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Trigonometric Functions
What is Trigonometry?
Define the six Trigonometric functions
How do you find the six Trigonometric Functions of an angle?
Trigonometric Function (or ratio) for 45° angle
Trigonometric Function for 30° & 60° angle
Trigonometric Functions (or ratios) for 0° and 90°
Which trignometry functions are even or odd?
What are sine and cosine?
Why is the cosine function even?
How to find the trigonometric functions of an angle of 18°?
The word trigonometry is derived from the three greek words
In the ancient sense trigonometry defines relation between elements of a triangle. In a triangle there are six basic elements, three sides and three angles. Any three lines segments will form a triangle iff they satisfy three triangular inequalities i.e. the sum of any two lines segment is greater than third side. In Euclidean geometry the sum of three angles of a triangle is 180°. These requirements impose limitations on the manner in which the relations between the elements are defined.
In trigonometry, we need to evaluate six basic trigonometric functions. They are named as - sine, cosine, tangent, cosecant, secant, and cotangent – for a single angle in a right angled triangle. They are also called as – Sin, Cos, Tan, Cosec, Sec and Cot respectively in short.
In a right angled triangle, we have base (or adjacent) (B), perpendicular (or opposite) (P) and hypotenuse (H). So on the basis of these sides, trigonometric functions are defined as follows:
Note: - sec, cosec and cot functions are reciprocal of cos, sin and tan respectively. For Example - if sin θ = 1/2 than cosece θ = 2, similarly for cos & sec and tan & cot functions.
In Basic trigonometry few angles like 0°,3°, 45°^{, }60° & 90° are considered as very basic angles. So we first here will try to learn finding the values of trigonometric functions for these angles. Between 0° and 90° is considered as the first quadrant. So these means here, we are only considering the first quadrant only.
Consider a right angled triangle ABC right angled at B (shown in figure). Angle BAC is θ. Assuming θ = 45°^{, }than angle ACB will also be equals to 45°. Thus, AB = BC. So by using phythagorus theorem, we can easily find that, AB : BC : AC = 1 : 1: √2.
Again, consider an equilateral triangle ABC. From vertex A, drop a perpendicular bisector on BC.
Now Triangle BAD and DAC will be congruent by RHS congruency.
So ∠BAD = ∠CAD = 30°
Now, in ∆ADB right angled at D.
BD = 1/2 AB = a (let) or BD : AB = 1 : 2
Now using phythogorus theorem in right triangle ADB, we get AD = a√3.
Also,
Consider a right angled triangle ABC right angled at B. Let us understand what happens if the ∠A tends to zero. From the below figure we can easily understand that, if ∠A is made zero, Side AC (Hypotenuse) will become equals to AB (Base) and side BC (perpendicular) will become zero.
Now, let’s take the situation where the angle A tends to become right angle.
See the below figure to understand the situation graphically. If, ∠A tends to approach 90°, Side AC (Hypotenuse) will become equals to BC (perpendicular) and side AB (base) will become zero.
Thus we can tabulate all these values for different angles of trigonometric functions.
*Note: In the above table n.d. represents not defined or infinity.
Functions are called as even functions iff
f(-x) = f(x)
Graphically, the functions are even iff they are symmetric about the y – axis.
In Trigonometry, cos (-x) = cos (x), sec (-x) = sec (x). And Thus cosine and secant functions are even functions. We may also look their graph to understand the same.
Functions are called as odd functions iff
f(-x) = – f(x)
Graphically, the functions are even iff they are odd symmetric about the y – axis.
In Trigonometry, sin (-x) = -sin (x), tan (-x) = -tan (x), cosec (x) = -cosec (x), cot (-x) = -cot (x) And Thus sine, tangent, cosecant and cotangent functions are odd functions. We may also look their graph to understand the same.
Sine function, in trigonometry, is defined for a right angled triangle for any acute angle α.
Mathematically, sine of any angle α is the ratio of length of perpendicular (or opposite) side to the length of hypotenuse.
Since the angle α can be any value between o to 360. And in general 0 + 2nπ ≤ α < 360 + 2nπ, where, n can be any integer value. And thus the value of sine repeats after every fixed interval of 2π.
Below is the graphical representation of sine function.
From the graph, It can be easily interpreted that, sine function is bounded function which means its value is also bounded between two fixed constants i.e. -1 and 1.
So, -1 ≤ sin α ≤1
Also, sin 0 = 0 & sin 90° = 1
Similarly, cosine functions are defined as the ratio of length of base to the length of hypotenuse side.
Since the length of hypotenuse is the greatest in any right angled triangle and thus the ratio of base to hypotenuse can never by greater than 1. Thus cosine function is also bounded between -1 and 1.
