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Given trigonometric functions of a special angle, evaluate using side lengths.

  1. Use the side lengths shown in [link] for the special angle you wish to evaluate.
  2. Use the ratio of side lengths appropriate to the function you wish to evaluate.

Evaluating trigonometric functions of special angles using side lengths

Find the exact value of the trigonometric functions of π 3 , using side lengths.

sin ( π 3 ) = opp hyp = 3 s 2 s = 3 2 cos ( π 3 ) = adj hyp = s 2 s = 1 2 tan ( π 3 ) = opp adj = 3 s s = 3 sec ( π 3 ) = hyp adj = 2 s s = 2 csc ( π 3 ) = hyp opp = 2 s 3 s = 2 3 = 2 3 3 cot ( π 3 ) = adj opp = s 3 s = 1 3 = 3 3
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Find the exact value of the trigonometric functions of π 4 , using side lengths.

sin ( π 4 ) = 2 2 , cos ( π 4 ) = 2 2 , tan ( π 4 ) = 1 ,
sec ( π 4 ) = 2 , c s c ( π 4 ) = 2 , cot ( π 4 ) = 1

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Using equal cofunction of complements

If we look more closely at the relationship between the sine and cosine of the special angles relative to the unit circle, we will notice a pattern. In a right triangle with angles of π 6 and π 3 , we see that the sine of π 3 , namely 3 2 , is also the cosine of π 6 , while the sine of π 6 , namely 1 2 , is also the cosine of π 3 .

sin π 3 = cos π 6 = 3 s 2 s = 3 2 sin π 6 = cos π 3 = s 2 s = 1 2

See [link]

A graph of circle with angle pi/3 inscribed.
The sine of π 3 equals the cosine of π 6 and vice versa.

This result should not be surprising because, as we see from [link] , the side opposite the angle of π 3 is also the side adjacent to π 6 , so sin ( π 3 ) and cos ( π 6 ) are exactly the same ratio of the same two sides, 3 s and 2 s . Similarly, cos ( π 3 ) and sin ( π 6 ) are also the same ratio using the same two sides, s and 2 s .

The interrelationship between the sines and cosines of π 6 and π 3 also holds for the two acute angles in any right triangle, since in every case, the ratio of the same two sides would constitute the sine of one angle and the cosine of the other. Since the three angles of a triangle add to π , and the right angle is π 2 , the remaining two angles must also add up to π 2 . That means that a right triangle can be formed with any two angles that add to π 2 —in other words, any two complementary angles. So we may state a cofunction identity : If any two angles are complementary, the sine of one is the cosine of the other, and vice versa. This identity is illustrated in [link] .

Right triangle with angles alpha and beta. Equivalence between sin alpha and cos beta. Equivalence between sin beta and cos alpha.
Cofunction identity of sine and cosine of complementary angles

Using this identity, we can state without calculating, for instance, that the sine of π 12 equals the cosine of 5 π 12 , and that the sine of 5 π 12 equals the cosine of π 12 . We can also state that if, for a certain angle t , cos   t = 5 13 , then sin ( π 2 t ) = 5 13 as well.

Cofunction identities

The cofunction identities in radians are listed in [link] .

cos t = sin ( π 2 t ) sin t = cos ( π 2 t )
tan t = cot ( π 2 t ) cot t = tan ( π 2 t )
sec t = csc ( π 2 t ) csc t = sec ( π 2 t )

Given the sine and cosine of an angle, find the sine or cosine of its complement.

  1. To find the sine of the complementary angle, find the cosine of the original angle.
  2. To find the cosine of the complementary angle, find the sine of the original angle.

Using cofunction identities

If sin t = 5 12 , find ( cos π 2 t ) .

According to the cofunction identities for sine and cosine,

sin t = cos ( π 2 t ) .

So

cos ( π 2 t ) = 5 12 .
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If csc ( π 6 ) = 2 , find sec ( π 3 ) .

2

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Questions & Answers

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Noor Reply
If auger is pair are the roots of equation x2+5x-3=0
Peter Reply
Wayne and Dennis like to ride the bike path from Riverside Park to the beach. Dennis’s speed is seven miles per hour faster than Wayne’s speed, so it takes Wayne 2 hours to ride to the beach while it takes Dennis 1.5 hours for the ride. Find the speed of both bikers.
MATTHEW Reply
420
Sharon
from theory: distance [miles] = speed [mph] × time [hours] info #1 speed_Dennis × 1.5 = speed_Wayne × 2 => speed_Wayne = 0.75 × speed_Dennis (i) info #2 speed_Dennis = speed_Wayne + 7 [mph] (ii) use (i) in (ii) => [...] speed_Dennis = 28 mph speed_Wayne = 21 mph
George
Let W be Wayne's speed in miles per hour and D be Dennis's speed in miles per hour. We know that W + 7 = D and W * 2 = D * 1.5. Substituting the first equation into the second: W * 2 = (W + 7) * 1.5 W * 2 = W * 1.5 + 7 * 1.5 0.5 * W = 7 * 1.5 W = 7 * 3 or 21 W is 21 D = W + 7 D = 21 + 7 D = 28
Salma
Devon is 32 32​​ years older than his son, Milan. The sum of both their ages is 54 54​. Using the variables d d​ and m m​ to represent the ages of Devon and Milan, respectively, write a system of equations to describe this situation. Enter the equations below, separated by a comma.
Aaron Reply
find product (-6m+6) ( 3m²+4m-3)
SIMRAN Reply
-42m²+60m-18
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-24m+3+3mÁ^2
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-6m(3mA²+4m-3)+6(3mA²+4m-3) =-18m²A²-24m²+18m+18mA²+24m-18 Rearrange like items -18m²A²-24m²+42m+18A²-18
Salma
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x=3-2y
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y=x+3/2
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given that (7x-5):(2+4x)=8:7find the value of x
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3x-12y=18
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A meditation garden is in the shape of a right triangle, with one leg 7 feet. The length of the hypotenuse is one more than the length of one of the other legs. Find the lengths of the hypotenuse and the other leg
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The Jones family took a 15 mile canoe ride down the Indian River in three hours. After lunch, the return trip back up the river took five hours. Find the rate, in mph, of the canoe in still water and the rate of the current.
cameron Reply
Shakir works at a computer store. His weekly pay will be either a fixed amount, $925, or $500 plus 12% of his total sales. How much should his total sales be for his variable pay option to exceed the fixed amount of $925.
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I'm guessing, but it's somewhere around $4335.00 I think
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12% of sales will need to exceed 925 - 500, or 425 to exceed fixed amount option. What amount of sales does that equal? 425 ÷ (12÷100) = 3541.67. So the answer is sales greater than 3541.67. Check: Sales = 3542 Commission 12%=425.04 Pay = 500 + 425.04 = 925.04. 925.04 > 925.00
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Jazmine trained for 3 hours on Saturday. She ran 8 miles and then biked 24 miles. Her biking speed is 4 mph faster than her running speed. What is her running speed?
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Source:  OpenStax, Precalculus. OpenStax CNX. Jan 19, 2016 Download for free at https://legacy.cnx.org/content/col11667/1.6
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