Exact Values of Trigonometric Functions - Questions With Answers
Find exact values of trigonometric functions without using a calculator. Questions with solutions and answers are presented. The trigonometric identities and formulas in this site may be used to solve the questions below.
Let us assume that we want to find the exact value of f(x), where f is any of the six trigonometric functions sine, cosine, tangent, cotangent, secant and cosecant. To find the exact value of f(x), we suggest the following steps:
1 - If the angle x is negative, we first use a formula for negative angles such as sin (- x) = - sin (x), cos (- x) = cos (x) and so on.
2 - Next we locate the terminal side of the angle in question, directly or using a positive coterminal angle t, which gives the sign of the trigonometric function.
3 - We find the reference angle Tr to the angle in question and use that fact that f(x) = + or - f(Tr). The sign + or - is determined using the quadrant found in step 2. If the angle in question or its coterminal angle are in quadrant 1, this last step is not needed.
Question 1
Find the exact value of sin (- Pi / 3).
Solution to Question 1:
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Use the identity for negative angles, to write
sin (- Pi / 3) = - sin (Pi / 3)
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Pi / 3 is in quadrant 1 and there is no need for either coterminal or reference angles. sin (- Pi / 3) is evaluated directly as follows
sin (- Pi / 3) = - sin (Pi / 3) = - sqrt (3) / 2
Question 2
Find the exact value of cos (- 390 o).
Solution to Question 2:
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We use the identity cos(-x) = cos(x) to write
cos (- 390 o) = cos (390 o)
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Since 390 o is greater than 360 o, we find a coterminal angle t, greater than zero and less than 360 o, to 390 o.
t = 390 - (360) = 30 o
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Note that since 390 o and angle t = 30 are coterminal, we can write
cos (390 o) = cos ( 30 o )
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Finally.
cos(- 330 o) = cos ( 330 o )
= cos (30) = sqrt (3) / 2
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Note that there was no need for reference angle since 30 o is in quadrant 1.
Question 3
Find the exact value of sec (3 Pi / 4).
Solution to Question 3:
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3 Pi / 4 has its terminal side in quadrant 2. In quadrant 2 the secant is negative. Hence
sec (3 Pi / 4) = - sec(Tr)
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where Tr is the reference angle to 3 Pi / 4 and is given by
Tr = Pi - 3 Pi / 4 = Pi / 4
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Hence
sec (3 Pi / 4) = - sec(Pi / 4) = - sqrt(2)
Question 4
Find the exact value of cot ( 840 o).
Solution to Question 4:
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840 o is positive and greater than 360 o, hence the need to first find the coterminal angle t
t = 840 o - 2 (360) o = 120 o
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We can write
cot ( 840 o) = cot (120 o)
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120 o is in quadrant 2 where the cotangent is negative, hence
cot ( 840 o) = cot (120 o) = - cot (Tr)
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120 o is in quadrant 2, hence its reference angle Tr is given by
Tr = 180 - 120 = 60 o
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Finally
cot ( 840 o) = - cot (60 o) = - sqrt(3) / 3
Question 5
Find the exact value of csc (- 7 Pi / 4).
Solution to Question 5:
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Negative angle identity gives
csc (- 7 Pi / 4) = - csc ( 7 Pi / 4 )
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The terminal angle of 7 Pi / 4 is in quadrant 4 where the cosecant is negative. The reference Tr angle of 7 Pi / 4 is given by
Tr = 2 Pi - 7 Pi / 4 = Pi / 4
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Hence
csc ( 7 Pi / 4 ) = - csc (Pi / 4) = - sqrt(2)
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Finally, substitute the above into csc (- 7 Pi / 4) = - csc ( 7 Pi / 4 ) to obtain
csc (- 7 Pi / 4) = sqrt(2)
Question 6
Find the exact value of cot (121 Pi / 3).
Solution to Question 6:
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We first note that
121 Pi / 3 = 120 Pi / 3 + Pi / 3
= 40 Pi + Pi / 3
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A positive coterminal angle t to 121 Pi / 3 may be calculated as follows
t = 121 Pi / 3 - 20 (2 PI) = 121 Pi / 3 - 40 Pi = Pi / 3
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The coterminal angle is in quadrant 1 and there is no need for the reference angle. Hence
cot (121 Pi / 3) = cot (Pi / 3) = sqrt (3) / 3
Question 7
Find the exact value of sec ( - 3810 o).
Solution to Question 7:
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Use negative angle identity
sec ( - 3810 o) = sec (3810 o)
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Note that
3810 o = 3600 o + 210 o
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The coterminal angle t to 3810 o may be calculated as follows
t = 3810 o - 10(360) o = 210 o
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The terminal side of the coterminal angle t is in quadrant 3 and where therefore sec (3810 o) is negative and given by
sec (3810 o) = - sec(Tr)
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where Tr is the reference angle to angle 210 o and is given by
Tr = 210 o - 180 o = 30 o
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Hence
sec ( - 3810 o) = - sec (30 o) = - 2 / sqrt(3)
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