1
$\begingroup$

I need some help solving some questions because I have no idea how to solve them, and some explanation would be appreciated.

The questions says:

Given $\cot\alpha=\frac{\sqrt{13}}{6}$ and $\alpha$ is in quadrant III, find the exact values of the remaining five trigonometric functions.

I need to find the value of $\sin\alpha$, $\cos\alpha$, $\tan\alpha$, $\csc\alpha$, $\sec\alpha$.

$\endgroup$
2
$\begingroup$

Hint: Suppose that $\alpha$ were in quadrant I. Then $$ \cot \alpha = \frac{\cos \alpha}{\sin \alpha} = \frac{\sqrt{1-\sin^2 \alpha}}{\sin \alpha}, $$ from which you can find $\sin \alpha$ by solving a quadratic equation. Given $\sin \alpha$, it is easy to compute the other trigonometric functions.

When $\alpha$ is in quadrant III, you have to introduce a minus somewhere.

$\endgroup$
  • $\begingroup$ I'm sorry I still don't understand, can you explain more? $\endgroup$ – AliAlM May 4 '15 at 2:59
  • $\begingroup$ No, now it's your time to solve the question using my hint. (Alternatively, if you wait a bit then someone else will probably provide a complete answer.) $\endgroup$ – Yuval Filmus May 4 '15 at 3:01
  • $\begingroup$ Thank you, I managed to find the asnwer. $\endgroup$ – AliAlM May 4 '15 at 3:16
1
$\begingroup$

the easy way to do this is to pick the point $(x,y)$ in the third quadrant so that $\cot \alpha = \frac x y = \frac{\sqrt{13}}6.$ one such point is $x = -\sqrt {13}, y = -6.$ you scale it down by $7 = \sqrt{6^2 + 13}$ to put it on the unit circle. the terminal point of the angle $\alpha$ is $$(x,y)=\left(-\frac{\sqrt{13}}7, -\frac67\right).\tag 1$$

now you can real all the trig ratios $\sin \alpha = y, \cos \alpha = x, \cdots$ from $(01).$

$\endgroup$
0
$\begingroup$

You are given that $\cot(\alpha) = \frac{\sqrt{13}}{6}$, and that the angle $\alpha$ is in third quadrant. Let's think of this using a right triangles. We know: $$\cot(\alpha) = \frac{\mbox{adjacent side}}{\mbox{opposite side}}$$ Also, since we are in the third quadrant, you can think of each of these sides as being negative, so technically, $\cot(\alpha) = \frac{-\sqrt{13}}{-6}$. This gives us the following picture.

diagram

Now, find the hypotenuse and find the other trig functions using the right triangle definitions.

$\endgroup$
0
$\begingroup$

Here are some useful tips

(S)OH (C)AH (T)OA: O = Opposite, A = Adjacent, H = Hypotenuse

Sin = O/H , Cos = A/H, Tan = O/A Csc = H/O , Sec = H/A, Cot = A/O

SOH stands for Sine equals Opposite over Hypotenuse.

CAH stands for Cosine equals Adjacent over Hypotenuse.

TOA stands for Tangent equals Opposite over Adjacent.

To determine where SIN, COS, or TAN positive or negative:

Remember this (ASTC) * (A)ll (S)tudents (T)ake (C)alculus *

Quadrant I = (A)ll meaning all of trig functions are positive

Quadrant II = (S)tudents meaning only Sine is positive(hint the S)

Quadrant III = (T)ake meaning only Tangent is positive(hint the T)

Quadrant IIII= (C)alculus meaning only Cosine is positive(hint the C)

These are for memorization. memorization is not the best technique in mathematics

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.