So, -1 ≤ cos α ≤1
Cosine function is also periodic function with period 2π. Below is the graph of cosine function
Also note from the figure that, cos0 =1 & cos90°^{ }= 0
Cosine functions are symmetric function symmetric about y – axis, that is
Thus from the definition of even function
In Trigonometry, cosine and secant functions are even functions.
Basic Trigonometric Identities:
sin^{2}θ + cos^{2}θ =1; where θ ∈R
1+ tan^{2}θ = sec^{2}θ or sec^{2}θ – tan^{2}θ =1
1+cot^{2}θ = cosec^{2}θ or cosec^{2}θ – cot^{2}θ =1
sin^{4}θ + cos^{4}θ = 1 - 2 sin^{2}θ cos^{2}θ
sin^{6}θ + cos^{6}θ =1 - 3 sin^{2}θ cos^{2}θ
Note:
(secθ – tanθ) is reciprocal of (secθ + tanθ) and vice versa
(cosecθ – cotθ) is reciprocal of (cosecθ + cotθ) and vice versa
Using above identities thousands of other identities can be proved. While proving identities you can use rationalization, factorization and many other similar mathematical operations.
Let θ stands for 18°, so that 2θ = 36° & 3θ = 54°.
Hence, 2θ = 90°^{ }- 3θ
and therefore
sin 2θ =sin (90°- 3θ ) =cos3θ
∴ 2sinθ cosθ = 4 cos^{2}θ – 3cosθ
Hence, either cosθ = 0, which gives θ = 90°, or
2 sin θ = 4 cos^{2}θ – 3 =1 – 4 sin^{2}θ
∴ 4 sin^{2}θ + 2sinθ =1
By solving this quadratic equation, we have
(In our case sin θ is necessarily a positive quantity. Hence we take the upper sign)
The remaining trigonometric ratio of 18° may be now found.
Since 72° is the complement of 18°, the value of the ratios for 72° may be obtained.
To find the trigonometric functions of an angle of 36°
Since cos2θ =1 – 2 sin^{2}θ,
The remaining trigonometric functions of 36° may now be found.
Also, 54° may be found.
For more conceptual clarity on periodicity and graphical representation please refer the video
Illustration 1:
Find all trigonometric functions of an angle in the third quadrant for which Solution: Before proceeding towards the solution, draw the figure so as to have a clear picture of the question. As in the Figure, cosθ= x/R = -5/6 and also we may consider x = -5 and R = 6. Since, x^{2 }+ y^{2 }= 6^{2 }= R^{2 }= 36 we find that y = -√R^{2 }- x^{2 }= -√11 Note that the negative signs indicate the third quadrant. And hence all the rest ratios follow
Illustration 2: Evaluate the value of the expression
3[sin^{4}(3π/2 - α) + sin^{4}(3π + α)] – 2 [sin^{6}(π/2 + α) + sin^{6}(5π - α)].
Solution: The given expression is
= 3[(cos^{4}α + sin^{4} α) – 2(cos^{6}α + sin^{6} α)]
= 3 [(1-2sin^{2}α cos^{2}α) -2(1-3sin^{2}α cos^{2}α)] = 1.
Illustration 3: Find the value of
3(sin x – cos x)^{4 }+ 6(sin x + cos x)^{2 }+ 4(sin^{6}x + cos^{6}x).
3(sin x – cos x)^{4 }+ 6(sin x + cos x)^{2 }+ 4(sin^{6}x + cos^{6}x)
= 3 (1 - sin 2x)^{2} + 6(1 + sin 2x) + 4[(sin^{2}x + cos^{2}x)^{3} -3sin^{2}x cos^{2}x (sin^{2}x + cos^{2}x)]
= 3 (1 - 2sin2x + sin^{2}2x) + (6 + 6 sin2x) + 4(1-3sin^{2}x cos^{2}x)
= 3 (1- 2sin2x + sin^{2}2x + 2 + 2sin2x) + 4 [1- ¾ sin^{2}2x]
= 13 + 3 sin^{2}2x – 3 sin^{2}2x = 13.
Illustration 4: If cos (a + b) = 4/5, sin (a-b) = 5/13 and a and b lie between 0 to π/4, find tan 2a.
Solution: It is given that cos (a + b) = 4/5, sin (a-b) = 5/13
It follows that sin (a + b) = 3/5, cos (a-b) = 12/13
Hence, tan (a+b) = 3/4 and tan (a-b) = 5/12
Hence, this implies tan [(a+b) + a-b] =
[tan (a+b) + tan (a-b)]/ [1 + tan(a+b)tan(a-b)]
= (3/4 + 5/12)/ (1 – 3/4. 5/12)
Hence, tan 2a = 14/12. 48/33 = 56/33.
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