Chapter 8
Calculation of The Trig. Ratios of Any Angle
Up to about mid 20th Century, the calculation of sine, cosine, tangent, and cotangent of angles from 0 to 90° were based on Trigonometric Tables. Such Tables had the trig ratios calculated in them at 1 degree intervals or smaller. The user would then interpolate between these values to get the trig ratio of other angles to decimal accuracies. For example, if it was desired to find sin36.7°, the user would locate the values for sin36° and sin37° in the Table, first. The user would then add 0.7 of the difference between those two values to the value of sin36° to get the numerical value for sin36.7°.
After the invention of scientific calculators, the tedious task of finding the trig ratios of angles other than 0, 30°, 45°, 60°, and 90° became very easy. Today, we just type the angle and then press the desired trig ratio or vice versa depending on the type of calculator.
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Exercises:
Find the trig ratios of the following angles:
1) a = 43°, 2) b = 2.9 radians, 3) c = 10π/9 radians, 4) d = 3981°,
5) e = 3,070 grads, 6) f = 3.9π radians, 7) g = 135.45°.
Calculation of an Angle from Its Trig Ratio (Inverse Trig Function)
Sometimes the trig ratio of an angle is known and we want to find the angle itself. Suppose we are given sinC=1/2 and we are asked to find angle C. The answer in this case is easy. The smallest possible angle is 30°; therefore, C = 30° or C = π/6.
The next possible solution (angle) is π  30° or π  π/6 = 5π/6 or 150°.
This is because sin(π  30°) = sin(30°) = 1/2. This way we can get infinite number of solutions (answers) by adding turns or (2π)'s to the angle.
We may say that " the angle which sine is 1/2 is 30°." We then write C = 30°.
This statement may be shown as Arcsin(1/2) = 30° that means: the arc that has a sine equal to 1/2 is 30°.
This can even be written simpler as sin^{1}(1/2) = 30°. "Arcsin" and sin^{1} mean the same thing.
sin^{1} is called "inverse sine." Similarly,
cos^{1} is called "inverse cosine,"
tan^{1} is called "inverse tangent," and
cot^{1} is called "inverse cotangent."
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Example 1: Write the following arcsin, arccos, ... in the simpler form of sin^{1},cos^{1}, ...
and find angles: x, y, z, and θ assuming that they are all less than 90°.
i) Arctang(1) = x, ii) Arcsin(√2/2) = y, iii) Arccos(1/2) = z, iv) Arccot(√3) = θ.
Solution:
i) Arctan(1) = x or, tan^{1}(1) = x → x = 45° because tan45° = 1.
ii) Arcsin(√2/2) = y or, sin^{1}(√2/2) = y → y = 45° because sin45° = √2/2.
iii) Arccos(1/2) = z or, cos^{1}(1/2) = z → z = 60° because cos60° = 1/2.
iv) Arccot(√3) = θ or, cot^{1}(√3) = θ → θ = 30° because cot30° = √3.
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Example 2: Write the following trig equalities in the form of inverse trig equalities:
a) tan60°=√3, b) cos60°=1/2, c) sin90°=1, and d) cot45°=1.
Solution:
a) tan(π/3) = √3 → π/3 = tan^{1}(√3)
b) cos(π/3) = 1/2 → π/3 = cos^{1}(1/2)
c) sin(π/2) = 1 → π/2 = sin^{1}(1)
d) cot(π/4) = 1 → π/4 = cot^{1}(1)
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Example 3: Repeat Example 2 by imposing the appropriate inverse function on both sides:
Solution: Part (a) is solved below. Other parts are left for students.
a) tan(π/3) = √3. Now, let's take the tan^{1} of both sides as shown:
tan^{1} [tan(π/3)] = tan^{1}[√3]
On the left side, tan^{1} undoes what tan does; in other words,
the inverse tangent of tangent of π/3 is π/3 itself. We may write:
π/3 = tan^{1}[√3].
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Exercises:
2) Assuming that all angles are acute (less than 90°), find each angle from its given trig ratio. Use a calculator, only if the result is an angle other than 0°, 30°, 45°, 60°, or 90°.
sinA = 1/2, cosB = √2/2, tan C = √3, tanD = 1,
sinx = √3/2, cosy = 1/2, tanz = √3/3, sinA = 0.545
cosB = 0.927, tanC = 2.100, cotD = 5.237,
4cosB = 7/5, and 1/sinA = 2.5.
