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, in-a-nutshell.
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 | |
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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 | $1$ |
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Reciprocals retain the sign ($+/-$) of the original number. Therefore, in all the quadrants:
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In Special Triangles and Common Trigonometric Values,
the ‘LocateShrink/SizeSigns’ 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 $\,\theta\,$:
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REFERENCE/QUADRANT: Lay off $\,\theta\,$ in the standard way:
Determine the quadrant for $\,\theta\,.$ |
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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:
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