Finding expressions for the five other trigonometric functions....

[TrigEatsMe]

So here's the question:

Suppose cos(θ) =x/4. Find expressions for the other five trigonometric functions in terms of x.

In our practice problems we never had a variable x used and we were able to use the pythagorean theorem to determine the final side of the triangle and simply figure out the relationship using soh-cah-toa.

The x is throwing me off and the fact that the question doesn't specify that it's a right triangle...but perhaps that's a given since they specified that it was an angle with a cosine in the fourth quadrant?

In any case, all I've got is that sec(θ)=4/x. :(

I have a feeling there's some identity or something I should be using -- even a hint would be greatly appreciated.

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Additionally -- it does not specify whether it's a 45/45/90 or 30/60/90 triangle which would have been helpful as well.

 

[Dethrone]

Hi TrigEatsMe, (Wave)

The fact that there is an $x$ actually doesn't make it any harder!

$$\cos\left({\theta}\right)=\frac{x}{4}=\frac{\text{adjacent}}{\text{hypotenuse}}$$

From this, we can construct a right angled triangle (it need not matter whether it's 30-60-90 or 45-45-90). Using Pythagorean theorem, and knowing that the adjacent side is $x$ and the hypotenuse is $4$, how can I find the opposite side?

Remember that if we have the values of all three sides, we can easily find the other ratios:
$$\sin\left({\theta}\right)=\frac{\text{opposite}}{\text{hypotenuse}}$$
$$\tan\left({\theta}\right)=\frac{\text{opposite}}{\text{adjacent}}$$

The inverse trig ratios follow from this. :D

 

[TrigEatsMe]

Thanks for such a quick response! I must be missing something, though.

a^2 + b^2 = c^2

so I have x^2 + (b^2) = 16

The question asks for me to provide the other five trig functions in terms of x. I'm not seeing how to find out what that opposite side is -- what is its relationship to the adjacent side unless it's a 45-45-90 or 30/60/90?

Sorry :(

 

[topsquark]

Picking up where Rido12 left off...take this from the basic definitions, and yes, we are going to use the Pythagorean Theorem, but in a different way. You have \(\displaystyle cos( \theta ) = x/4\). We also know that \(\displaystyle sin^2( \theta ) + cos^2 ( \theta ) = 1\), so if we know \(\displaystyle cos( \theta )\) we can get \(\displaystyle sin( \theta )\). Give it a try.

-Dan

 

[SuperSonic4]

Thanks for such a quick response! I must be missing something, though.

a^2 + b^2 = c^2

so I have x^2 + (b^2) = 16

The question asks for me to provide the other five trig functions in terms of x. I'm not seeing how to find out what that opposite side is -- what is its relationship to the adjacent side unless it's a 45-45-90 or 30/60/90?

Sorry :(

You're getting there and you want to remove the variable \(\displaystyle b\). Can you find an expression for \(\displaystyle b\) in terms of \(\displaystyle 16 \text{ and } x^2\)

Spoiler

\(\displaystyle b = \sqrt{16-x^2}\)

Here is a picture in case a visual aid helps

I used \(\displaystyle A\) instead of \(\displaystyle \theta\) to save my poor drawing hand!

Once you know all three sides you can work out the other trig ratios. Start with \(\displaystyle \sin(\theta)\) and \(\displaystyle \tan(\theta)\) and use those to find the others. Your answer will be in terms of \(\displaystyle x\) but that's expected and correct in this case

 

[TrigEatsMe]

Thanks for all your help, folks -- I was able to get it submitted in time and assuming I got the answer correct I wills post what I submitted sometime tomorrow. If I didn't...well, I'll post it anyway so someone can correct me. ;)