need help with basic trig: If tan = -2 and sec = 5 radical. find...

[jeremiahcp]

Hello I am a college student in math 171, I transferred schools and in the school I came from they did not teach any trig in 151, so I have no exposure to it. However in this school they expect 171 students to know basic trig.

So If I can get some help getting up to speed that'd be great. They have a learning center but, for some reason it is not open on the weekends. So I am really on my own, trying to teach myself trig and it is not going anywhere.

Here is the problem I am having issues with.

If tan = -2 and sec = 5 radical.

find

sin =

cos =

cot =

csc =

I have gotten as far as figuring out what soh - cah - toa means, but that is about it, so any help would be great.

 

[ksdhart2]

Before we get started, I'm assuming that by "5 radical" you mean \(\displaystyle \sqrt{5}\). If that is incorrect, please reply with any necessary corrections. Now, since you say you're familiar with the mnemonic device SOH CAH TOA, you know how the six basic trig functions relate to a right triangle. So, it follows that with any trig problem a good starting place would be to draw a right triangle. Let's place the unknown angle (we'll call it theta or \(\displaystyle \theta\)) at the bottom right angle of the triangle, like so:

Now, since Tangent is Opposite over Adjacent, and Secant is 1 over Cosine or Hypotenuse over Adjacent, we know that:

\(\displaystyle \dfrac{\text{Opposite}}{\text{Adjacent}}=-2\) and \(\displaystyle \dfrac{\text{Hypotenuse}}{\text{Adjacent}}=\sqrt{5}\)

So what happens if we say the value of Adjacent is x? What then is the value of Opposite, in terms of x? And what is the value of Hypotenuse, in terms of x? Then, since Sine is Opposite over Hypotenuse, what is its value? Then use the mnemonic device and note that Secant = 1/Cosine, Cosecant = 1/Sine, and Cotangent = 1/Tangent, to calculate the remaining trig functions.

 

[jeremiahcp]

Thanks for the help, ksdhart2.

I am gonna try two here, please let me know if I am on the right track.

sin =2/√5, and cos =2O/ √5A

 

[ksdhart2]

Your answer for sine is very close, but the sign is wrong (it should be -2 in the numerator, make sure you know why that is). And your answer for cosine is not correct at all. Let's examine the relationship a bit further. If we continue with the notation of letting the adjacent side be x, then the given value for tangent tells us that the opposite side is -2x. Similarly, the given value of secant tells us that the hypotenuse is \(\displaystyle \sqrt{5} \cdot x\). We also know that cosine is "Adjacent over Hypotenuse" or:

\(\displaystyle cos(\theta)=\dfrac{x}{\sqrt{5} \cdot x}=\text{?}\)

Do you see how much easier this type of problem becomes when all sides are put in terms of x? What do you get when you simplify the above expression?

 

[jeremiahcp]

Ok I so I use tan = -2 to find the opposite (O) and sec = √5 to find the hypotenuse (H) than I can assign those values to my sides of the triangle in relation to theta?

Like this

Then use that to solve the rest?

So

cos = x/√5x 1/√5

cot =- 2

csc = √5

?