1. ## changing the term

x=3sinu

How can I write 1/3 sec u in terms of x?

2. ## Re: changing the term

So if I am understanding correctly, you want to write

$\displaystyle \frac{1}{3}sec(u)$ in terms of x where you use the equation $\displaystyle x=3sin(u)$

Well, we know that $\displaystyle \sec(u) = \frac{1}{\cos(u)}$, thus $\displaystyle \frac{1}{3}sec(u) = \frac{1}{\3cos(u)}$

We also know the mother of all identities is $\displaystyle \sin^2(u) + \cos^2(u) = 1$ which can be solved for $\displaystyle \cos(u)$ as $\displaystyle \cos(u) = \pm\sqrt{1-\sin^2(u)}$

Now, using the equation $\displaystyle x=3sin(u)$ do you think you can finish this?

3. ## Re: changing the term

hi Youkla

i think it should be $\displaystyle \pm\sqrt{1-\sin^2{x}}$

4. ## Re: changing the term

Originally Posted by anonimnystefy
hi Youkla

i think it should be $\displaystyle \pm\sqrt{1-\sin^2{x}}$
In the words of Storage Wars..."Yuuuuuuuup!"

Thanks

5. ## Re: changing the term

Hello, Stuck Man!

$\displaystyle \text{Given: }\:x \:=\:3\sin u$

$\displaystyle \text{Write }\tfrac{1}{3}\sec u\text{ in terms of }x.$

We are given: .$\displaystyle x \:=\:3\sin u \quad\Rightarrow\quad \sin u \:=\:\frac{x}{3} \:=\:\frac{opp}{hyp}$

That is, $\displaystyle u$ is an angle in a right triangle with: $\displaystyle opp = x,\;hyp = 3$
Using Pythagorus, we find that: .$\displaystyle adj = \sqrt{9-x^2}$

Hence: .$\displaystyle \sec u \:=\:\frac{hyp}{adj} \:=\:\frac{3}{\sqrt{9-x^2}}$

Therefore: .$\displaystyle \frac{1}{3}\sec u \;=\;\frac{1}{3}\cdot\frac{3}{\sqrt{9-x^2}} \;=\; \frac{1}{\sqrt{9-x^2}}$

6. ## Re: changing the term

I don't think the plus/minus sign is justified. A root of that expression can't be negative. Thanks for the help. Sometimes I have wondered why people drop the plus/minus sign.

7. ## Re: changing the term

Originally Posted by Stuck Man
I don't think the plus/minus sign is justified. A root of that expression can't be negative. Thanks for the help. Sometimes I have wondered why people drop the plus/minus sign.
Actually, the " ± " is justified. The cosine function (and thus the secant function) can have the same or opposite sign as the sine function, depending upon the quadrant.

Also, when Soroban found that $\displaystyle adj = \sqrt{9-x^2}$, he found it by taking the square root of $\displaystyle (adj)^2$, so it should have been $\displaystyle adj = \pm\sqrt{9-x^2}$, if you prefer to look at it this way.