Trig will be the death of me

[jtw2e2]

I attempted to solve this with the exact pattern we were given for a similar problem (cos(tan[sup:4kcj2tre]-1[/sup:4kcj2tre](5x)), which I got correct. We were instructed to draw a right triangle with acute angle ?. I used the Pythagorean theorem to find the length of the hypotenuse and then solved the original in terms of x.

When I practiced the exact same approach I was greeted with the red x of death from my homework website.
Work:

let: sin[sup:4kcj2tre]-1[/sup:4kcj2tre](?5/12)= ?
then: sin? = ?5/12

Therefore:

So a[sup:4kcj2tre]2[/sup:4kcj2tre] + (?5)[sup:4kcj2tre]2[/sup:4kcj2tre] = 12[sup:4kcj2tre]2[/sup:4kcj2tre]
a=?(139)

tan = o/a = ?(5)/?(139)

Would someone please explain where I went wrong and why? HW due tonight with test tomorrow! I know it seems like I'm posting a lot. I'm not a slacker. For the past week I've spent at least 4+ hours a day studying, along with visiting teacher's office daily and going to tutorial center. I just can't seem to "get it."

 

[stapel]

Are you maybe supposed to rationalize the denominator...? :?:

 

[jtw2e2]

Re:

Are you maybe supposed to rationalize the denominator...? :?:

Tried it unsuccessfully. The answer it wants will be in the form: 1/x?b

 

[soroban]

Hello, jtw2e2!

I see nothing wrong with your reasoning, your work, and your answer.

\(\displaystyle \text{Find the exact value: }\;\tan\left[\sin^{-1}\left(\frac{\sqrt{5}}{12}\right)\right]\)

\(\displaystyle \text{The answer is indeed: }\;\frac{\sqrt{5}}{\sqrt{139}}\;=\;\frac{\sqrt{695}}{139}\)

\(\displaystyle Who\text{ said the answer is in the form: }\:\frac{1}{x\sqrt{b}}\:?\)
\(\displaystyle \text{That's really stupid!}\)


\(\displaystyle \text{We can do it, but it's embarrassing . . .}\)

\(\displaystyle \text{We have: }\;\frac{\sqrt{5}}{\sqrt{139}}\)

\(\displaystyle \text{Multiply by }\frac{\sqrt{5}}{\sqrt{5}}\!:\;\;\frac{\sqrt{5}}{\sqrt{139}}\cdot\frac{\sqrt{5}}{\sqrt{5}} \;=\;\frac{5}{\sqrt{695}}\)

\(\displaystyle \text{Divide top and bottom by 5}:\;\frac{1}{\frac{1}{5}\sqrt{695}} \quad\hdots\;\;There!\)

 

[jtw2e2]

stapel and soroban,

Thanks very much. It did indeed take the answer after I rationalized the denominator. The place where I got the idea it should be in that silly form was the practice problem for this problem. :shock: The 1 in the numerator of the silly form would have made more sense if the value in the original problem was the sin[sup:1hb0rs9o]-1[/sup:1hb0rs9o] of some whole number and not a fraction. Then I would have had to put that whole number over 1, and "a" from my triangle would have been equal to 1.

Hope that made sense. Thanks again for your help. :mrgreen: