Trigonometry name itself says that it is a subject that deals with the geometry of triangles and it is very useful for situations when needed to find when there are some sides given and we need the relations between the sides or angles between the sides. In Trigonometry we have different ratios that are sin A, cos A, tan A, cot A, sec A, cosec A with the help of which, the relation between the sides and the angle between the sides of the triangle can be obtained.
Trigonometric Functions
The trigonometric functions define the relation between the sides and angles and the examples are sin A, cos A, tan A, cot A, sec A, cosec A. These six functions show the relation between the sides of the right angles triangle i.e base, height, and hypotenuse. Examples are,
- sin(A + B) = sin A cos B + cos A sin B
- sin(A – B) = sin A cos B – cos A sin B
- cos (A + B) = cos A cosB – sin AsinB
- cos (A – B) = cos A cos B + sin A sin B
- tan (A + B) = tan A + tan B/ 1 – tanA tanB
- tan (A – B) = tan A – tan B/ 1+ tan Atan B
Trigonometric Identities
The relation between different trigonometric functions is a trigonometric identity. The identities are very useful to test the inequality in the trigonometric equations. Identities are also used to simplify complex trigonometric equations. The six trigonometric ratios are related to each other such that sin is the reciprocal of cosec, and so on. Let’s take a look at some basic trigonometric identities,
- Tan A = sin A/cos A
- sin A = 1/cosec A
- cos A = 1/sec A
- Tan A = 1/cot A
Double Angle Identities
- sin 2A = 2sin Acos A
- cos2A = cos²A – sin²A
- tan 2A = 2 tan A / (1 – tan ²A)
Triple Angle Identities
- sin 3A = 3sin A – 4 sin³A
- cos3A = 4cos³A – 3 cosA
- tan 3A = (3tan A – tan³A)/(1 – 3tan²A)
Prove that
There are basic identities that are required in order to solve the above problem statement, lets look at some of the basic identities of the 6 trigonometric functions that are required in this case,
Prerequisite identities used in the proof
- sec2A – tan2A =1
- cosec A = 1/sin A
- cos A/sin A = cot A
- cot A/2 =
- sin2A = 1 – cos2A
- a2 – b2 = (a + b)(a – b)
- (a – b)/c = a/c – b/c
- sec A = 1/cos A
Given Trigonometric equation
LHS =
RHS = cosec A – cot A
Deriving Proof from LHS side
Given LHS
Step 1
Multiplying with (sec A – 1)/(sec A – 1) which is equal to 1 to bring the degree of 1 which is present on RHS.
=
Step 2
Simplifying the equation obtained in step – 2 by further
=
=
= (sec A – 1)/tan A
Step 3
Breaking the equation in step – 2 in general form
= (sec A/tan A) – (1/tan A)
= ((1/cos A)/(sin A /cos A)) – (1/tan A)
= (1/sin A) – (1/tan A)
Step 4
Substituting with standard formulas in the equation obtained in step – 3
= cosec A – cot A
From step – 4 it can be concluded that LHS = cosec A – cot A which is equal to RHS and thus,
cosec A – cot A = cosec A – cot A
LHS = RHS
Hence Proved.
Deriving Proof from the RHS side
Given RHS: cosec θ – cot θ
Step 1
Simplifying the equation by substituting standard formulas
(1/sin A) – (cos A/sin A)
= (1 – cos A)/(sin A)
Step 2
Simplifying the denominator
= (1 – cos A)/(√1 – cos 2A)
=
=
Step 3
(1 – cos A) in numerator and denominator of the equation in step – 2 gets cancelled so it becomes,
=
=
=
From the step 3 it can be concluded that RHS =
which is equal to LHS and thus, LHS = RHS
Hence Proved.
Sample Problems
Question 1: Solve the trigonometric identity:
Solution:
Using identity 2 and 3 to simplify the identity
=
Now using our derived identity
=
= √(1)
= 1
Question 2: Solve the trigonometric identity: (cosec θ – cot θ) × 16 cot A/2
Solution:
Using the identity 4 to simplify the identity
= 16 × (cosec θ – cot θ) × (√((1+cos A)/(1-cos A))
Taking cos A common in Numerator and Denominator in the square root and using the identity 8
= 16 × (cosec θ – cot θ) × √{(sec A+1)/(sec A- 1)
By using the above proved identity
= 16 × (cosec θ – cot θ) × 1/(cosec θ – cot θ)
= 16