Thread: trig inverse...

1. trig inverse...

i have two pre calculus problems.please help!
problem 1: if $\cos ^{-1} x + \cos^{-1}y + \cos^{-1}z=\pi$ then prove that $x^{2}+y^{2}+z^{2}+2xyz=1$

problem 2:evaluate $\lim_{n\to 2} \frac{x^{2}-4}{\sqrt{x+2}-\sqrt{3x-2}}$

2. Originally Posted by earthboy
i have two pre calculus problems.please help!
What are the problems?

3. what i have done:

problem 1:
i took $\cos ^{-1} x , \cos^{-1}y , \cos^{-1}z$ as A,B and C respectively.now $A+B+C=\pi$ now we have to show $\cos ^{2} A + \cos^{2}B + \cos^{2}C + 2 \cos A\cos B\cos C = 1$ but i keep getting stuck after thatplease help!
problem 2: is the answer 0? i factored the expression into $\frac{(2-x)(\sqrt{x+2}+\sqrt{3x-2})}{2}$ and substituted 2 in it.

thanks in advance!

4. Originally Posted by e^(i*pi)
What are the problems?
sorry i pressed the submit button by mistake...edited

5. is the answer to the limit problem 0 ?
please reply fast.

6. Hello, earthboy!

$\displaystyle\text{2. Evaluate: }\;\lim_{x\to 2} \frac{x^{2}-4}{\sqrt{x+2}-\sqrt{3x-2}}$

Multiply numerator and denominator by the conjugate:

$\displaystyle \frac{x^2-4}{\sqrt{x+2} - \sqrt{3x-2}} \cdot\frac{\sqrt{x+2} + \sqrt{3x-2}}{\sqrt{x+2} + \sqrt{3x-2}} \;=\;\frac{(x^2-4)(\sqrt{x+2} + \sqrt{3x-2})}{(x+2) - ( 3x-2)}$

. . . . $\displaystyle =\;\frac{(x^2-4)(\sqrt{x+2} + \sqrt{3x-2})} {-2x + 4} \;=\; \frac{(x-2)(x+2)(\sqrt{x+2} + \sqrt{3x-2})}{-2(x-2)}$

. . . . $\displaystyle =\;\frac{(x+2)(\sqrt{x+2} + \sqrt{3x-2})}{-2}$

$\displaystyle \text{Therefore: }\;\lim_{x\to2}\frac{(x+2)(\sqrt{x+2} + \sqrt{3x-2})}{-2} \;=\;\frac{4(\sqrt{4} + \sqrt{4})}{-2}$

. . . . . . . . . . $\displaystyle =\;\frac{4(2+2)}{-2} \;=\;\frac{16}{-2} \;=\;-8$

7. Originally Posted by earthboy
problem 1:
i took $\cos ^{-1} x , \cos^{-1}y , \cos^{-1}z$ as A,B and C respectively.now $A+B+C=\pi$ now we have to show $\cos ^{2} A + \cos^{2}B + \cos^{2}C + 2 \cos A\cos B\cos C = 1$ but i keep getting stuck after thatplease help!

$\cos ^{2} A + \cos^{2}B + \cos^{2}C + 2 \cos A\cos B\cos C$
$= \cos ^{2} A + \cos^{2}B + \cos^{2}(\pi - A - B) + 2 \cos A\cos B\cos(\pi - A - B)$
$= \cos ^{2} A + \cos^{2}B + \cos^{2}(A + B) - 2 \cos A\cos B\cos(A + B)$
$= \cos ^{2} A + \cos^{2}B + (\cos A \cos B - \sin A \sin B)^2 - 2 \cos A\cos B(\cos A \cos B - \sin A \sin B)$
$= \cos ^{2} A + \cos^{2}B + \cos^2 A \cos^2 B - 2\cos A \cos B \sin A \sin B + \sin^2 A \sin^2 B - 2 \cos^2 A\cos^2 B + 2\cos A \cos B \sin A \sin B$
$= \cos ^{2} A + \cos^{2}B + \cos^2 A \cos^2 B + \sin^2 A \sin^2 B - 2 \cos^2 A\cos^2 B$
...

Now use $\sin^2 x = 1 - \cos^2 x$