Calculation of all Possible Angles
Recall our first example of sinC = 1/2. If the problem states that angle C is less than 90°, there is only one possible answer to the problem and that is C = 30° or C = π/6.
If the problem does not state anything about where the angle ends, we need to point out ALL POSSIBLE SOLUTIONS.
(I) All Possible Angles When
sin (θ) = β
Anytime the sine of an angle like θ equals a number like β, there are two possible solution sets each of which contains an infinite number of angles. Each of such angles has a sine equal to β. The two possible solution sets are:
The result of sin (θ) = β is the following: 
either k(2π) + θ where k = 0, 1,2,3,...,∞ (1) or, k(2π) + (π  θ) where k = 0, 1,2,3,...,∞ (2) Note that sin (π  θ) = sinθ. Recall that in the 2nd Quad., sine is positive. 
Let's examine a few angles and give k values of 0, 1, 2, and 3, for example.
If k = 0, (1) gives us 0 + θ, and (2) gives us (π  θ). We already know sin(π  θ) = sinθ.
If k = 1, (1) gives us 2π +θ, and (2) gives us 2π +(π  θ).
If k = 2, (1) gives us 4π +θ, and (2) gives us 4π +(π  θ).
If k = 3, (1) gives us 6π +θ, and (2) gives us 6π +(π  θ), and so forth.....
The above 8 angles have a sine value equal to β as well because they either end at where θ ends or end where π  θ ends.
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Example 4: If sin(θ) = 1/2, find all possible θ's.
We already know that sin30° =1/2, or sin (π/6) = 1/2; therefore, θ = π/6. The two solution sets are therefore: 
either k(2π) + π/6 where k = 0, 1,2,3,...,∞ (3) k(2π) + π/6 or, k(2π) + (π  π/6) where k = 0, 1,2,3,...,∞ (4) k(2π) + 5π/6 Note that (π  π/6) is like (6π/6  π/6) that equals 5π/6. 
If k = 0, (3) gives us 0 + π/6, and (4) gives us 0 + (π  π/6)
If k = 1, (3) gives us 2π +π/6, and (4) gives us 2π + (π  π/6)
If k = 2, (3) gives us 4π +π/6, and (4) gives us 4π + (π  π/6)
If k = 3, (3) gives us 6π +π/6, and (4) gives us 6π + (π  π/6), and so forth... .
The above 8 angles have a sine value equal to 1/2 as well because they either end at where π/6 ends or end where 5π/6 ends.
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Example 5: If sin(θ) = √2/2, find all possible θ's.
We already know that sin45° =√2/2, or sin (π/4) = √2/2; therefore, θ = π/4. The two solution sets are therefore: 
either k(2π) + π/4 where k = 0, 1,2,3,...,∞ (5) k(2π) + π/4 or, k(2π) + (π  π/4) where k = 0, 1,2,3,...,∞ (6) k(2π) + 3π/4 Note that (π  π/4) is like (4π/4  π/4) that equals 3π/4. 
If k = 0, (5) gives us 0 +π/4, and (6) gives us 0 +(π  π/4)
If k = 1, (5) gives us 2π +π/4, and (6) gives us 2π +(π  π/4)
If k = 2, (5) gives us 4π +π/4, and (6) gives us 4π +(π  π/4)
If k = 3, (5) gives us 6π +π/4, and (6) gives us 6π +(π  π/4), and so forth... .
The above 8 angles have a sine value equal to √2/2 as well because they either end at where π/4 ends or end where 3π/4 ends.
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Example 6: If sinθ = 0.76, find all possible θ's. Suppose you do not have a calculator and want to use the sin^{1} function to express your answers.
Since sinθ =
0.76,
imposing sin^{1}
to both sides of it, results in: sin^{1}[sinθ] = sin^{1}0.76, or θ = sin^{1}0.76 The two solution sets are therefore: 
Either k(2π) + sin^{}^{1}0.76 where k = 0, 1,2,3,...,∞ (7) or, k(2π) + π  sin^{}^{1}0.76 where k = 0, 1,2,3,...,∞ (8)

If k = 0, (7) gives us 0 + sin^{}^{1}0.76, and (8) gives us 0 + (π  sin^{}^{1}0.76)
If k = 1, (7) gives us 2π + sin^{}^{1}0.76, and (8) gives us 2π +(π  sin^{}^{1}0.76)
If k = 2, (7) gives us 4π + sin^{}^{1}0.76, and (8) gives us 4π +(π  sin^{}^{1}0.76)
If k = 3, (7) gives us 6π + sin^{}^{1}0.76, and (8) gives us 6π +(π  sin^{}^{1}0.76), and so on... .
