With the work done in prior sections (particularly those listed below),
we have everything needed to efficiently find exact values for
things like $\displaystyle\,\cos\frac{81\pi}{4}\,$ and $\,\csc (2640^\circ)\,.$
All the necessary tools/ideas are repeated below, inanutshell.
Having trouble following the brief discussion on this page?
If so, review
the links above (in order)—they offer a much slower and kinder approach.
two special triangles  reciprocal relationships for trig functions  



Sine Opposite Hypotenuse Cosine Adjacent Hypotenuse Tangent Opposite Adjacent 
For example: $\displaystyle\csc = \frac{1}{\sin}$ 
angle/number  sine $\displaystyle\sin = \frac{\text{OPP}}{\text{HYP}}$ 
cosine $\displaystyle\cos = \frac{\text{ADJ}}{\text{HYP}}$ 
tangent $\displaystyle\tan = \frac{\text{OPP}}{\text{ADJ}}$ 
cotangent (reciprocal of tangent) 
secant (reciprocal of cosine) 
cosecant (reciprocal of sine) 
$0^\circ = 0 \text{ rad}$  $0$  $1$  $0$  not defined  $1$  not defined 
$\displaystyle 30^\circ = \frac{\pi}{6} \text{ rad}$  $\displaystyle\frac 12$  $\displaystyle\frac{\sqrt 3}2$  $\displaystyle\frac1{\sqrt 3} = \frac{\sqrt 3}{3} $  $\sqrt 3$  $\displaystyle\frac{2}{\sqrt 3} = \frac{2\sqrt 3}{3}$  $2$ 
$\displaystyle 45^\circ = \frac{\pi}{4} \text{ rad}$  $\displaystyle\frac 1{\sqrt 2} = \frac{\sqrt 2}{2}$  $\displaystyle\frac 1{\sqrt 2} = \frac{\sqrt 2}{2}$  $1$  $1$  $\sqrt 2$  $\sqrt 2$ 
$ \displaystyle 60^\circ = \frac{\pi}{3} \text{ rad}$  $\displaystyle\frac{\sqrt 3}{2}$  $\displaystyle \frac{1}{2}$  $\displaystyle \sqrt 3$  $\displaystyle\frac1{\sqrt 3} = \frac{\sqrt 3}{3} $  $2$  $\displaystyle\frac{2}{\sqrt 3} = \frac{2\sqrt 3}{3}$ 
$\displaystyle 90^\circ = \frac{\pi}{2} \text{ rad}$  $1$  $0$  not defined  $0$ ($\cot := \frac{\cos}{\sin}$) 
not defined  $1$ 
Reciprocals retain the sign ($+/$) of the original number. Therefore, in all the quadrants:

In Special Triangles and Common Trigonometric Values,
the ‘Locate–Shrink/Size–Signs’ method
was introduced for finding
trigonometric values of special angles.
With additional tools and terminology now at hand, that discussion is
presented more generally and efficiently here.
REDUCE: [This step is optional. If your angle isn't too big (say, $\,510^\circ\,$ or $\,\frac{7\pi}{3}\,$), then it may be easy for you to find its reference angle and quadrant, without ‘reducing’ it first. Your choice!] As discussed in Reference Angles, remove any extra rotations from $\,\theta\,$:


REFERENCE/QUADRANT: Lay off $\,\theta\,$ in the standard way:
Determine the quadrant for $\,\theta\,.$ 

SIZE/SIGN: Use the reference angle/number to find the correct SIZE of the desired trigonometric value. Use the quadrant to find the correct SIGN of the desired trigonometric value. 
In this first example, the angles aren't too big, so the optional (reduce) step is skipped.
Find: $\,\sec 510^\circ\,$  Find: $\displaystyle\,\cot (\frac{7\pi}{3})\,$  
REDUCE 
(skipped) [Dr. Burns would work with $\,510^\circ  360^\circ = 150^\circ\,.$ ] 
(skipped) [Dr. Burns would work with $\,\frac{7\pi}3 + \frac{6\pi}3 = \frac{\pi}3\,.$ ] 
REFERENCE/QUADRANT 
$\,510^\circ\,$ is in quadrant II; the reference angle is $\,30^\circ\,$ 
$\displaystyle\,\frac{7\pi}{3}\,$ is in quadrant IV; the reference angle is $\,\displaystyle\frac{\pi}{3}\,$ 
SIZE/SIGN 
SIZE: $\displaystyle\sec 30^\circ = \frac{2}{\sqrt 3}\,$ SIGN: In quadrant II, the secant is negative. Thus: $\displaystyle\,\sec 510^\circ = \frac{2}{\sqrt 3}\,$ 
SIZE: $\displaystyle\cot\frac{\pi}{3} = \frac{1}{\sqrt 3}\,$ SIGN: In quadrant IV, the cotangent is negative. Thus: $\displaystyle\,\cot(\frac{7\pi}{3}) = \frac{1}{\sqrt 3}\,$ 
In this final example, the angles are very big, so get rid of extra rotations (reduce) in the first step:
Find: $\,\csc(2640^\circ)\,$  Find: $\displaystyle\,\cos (\frac{81\pi}{4})\,$  
REDUCE 
$\displaystyle \frac{\theta}{360^\circ} = \frac{2640^\circ}{360^\circ} \approx 7$ $2640^\circ + 7\cdot 360^\circ = 120^\circ$ Work with $\,120^\circ\,$ instead of $\,2640^\circ\,.$ 
$\displaystyle \frac{\theta}{2\pi} = \frac{81\pi/4}{2\pi} \approx 10$ $\displaystyle\frac{81\pi}{4}  10\cdot 2\pi = \frac{81\pi}{4}  \frac{80\pi}{4} = \frac{\pi}4$ Work with $\displaystyle\,\frac{\pi}4\,$ instead of $\displaystyle\,\frac{81\pi}4\,.$ 
REFERENCE/QUADRANT 
$\,120^\circ\,$ is in quadrant III; the reference angle is $\,60^\circ\,$ 
$\,\frac{\pi}4\,$ is in quadrant I; the reference angle is $\,\displaystyle\frac{\pi}{4}\,$ 
SIZE/SIGN 
SIZE: $\displaystyle\csc 60^\circ = \frac{2}{\sqrt 3}\,$ SIGN: In quadrant III, the cosecant is negative. Thus: $\displaystyle\,\csc(2640^\circ) = \frac{2}{\sqrt 3}\,$ 
SIZE: $\displaystyle\cos\frac{\pi}{4} = \frac{1}{\sqrt 2}\,$ SIGN: In quadrant I, the cosine is positive. Thus: $\displaystyle\,\cos\frac{81\pi}{4} = \frac{1}{\sqrt 2}\,$ 
On this exercise, you will not key in your answer. However, you can check to see if your answer is correct. 
PROBLEM TYPES:
