how to find the exact value of each trigonometric function

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 exist used to solve the questions below.

Let united states assume that nosotros want to find the verbal value of f(10), where f is any of the six trigonometric functions sine, cosine, tangent, cotangent, secant and cosecant. To discover the verbal 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 (- ten) = cos (x) and and then on.
two - Next nosotros locate the final side of the bending in question, directly or using a positive coterminal bending t, which gives the sign of the trigonometric function.
3 - Nosotros 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 constitute in footstep 2. If the angle in question or its coterminal angle are in quadrant i, this terminal step is not needed.

Question 1

Observe the exact value of sin (- Pi / 3).
Solution to Question ane:

Use the identity for negative angles, to write
sin (- Pi / three) = - sin (Pi / 3)
Pi / 3 is in quadrant one and there is no need for either coterminal or reference angles. sin (- Pi / 3) is evaluated straight as follows
sin (- Pi / 3) = - sin (Pi / three) = - sqrt (three) / ii

Question 2

Find the exact value of cos (- 390 ^o ).
Solution to Question 2:

We use the identity cos(-x) = cos(x) to write
cos (- 390 ^o) = cos (390 ^o)
Since 390 ^o is greater than 360 ^o, we notice a coterminal angle t, greater than zero and less than 360 ^o, to 390 ^o.
t = 390 - (360) = 30 ^o
Annotation that since 390 ^o and angle t = 30 are coterminal, we can write
cos (390 ^o) = cos ( thirty ^o )
Finally.
cos(- 330 ^o) = cos ( 330 ^o )
= cos (30) = sqrt (iii) / 2
Note that there was no need for reference bending since xxx ^o is in quadrant 1.

Question iii

Notice the exact value of sec (3 Pi / 4).

Solution to Question 3:

3 Pi / 4 has its terminal side in quadrant 2. In quadrant 2 the secant is negative. Hence
sec (3 Pi / 4) = - sec(Tr)
where Tr is the reference bending to iii Pi / 4 and is given by
Tr = Pi - 3 Pi / 4 = Pi / four
Hence
sec (iii Pi / 4) = - sec(Pi / 4) = - sqrt(two)

Question iv

Detect the exact value of cot ( 840 ^o ).
Solution to Question iv:

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
We can write
cot ( 840 ^o) = cot (120 ^o)
120 ^o is in quadrant ii where the cotangent is negative, hence
cot ( 840 ^o) = cot (120 ^o) = - cot (Tr)
120 ^o is in quadrant 2, hence its reference angle Tr is given past
Tr = 180 - 120 = 60 ^o
Finally
cot ( 840 ^o) = - cot (60 ^o) = - sqrt(3) / 3

Question 5

Find the exact value of csc (- 7 Pi / four).
Solution to Question 5:

Negative angle identity gives
csc (- 7 Pi / four) = - csc ( vii Pi / four )
The concluding angle of vii Pi / 4 is in quadrant four where the cosecant is negative. The reference Tr angle of vii Pi / iv is given past
Tr = 2 Pi - 7 Pi / 4 = Pi / four
Hence
csc ( 7 Pi / iv ) = - csc (Pi / 4) = - sqrt(2)
Finally, substitute the above into csc (- 7 Pi / 4) = - csc ( 7 Pi / 4 ) to obtain
csc (- seven Pi / iv) = sqrt(2)

Question 6

Find the exact value of cot (121 Pi / 3).
Solution to Question 6:

Nosotros kickoff note that
121 Pi / 3 = 120 Pi / 3 + Pi / three
= 40 Pi + Pi / three
A positive coterminal bending t to 121 Pi / 3 may exist calculated every bit follows
t = 121 Pi / three - 20 (ii PI) = 121 Pi / 3 - 40 Pi = Pi / 3
The coterminal angle is in quadrant 1 and at that place is no need for the reference bending. Hence
cot (121 Pi / 3) = cot (Pi / 3) = sqrt (iii) / three

Question 7

Find the exact value of sec ( - 3810 ^o ).
Solution to Question 7:

Utilize negative angle identity
sec ( - 3810 ^o) = sec (3810 ^o)
Note that
3810 ^o = 3600 ^o + 210 ^o
The coterminal angle t to 3810 ^o may exist calculated as follows
t = 3810 ^o - 10(360) ^o = 210 ^o
The terminal side of the coterminal angle t is in quadrant 3 and where therefore sec (3810 ^o) is negative and given past
sec (3810 ^o) = - sec(Tr)
where Tr is the reference bending to angle 210 ^o and is given past
Tr = 210 ^o - 180 ^o = 30 ^o
Hence
sec ( - 3810 ^o) = - sec (30 ^o) = - 2 / sqrt(three)