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(II) All Possible Angles For
cos(θ) = β
Anytime the cosine of an angle like θ is equal to a number like β, there are two solution sets that each contains infinite angles the cosine of each equals β. These two solution sets are:
The result of cos(θ) = β is the following: 
either k(2π) + θ where k = 0, 1,2,3,...,∞ (9) or, k(2π)  θ where k = 0, 1,2,3,...,∞ (10) The COMPACT FORM of the above two is : k(2π) ± θ. In the 4th Quadrant, cosine is positive, and cos(θ) = cosθ. 
Let's examine a few angles and give k values of 0, 1, 2, and 3, for example.
If k = 0, (9) gives us 0 + θ, and (10) gives us 0  θ. We already know cos(θ) = cosθ.
If k = 1, (9) gives us 2π +θ, and (10) gives us 2π  θ.
If k = 2, (9) gives us 4π +θ, and (10) gives us 4π  θ.
If k = 3, (9) gives us 6π +θ, and (10) gives us 6π  θ, and so forth... .
The above 8 angles have the same cosine value each equal to β.
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Example 7: If cos(θ) = 1/2, find all possible θ's.
We already know that cos60° =1/2, or cos(π/3) = 1/2; therefore, θ = π/3. The two solution sets are therefore: 
either k(2π) + π/3 where k = 0, 1,2,3,...,∞ (11) or, k(2π)  π/3 where k = 0, 1,2,3,...,∞ (12) The COMPACT FORM of the above two is : k(2π) ± π/3. Note that cos(π/3) = cos(π/3). 
If k = 0, (11) gives us 0 + π/3, and (12) gives us 0  π/3
If k = 1, (11) gives us 2π +π/3, and (12) gives us 2π  π/3
If k = 2, (11) gives us 4π +π/3, and (12) gives us 4π  π/3
If k = 3, (11) gives us 6π +π/3, and (12) gives us 6π  π/3, and so forth... .
The cosine of each of the above 8 angles is 1/2.
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Example 8: If cosθ = 0.31, find all possible θ's. Suppose you do not have a calculator and want to use the cos^{1} function to express your solution(s).
Solution:
Since cosθ =
0.31,
imposing cos^{1}
to both sides of it, results in: cos^{1}[cosθ] = cos^{}1(0.31) or θ = cos^{1}(0.31) The two solution sets are therefore: 
Either k(2π) + cos^{}^{1}(0.31) where k = 0, 1,2,3,...,∞ (13) or, k(2π)  cos^{}^{1}(0.31) where k = 0, 1,2,3,...,∞ (14) The COMPACT FORM of the above two is : k(2π) ± cos^{}^{1}(0.31). 
If k = 0, (13) gives us 0 + cos^{}^{1}(0.31), and (14) gives us 0  cos^{}^{1}(0.31).
If k = 1, (13) gives us 2π + cos^{}^{1}(0.31), and (14) gives us 2π  cos^{}^{1}(0.31).
If k = 2, (13) gives us 4π + cos^{}^{1}(0.31), and (14) gives us 4π  cos^{}^{1}(0.31).
If k = 3, (13) gives us 6π + cos^{}^{1}(0.31), and (14) gives us 6π  cos^{}^{1}(0.31), and so on ....
The above 8 angles have the same cosine the value each equals 0.31.
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(III) All Possible Angles For
tan(θ) = β
Anytime the tangent of an angle like θ equals a number like β, there is only one solution set that contains infinite angles the tangent of each equals β. The solution set is:
The result of tan(θ) = β is the following: 
k(π) + θ where k = 0, 1,2,3,...,∞ (15) In the 3rd Quadrant, tangent is positive, and tan (π + θ) = tanθ. 
Let's examine a few angles and give k values of 0, 1, 2, and 3, for example.
If k = 0, (15) gives us 0 + θ.
If k = 1, (15) gives us π + θ. We already know tan (π + θ) = tanθ.
If k = 2, (15) gives us 2π +θ.
