# Need help on Review

• Apr 5th 2009, 03:47 PM
goldenroll
Need help on Review
Can someone help with these problems I got a test on Thursday and I don't know how to do these. I can't use a cal by the way.

1. What is tan(2π/3)?
I know that tan is y/x so it is -1/2, sqroot 3/2 so how would I put this in an equation?

2. If t^(2) = ln(1 + x) − ln(1 − x), what is x in terms of t?

3. What are all solutions of the equation: sin^(2) x + 2 sin x = 5/4?

4. If f (x) = √4x^(2) + 8, x
≥ 0, what is f ^(− 1) (x)?

5. sin(2 cos^(−1) x) is equivalent to

6. If f (x) = 1+sin x^(2) −sin x , −pie/2 ≤ x ≤ pie/2 , what is f^(−1) (x)?

7. Whats equivalent to 4a/(2/b) + (1/2)

8. √12 x Squareroot y/4

9. What is the domain of f (x) = 2 − ln(x^(2) − 9)?

10. A water tank is initially 3/5 full. After adding 5 gallons of water, it is 2/3 full. What is the capacity of the tank in gallons? I got 75 for this am I right? 3/5= 18/30, 2/3=20/30 and 2= 5 gallons so 2/30= 15x5=75
• Apr 5th 2009, 10:53 PM
Hello goldenroll
Quote:

Originally Posted by goldenroll
Can someone help with these problems I got a test on Thursday and I don't know how to do these. I can't use a cal by the way.

1. What is tan(2π/3)?
I know that tan is y/x so it is -1/2, sqroot 3/2 so how would I put this in an equation?

Use $\displaystyle \tan \pi/3 = \sqrt3$, and then $\displaystyle \tan 2\pi /3 = -\tan(\pi - \pi /3)$

Quote:

2. If t^(2) = ln(1 + x) − ln(1 − x), what is x in terms of t?
$\displaystyle \ln(1+x) -\ln(1-x) = \ln\left(\frac{1+x}{1-x}\right) = t^2$

Raise $\displaystyle e$ to the power of both sides:

$\displaystyle \frac{1+x}{1-x} = e^{t^2}$

Solve for $\displaystyle x$
.

Quote:

3. What are all solutions of the equation: sin^(2) x + 2 sin x = 5/4?
Write the equation as

$\displaystyle 4\sin^2x+8\sin x - 5 =0$

This is a quadratic in $\displaystyle \sin x$. Factorise and solve.

Quote:

4. If f (x) = √4x^(2) + 8, x ≥ 0, what is f ^(− 1) (x)?
I'm not quite sure what the square root sign covers here. I assume you mean:

$\displaystyle f(x) = \sqrt{4x^2 + 8}$

So, to find $\displaystyle f^{-1}$, let $\displaystyle f(x)=y$. Then

$\displaystyle y = \sqrt{4x^2 + 8}$

Now make $\displaystyle x$ the subject of this equation, to give

$\displaystyle x = \sqrt{\frac{y^2-8}{4}}=f^{-1}(y)$

So $\displaystyle f^{-1}(x) =$
$\displaystyle \sqrt{\frac{x^2-8}{4}}$

Quote:

5. sin(2 cos^(−1) x) is equivalent to
Let $\displaystyle \cos^{-1}x = y$.

$\displaystyle \Rightarrow \cos y = x$ and $\displaystyle \sin y = \sqrt{1-x^2}$

$\displaystyle \Rightarrow \sin(2\cos^{-1}x) = \sin 2y = 2\sin y \cos y = ?$
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