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 [math]cos( \theta ) = x/4[/math]. We also know that [math]sin^2( \theta ) + cos^2 ( \theta ) = 1[/math], so if we know [math]cos( \theta )[/math] we can get [math]sin( \theta )[/math]. 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 $$b$$. Can you find an expression for $$b$$ in terms of $$16 \text{ and } x^2$$

Spoiler

[math]b = \sqrt{16-x^2}[/math]

Here is a picture in case a visual aid helps

I used $$A$$ instead of $$\theta$$ to save my poor drawing hand!

Once you know all three sides you can work out the other trig ratios. Start with $$\sin(\theta)$$ and $$\tan(\theta)$$ and use those to find the others. Your answer will be in terms of $$x$$ 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. ;)