & \Longrightarrow & \cos\beta \;=\; x\sqrt{3} & \Longrightarrow & \sin\beta \;=\; \sqrt{1-3x^2}<br /> \end{array}$ .[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: . $x \;=\;\pm\frac{1}{2}$
November 15th 2009, 12:53 PM
skeeter

$x = -\frac{1}{2}$ is an extraneous solution
November 15th 2009, 01:55 PM
Soroban

Absolutely right, skeeter . . . *blush*

Thanks for the correction.
November 15th 2009, 04:21 PM
xwrathbringerx

Hmmm how exactly do u find the domain for x, skeeter and Soroban?
November 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?
November 15th 2009, 04:28 PM
xwrathbringerx

Oops i mean domain for x. (Doh)

Sori
November 15th 2009, 05:01 PM
skeeter

start a new thread.
November 15th 2009, 06:16 PM
Sunyata

ARGH I get a massive equation for the 2nd question which I cannot solve...
November 15th 2009, 07:19 PM
xwrathbringerx

Ummm for the 2nd question, am i MEANT to get 7744x^10 - 3008x^6 + 176 x^5 + 240x^4 - 8x^2 - 8x + 1 = 0 because I keep on getting it, which I've been trying to solve for ages (Sleepy)