Trigonometric ratios of some specific angles: In Mathematics, trigonometry is a branch that deals with the sides and angles of a right triangle. The three basic trigonometric functions are sine (sin), cosine (cos) and tangent (tan). The other trigonometric functions which are derived from the basic trigonometric functions are cosecant (cosec), secant (sec) and cotangent (cot). In this article, we will learn how to derive the trigonometric ratios of some specific angles with a complete explanation.
Table of Contents:
Trigonometric Ratios of Some Specific Angles – Definition
Trigonometric ratios of some specific angle are defined as the ratio of the sides of a right-angle triangle with respect to any of its acute angles. Trigonometric ratios of some specific angles include 0°, 30°, 45°, 60° and 90°. Now, let us learn how to find the trigonometric ratios of these angles in detail.
Trigonometric Ratios of 45°
Consider a right-angle triangle ABC as shown in the figure.
Here, a triangle ABC is right-angled at B. (i.e) ∠B = 90°.
If one of the angles is 45°, then the remaining angle should be 45°, as the sum of the interior angles of a triangle is 180°.
Therefore, ∠A = ∠C = 45°.
Also, AB = BC.
Now, assume that AB = BC = a
By using Pythagoras theorem, we can say that AC2 = AB2 + BC2
AC2 = a2 + a2
AC2 = 2a2
Therefore, AC = a√2.
Therefore,
sin 45° = BC/AC = a/ a√2 = 1/√2.
cos 45° = AB/AC = a/ a√2 = 1/√2.
tan 45° = BC/AB = a/a = 1
As, cosecant, secant and cotangent are the reciprocal of sine, cosine and tangent, respectively, then we can write:
cosec 45° = 1/ sin 45° = √2
sec 45° = 1/ cos 45° = √2
Cot 45° = 1/ tan 45° = 1.
Trigonometric Ratios of 30° and 60°
Consider an equilateral triangle ABC. We know that in an equilateral triangle all the angles should be equal to 60°. Therefore,
∠A = ∠B = ∠C = 60°.
We have to calculate the trigonometric ratios of 30° and 60°. Now, draw the perpendicular line AD from vertex A to the side BC as shown below.
So, we can observe that, ∆ ABD ≅ ∆ ACD
Also, BD = DC.
Hence, by using the CPCT (Corresponding Parts of Congruent Triangles) rule, we can write ∠ BAD = ∠ CAD.
So, the triangle ADB is a right triangle, which is right-angled at D and with ∠ BAD = 30° and ∠ABD = 60°.
To find the trigonometric ratio, we need to know the measurements of the side length of a triangle.
Assume that, AB = 2a.
Also, BD = ½ BC = ½ (2a) = a (Since, all the sides are equal in equilateral triangle)
Now, by using Pythagoras theorem, AB2 = BD2 + AD2
Therefore, AD2 = AB2– BD2
AD2 = (2a)2 – a2
AD2 = 4a2-a2
AD2 = 3a2
Hence, AD = a√3.
So, the trigonometric ratios of 30° are:
sin 30° = BD/AB = a/2a = ½
cos 30° = AD/AB = a√3 / 2a = √3/2
tan 30° = BD/AD = a/a√3 = 1/√3.
cosec 30° = 1/sin 30° = 2
sec 30° = 1/cos 30°= 2/√3
cot 30° = 1/tan 30° = √3
Similarly, the trigonometric ratios of 60° are:
Sin 60° = a√3/(2a) = √3/2.
cos 60° = 1/2
tan 60° = √3
cosec 60° = 2/√3
sec 60° = 2
cot 60° = 1/√3
Trigonometric Ratios of 0° and 90°
Consider the same triangle ABC which is right-angled at B. Let us see what happens when the trigonometric ratio of angle A is made smaller and smaller till it becomes zero.
It is observed that ∠A is very close to 0°, BC also get close to 0. Hence, Sin A = BC/AC is close to 0.
Similarly, if ∠A is very close to 0°, AC is the same as AB, then Cos A = AB/AC = 1
Hence, we can define: sin 0° = 0 and cos 0° = 1
With these two observations, we can derive the other ratios of 0°.
tan 0° = sin 0°/cos 0° = 0/1 = 0
cosec 0° = 1/sin 0° = Not defined
sec 0° = 1/cos 0° = 1/1 = 1
cot 0° = 1/tan = 1/0° = Not defined
Similarly, consider the triangle ABC which is right-angled at B. What happens when the trigonometric ratio of angle A is made larger and larger till it becomes 90°.
It is observed that if ∠C is very close to 0°, ∠A is very close to 90°, and AC = BC, so sin A is very close to 1. Similarly, when ∠A is very close to 90°, ∠C is very close to 0°, and AB =0, so cos A is very close to 0.
Hence, we can define: sin 90° = 1 and cos 90° = 0.
Therefore,
tan 90° = sin 90°/cos 90° = 1/0 = Not defined
cosec 90° = 1/sin 90° = 1/1 = 1
sec 90° = 1/cos 90° = 1/0 = Not defined
cot 90° = 1/tan 90° = 1/ Not defined(∞) = 0.
Video Lesson on Trigonometry
| Also, read: |
|---|
Trigonometric Ratios of Some Specific Angles Examples
Example 1:
Determine the value of A and B, if sin (A-B) = ½, cos(A+B) = ½, and 0° < A + B ≤ 90°, where A>B.
Solution:
Given that: sin (A-B) = ½
cos(A+B) = ½, and 0° < A + B ≤ 90°, where A>B.
sin (A- B) = ½
Therefore, A-B = Sin-1(½) = 30°….(1)
cos (A+B) = ½
A+B = cos-1(½)
A+B = 60°…(2)
On solving the equation (1) and (2), we get
A = 45° and B = 15°.
Therefore, the values of A and B are 45° and 15°, respectively.
Example 2:
Evaluate the expression sin 60° cos 30° + sin 30° cos 60°.
Solution:
Given Expression: sin 60° cos 30° + sin 30° cos 60°
sin 30° = 1/2
cos 30°= √3/2
sin 60° = √3/2
cos 60°= 1/2
Now, substitute the values of sin 30°, cos 30°, sin 60°, cos 60°,
sin 60° cos 30° + sin 30° cos 60° = (√3/2)(√3/2) + (½)(½)
= (¾) + (¼)
= (3+1)/4
= 4/4 = 1
Therefore, sin 60° cos 30° + sin 30° cos 60° = 1.
Practice Problems
Solve the following problems:
- Find the value of A and B, if tan (A+B) = √3, and tan (A- B) = 1/√3, and 0°< A + B ≤ 90°, where A > B.
- Evaluate the expression 2 tan2 45° + cos2 30° – sin2 60°.
- Is sin θ = cos θ for all values of θ. Justify your answer.
Stay tuned with BYJU’S – The Learning App and also learn all Maths-related concepts by exploring more exciting videos.
Comments