Need help on Review

Hello goldenroll

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

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

$\ln(1+x) -\ln(1-x) = \ln\left(\frac{1+x}{1-x}\right) = t^2$

Raise $e$ to the power of both sides:

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

Solve for $x$ .

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

Write the equation as

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

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

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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:

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

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

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

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

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

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

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5. sin(2 cos^(−1) x) is equivalent to

Let $\cos^{-1}x = y$ .

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

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

Grandad