



Trigonometry: Trigonometric Functions 
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Trigonometric Functions
In this lesson you'll learn the fundamentals and how to determine the value of Trigonometric Functions. E.g. you are asked to find the angle, if the value of a trigonometric ratio is given. Illustrating it with an example, find the value of , if tan = 0.38386. Using the calculator or Trig Table, you'll find that = tan^{1} 0.38386 = 21. You can check if this is correct, tan 21 equals to 0.38386.
Notice that tan 201 is also 0.38386. Similarly tan 381 and some negative value like tan ( 159) shall also equal to 0.38386.
Periodic Functions
You know that the trigonometric functions are periodic. It means that their values reoccur over and over again. In other words there are many values of , as you have seen in the above equation tan = 0.38386. Actually the number is infinite i.e. it has infinite number of solutions.
Notice that how the graph of y = tan looks like. You'll see a regular pattern that repeats every 180 (Figure: 1). There are also "gaps" in the curve (at 90, 270 etc) where tan has no value. (You may check it your calculator  what it displays if you try to find the value of tan 90, say.)







In figure: 2 below, you'll notice that if you draw line which shows the yvalue
as 0.38386.Then drop the graph intersection for this value down on to the axis. It is observed that tan 21 = 0.38386 and also other values can be seen. E.g. tan 201 is 0.3838, tan 381 and some negative value like tan( 159) shall also are 0.3838.
Determining all such angles:
In order to find method such that you can find all the values of that result in a given trigonometric equation, one needs to understand the reference angle concept and applying it to solve the problems.
Reference Angle:
You may learn about reference angles taking an example of an angle (> 90), which will have a corresponding acute angle . This corresponding angle is called Reference
Angle.
Note that in Quadrant II, can be written as = (180  ). (Refer Fig: 3). E.g. say in second quadrant = 132. The reference angle thus measures = 48.
In furtherance for Quadrant III, it may be written as = (180 + ). (Refer Fig: 4)
For example, in third quadrant = 235 = (180 + 55). Here reference angle () measures 55.
Next is Quadrant (IV), where it will be in the form of = (360  ). (Refer Fig: 5)
It can be understood taking the example, say in fourth quadrant = 320 = (360  40). Here reference angle () measures 40.
Remember: Notice that for angles greater than 2 radians, subtract 2 from them, and then use the steps above to calculate the accompanying reference angle. Once you become familiar with the values of certain trigonometric functions at certain common angles e.g. /6, /4 etc, you will be able to use reference angles to work out the values of these functions at an infinite number of other angles.
Here is an illustration to explain it further: Given find angles whose cosine is 0.7.
To solve it, follow the steps given below:
From the earlier learning, you know that there may be an infinite number of answers to this problem and you may choice to pick up the angles whose cosine is 0.7.
First step is to use the calculator or Trig Tables, you get one answer say: cos10.7 = 45.57. You'll notice that this is an acute angle, you can use it as the reference angle and write = 45.57.
Now proceed to get another answer. Recall from the trigonometric function table/representative graph that the two places where cosine is positive are: Quadrant I and IV. The corresponding angle in the fourth quadrant will be:
360  = 360  45.57 = 314.43. You can verify it now i.e. cos 45.57 = 0.7 and cos 314.43 = 0.7
It may be noted that there are an infinite number of angles where the cosine of the angle is 0.7. See carefully in Fig: 6, (where these values are appearing i.e. the values where the blue arrows are pointing).
Remember to use preferably 'arcsin' instead of what you see in on your calculator i.e. 'sin^{1}'?' Otherwise may get confused with sin^{1}a
and csc a, as they are not the same (sin^{1} a means 'the angle whose sine is a', whereas csc a means '1/sin a').
So you have seen above how to solve for angle. If it is a multiple of /6 (30) or /4 (45), easy to solve, otherwise you need to write the solution as an arcfunction.
While learning further about Trig equations, you may have seen one important difference from other types of equations i.e. Trig functions is periodic. It means that they repeat their values over and over. Therefore a trig equation has an infinite number of solutions as seen above.
Just think of an equation like sin a = 1, for which /2 is a solution, but the sine function repeats its values every 2. Therefore /2 2, /2 4, and so on are also good solutions. Now generalize and write it: a = /2 + 2n,
where n is any integer, positive, negative, or zero.
Remember: the tangent and cotangent functions repeat all their values every radians, so the solution to tan b = 1 is b = /4 + n,
and not +2n.
Suppose you are to find solution to the equation: sec (3a) = 2.5 i.e. what are the possible values of the angle 3a?
You may find it difficult to solve it for the secant function. But you know that
1/sec (3a) = cos((3a)). By rewriting the equation you get: 1/sec((3a))
= 1/2.5 or cos((3a)) = 0.4
Now you can find for what angles it holds true?
When you write arccos(0.4), it means the angle is in Quadrant I for which cosine equals to 0.4. In some references you may find it as cos^{1}(0.4) in place of arccos(0.4). You should prefer to use arccos notation for reasons given in the earlier illustration.
Initially it can be written as (3a) = arccos(0.4) + 2n.
You know that any angle in Quadrant I has a reflection in quadrant IV with the same cosine value. Therefore, you need to consider both the angles:
(3a) = arccos(0.4) + 2n
or alternatively, 2  arccos(0.4) + 2n