If k = 3, (15) gives us 3π +θ, and so forth.....
The above 4 angles have the same tangent value each of which equals β.
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Example 9: If tan(θ) = 1, find all possible θ's.
We already know that tan45° =1, or tan(π/4) = 1; therefore, θ = π/4. The solution set is therefore: 
k(π) + π/4 where k = 0, 1,2,3,...,∞ (16) Note that tan(π + π/4) = tan(π/4). 
If k = 0, (16) gives us 0 + π/4.
If k = 1, (16) gives us 1π +π/4.
If k = 2, (16) gives us 2π +π/4.
If k = 3, (16) gives us 3π +π/4, and so on ... .
The tangent of each of the above 4 angles is 1.
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Example 10: If tanθ = 4.3, find all possible θ's. Suppose you do not have a calculator and want to use the tan^{}1 function to express your answers.
Solution:
Since tanθ =
4.3,
imposing tan^{1}
to both sides of it, results in: tan^{1}[tanθ] = tan^{1}4.3 or θ = tan^{1}4.3 The solution set is therefore: 
kπ + tan^{}^{1}4.3 where k = 0, 1,2,3,...,∞ (17)

If k = 0, (17) gives us 0 + tan^{}^{1}4.3.
If k = 1, (17) gives us 1π + tan^{}^{1}4.3.
If k = 2, (17) gives us 2π + tan^{}^{1}4.3.
If k = 3, (17) gives us 3π + tan^{}^{1}4.3.
The above 4 angles have the same tangent value each of which equals 4.3.
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(IV) All Possible Angles For
cot(θ) = β
Anytime the cotangent of an angle like θ equals a number like β, there is only one solution set that contains infinite angles the cotangent of each equals β. The solution set is:
The result of cot(θ) = β is the following: 
k(π) + θ where k = 0, 1,2,3,...,∞ (18) In the 3rd Quadrant, cotangent is positive, and cot (π + θ) = cotθ. 
Let's examine a few angles and give k values of 0, 1, 2, and 3, for example.
If k = 0, (18) gives us 0 + θ.
If k = 1, (18) gives us π + θ. We already know cot (π + θ) = cotθ.
If k = 2, (18) gives us 2π +θ.
If k = 3, (18) gives us 3π +θ, and so forth... .
The above 4 angles have the same cotangent value each of which equals β.
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Example 11: If cot(θ) = 1, find all possible θ's.
We already know that cot45° =1, or cot(π/4) = 1; therefore, θ = π/4. The solution set is therefore: 
kπ + π/4 where k = 0, 1,2,3,...,∞ (19) Note that cot(π + π/4) = cot(π/4). 
If k = 0, (19) gives us 0 + π/4.
If k = 1, (19) gives us 1π +π/4.
If k = 2, (19) gives us 2π +π/4.
If k = 3, (19) gives us 3π +π/4, and so forth ... .
The cotangent of each of the above 4 angles is 1.
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Example 12: If cotθ = 9.6, find all possible θ's. Suppose you do not have a calculator and want to use the cot^{}^{1} function to express your solution (s).
Solution:
Since cotθ =
9.6,
imposing cot^{1}
to both sides of it, results in: cot^{1}[cotθ] = cot^{1}9.6 or θ = cot^{1}9.6 The solution set is therefore: 
kπ + tan^{}^{1}9.6 where k = 0, 1,2,3,...,∞ (20)

If k = 0, (20) gives us 0 + cot^{}^{1}9.6.
If k = 1, (20) gives us 1π + cot^{}^{1}9.6.
If k = 2, (20) gives us 2π + cot^{}^{1}9.6.
If k = 3, (20) gives us 3π + cot^{}^{1}9.6.
The above 4 angles have the same cotangent value each of which equals 9.6.
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Exercises:
Find all possible angles that make each of the following equalities true. Some may require two solution sets and some just one set. Use a calculator only if the result is an angle other than 0°, 30°, 45°, 60°, or 90°.
8) sinA = 1/2, 9) cosB = √2/2, 10) tan C = √3, 11) cot D = 1,
12) sinx = √3/2, 13) cosy = 1/2, 14) tanz = √3/3, 15) sinA = 0.545,
16) cosB = 0.927, 17) tan C = 2.100, 18) cot D = 5.237,
19) 4cos B = 7/5, and 20) 1/sinA = 2.5 .