[MOVED] Trig functions: Write cot(t) in terms of sec(t) if

[snakeyesxlaw]

problem #1)

Write cot (t) in terms of sec (t) if the terminal point determined by t is in the third quadrant.

NOTE: Use S in your answer to stand for sec (t) (i.e. if the answer were 2 sec (t), you would enter 2S).

cot (t) = ?

okay, so:

cot t = cost t / sin t or 1 / tan t

sec t = 1 / cos t

the cot t in terms of sec t?

any suggestions?

Problem #2)

Suppose sin t = 5/11 and tan t < 0. Find each of the following:

sin (- t) =
cos (2 pi - t) =
tan t =
csc ( t + 10 pi) =

 

[soroban]

Re: Trig functions

Hello, snakeyesxlaw!

(1) Write \(\displaystyle \cot(t)\) in terms of \(\displaystyle \sec(t)\)
if the terminal point determined by \(\displaystyle t\) is in the third quadrant.

Identity: \(\displaystyle \:\sec^2\theta\:=\:\tan^2\theta\,+\,1\;\;\Rightarrow\;\;\tan\theta \:=\:\pm\sqrt{\sec^2\theta\,-\,1}\;\)

We have: \(\displaystyle \L\:\cot(t) \:=\:\frac{1}{\tan(t)} \:=\:\frac{1}{\pm\sqrt{\sec^2(t)\,-\,1}}\)

Since \(\displaystyle \cot(t)\) is positive in Quadrant 3: \(\displaystyle \L\:\cot(t) \:=\:\frac{1}{\sqrt{S^2\,-\,1}}\)

(2) Suppose \(\displaystyle \,\sin(t)\,=\,\frac{5}{11}\,\) and \(\displaystyle \,\tan(t)\,<\,0\)

Sine is positive in Quadrants 1 and 2.
Tangent is negative in Quadrants 2 and 4.
. . Hence, \(\displaystyle t\) is in Quadrant 2.

We are given: \(\displaystyle \:\sin(t) \:=\:\frac{5}{11}\:=\:\frac{opp}{hyp}\)
Angle \(\displaystyle t\) has: \(\displaystyle \,opp\,=\,5,\;hyp\,=\,11\)
Using Pythagorus, \(\displaystyle adj \,=\,\pm\sqrt{96} \,=\,\pm4\sqrt{6}\)
Since \(\displaystyle t\) is in Quadrant 2: \(\displaystyle \,adj \,=\,-4\sqrt{6}\)

Hence, we have: \(\displaystyle \:\begin{Bmatrix}\sin(t) & = & \frac{5}{11} \\ \\
\cos(t) & = & -\frac{4\sqrt{6}}{11} \\ \\ \tan(t) & = & -\frac{5}{4\sqrt{6}}\end{Bmatrix}\)

Find each of the following:

\(\displaystyle (a)\;\sin (- t)\)

We're expected to know that: \(\displaystyle \,\sin(-\theta) \:=\:-\sin(\theta)\)

Therefore: \(\displaystyle \:\sin(-t) \:=\:-\sin(t) \:=\:-\frac{5}{11}\)

\(\displaystyle (b)\;\cos(2\pi\,-\,t)\)

Since \(\displaystyle t\) is in Quadrant 2, then \(\displaystyle 2\pi\,-\,t\) is in Quadrant 3.
Hence: \(\displaystyle \:\cos(2\pi\,-\,t) \:=\:\cos(t)\)

Got it?


You should be able to do the last two now . . .