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Curated and Reviewed byLesson Planet: Curated OER Educator Rating Answer KeyJoin to access all included materialsIn this special angles learning exercise, students find the value of given trigonometric functions. This four-page learning exercise contains 30 problems. Answers are provided on the last two pages. Resource Details Your browser does not support the audio element. LESSON READ-THROUGH by Dr. Carol JVF Burns (website creator) Follow along with the highlighted text while you listen! Thanks for your support! With the work done in prior sections (particularly those listed below), we have everything needed to efficiently find exact values for things like $displaystyle,cosfrac81pi4,$ and $,csc (-2640^circ),$. All the necessary toolsideas 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 SOHCAHTOA Sine Opposite Hypotenuse Cosine Adjacent Hypotenuse Tangent Opposite Adjacent For example: $displaystylecsc = frac1sin$ anglenumber sine$displaystylesin = fractextOPPtextHYP$ cosine$displaystylecos = fractextADJtextHYP$ tangent$displaystyletan = fractextOPPtextADJ$ cotangent(reciprocalof tangent) secant(reciprocalof cosine) cosecant(reciprocalof sine) $0^circ = 0 text rad$ $0$ $1$ $0$ not defined $1$ not defined $displaystyle 30^circ = fracpi6 text rad$ $displaystylefrac 12$ $displaystylefracsqrt 32$ $displaystylefrac1sqrt 3 = fracsqrt 33 $ $sqrt 3$ $displaystylefrac2sqrt 3 = frac2sqrt 33$ $2$ $displaystyle 45^circ = fracpi4 text rad$ $displaystylefrac 1sqrt 2 = fracsqrt 22$ $displaystylefrac 1sqrt 2 = fracsqrt 22$ $1$ $1$ $sqrt 2$ $sqrt 2$ $ displaystyle 60^circ = fracpi3 text rad$ $displaystylefracsqrt 32$ $displaystyle frac12$ $displaystyle sqrt 3$ $displaystylefrac1sqrt 3 = fracsqrt 33 $ $2$ $displaystylefrac2sqrt 3 = frac2sqrt 33$ $displaystyle 90^circ = fracpi2 text rad$ $1$ $0$ not defined $0$($cot := fraccossin$) not defined $1$ Reciprocals retain the sign ($+-$) of the original number. Therefore, in all the quadrants: Cosecant has the same sign as sine. Secant has the same sign as cosine. Cotangent has the same sign as tangent. In Special Triangles and Common Trigonometric Values, the Locate–ShrinkSize–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. the RRQSS (Reduce-ReferenceQuadrant-SizeSign) Technique REDUCE: This step is optional. If your angle isnt too big (say, $,510^circ,$ or $,-frac7pi3,$), 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 $,theta,$: $theta,$ in DEGREES $theta,$ in RADIANS How many extra rotations (if any) in $,theta,$? To answer this question, compute $displaystyle,n := frac|theta|360^circ,$, rounded to the nearest whole number. If $,theta,$ is positive, then replace $,theta,$ by $,theta - ncdot 360^circ,$. If $,theta,$ is negative, then replace $,theta,$ by $,theta + ncdot 360^circ,$. Now, your anglenumber is manageable: it is between $,-180^circ,$ and $,180^circ,$. But, it has the same terminal point, so all the trigonometric values are the same! How many extra rotations (if any) in $,theta,$? To answer this question, compute $displaystyle,n := frac|theta|2pi,$, rounded to the nearest whole number. If $,theta,$ is positive, then replace $,theta,$ by $,theta - ncdot 2pi,$. If $,theta,$ is negative, then replace $,theta,$ by $,theta + ncdot 2pi,$. Now, your anglenumber is manageable: it is between $,-pi,$ and $,pi,$. But, it has the same terminal point, so all the trigonometric values are the same! REFERENCEQUADRANT: Lay off $,theta,$ in the standard way: start at the positive $,x,$-axis positive angles are swept out in a counterclockwise direction; start by going up negative angles are swept out in a clockwise direction; start by going down Determine the reference anglenumber for $,theta,$. Determine the quadrant for $,theta,$. SIZESIGN: Use the reference anglenumber to find the correct SIZE of the desired trigonometric value. Use the quadrant to find the correct SIGN of the desired trigonometric value. EXAMPLES In this first example, the angles arent too big, so the optional (reduce) step is skipped. Find: $,sec 510^circ,$ Find: $displaystyle,cot (-frac7pi3),$ REDUCE (skipped) Dr. Burns would work with $,510^circ - 360^circ = 150^circ,$. (skipped) Dr. Burns would work with $,-frac7pi3 + frac6pi3 = -fracpi3,$. REFERENCEQUADRANT $,510^circ,$ is in quadrant II; the reference angle is $,30^circ,$ $displaystyle,-frac7pi3,$ is in quadrant IV; the reference angle is $,displaystylefracpi3,$ SIZESIGN SIZE: $displaystylesec 30^circ = frac2sqrt 3,$ SIGN: In quadrant II, the secant is negative. Thus: $displaystyle,sec 510^circ = -frac2sqrt 3,$ SIZE: $displaystylecotfracpi3 = frac1sqrt 3,$ SIGN: In quadrant IV, the cotangent is negative. Thus: $displaystyle,cot(-frac7pi3) = -frac1sqrt 3,$ 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: $displaystyle,cos (frac81pi4),$ REDUCE $displaystyle frac|theta|360^circ = frac2640^circ360^circ approx 7$ $-2640^circ + 7cdot 360^circ = -120^circ$ Work with $,-120^circ,$ instead of $,-2640^circ,$. $displaystyle frac|theta|2pi = frac81pi42pi approx 10$ $displaystylefrac81pi4 - 10cdot 2pi = frac81pi4 - frac80pi4 = fracpi4$ Work with $displaystyle,fracpi4,$ instead of $displaystyle,frac81pi4,$. REFERENCEQUADRANT $,-120^circ,$ is in quadrant III; the reference angle is $,60^circ,$ $,fracpi4,$ is in quadrant I; the reference angle is $,displaystylefracpi4,$ SIZESIGN SIZE: $displaystylecsc 60^circ = frac2sqrt 3,$ SIGN: In quadrant III, the cosecant is negative. Thus: $displaystyle,csc(-2640^circ) = -frac2sqrt 3,$ SIZE: $displaystylecosfracpi4 = frac1sqrt 2,$ SIGN: In quadrant I, the cosine is positive. Thus: $displaystyle,cosfrac81pi4 = frac1sqrt 2,$

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