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

snakeyesxlaw

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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) =
 
Re: Trig functions

Hello, snakeyesxlaw!

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

Identity: sec2θ=tan2θ+1        tanθ=±sec2θ1  \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 cot(t)\displaystyle \cot(t) is positive in Quadrant 3: \(\displaystyle \L\:\cot(t) \:=\:\frac{1}{\sqrt{S^2\,-\,1}}\)




(2) Suppose sin(t)=511\displaystyle \,\sin(t)\,=\,\frac{5}{11}\, and tan(t)<0\displaystyle \,\tan(t)\,<\,0

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

We are given: sin(t)=511=opphyp\displaystyle \:\sin(t) \:=\:\frac{5}{11}\:=\:\frac{opp}{hyp}
Angle t\displaystyle t has: opp=5,  hyp=11\displaystyle \,opp\,=\,5,\;hyp\,=\,11
Using Pythagorus, adj=±96=±46\displaystyle adj \,=\,\pm\sqrt{96} \,=\,\pm4\sqrt{6}
Since t\displaystyle t is in Quadrant 2: adj=46\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:

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

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

Therefore: sin(t)=sin(t)=511\displaystyle \:\sin(-t) \:=\:-\sin(t) \:=\:-\frac{5}{11}


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

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

Got it?


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

 
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