How to solve trigonometric problems based on complimentary anngles?(concept-3)

IB Maths tutors give great importance to Trigonometry.

Trigonometry is one of the fascinating branches of Mathematics. It deals with the relationships among the sides and angles of a triangle.Word trigonometry was originated from the Greek word, where, ‘TRI‘ means Three‘GON‘ means sides and the ‘METRON’ means to measure. It’s an ancient and probably most widely used branch Mathematics. For basic learning, IB Maths Tutors divide trigonometry in two part:-

1. Trigonometry based on right triangles

2. Trigonometry based on non-right triangles.

Here, we are discussing trigonometry based on non-right triangles only.

In the third article of this series, we will discuss problems based on complementary angles

In the third article of this series, we will discuss problems based on complementary angles

<img src="right triangle.jpg" alt="right triangle">

In this right triangle Sin A=BC/AC & Cos C=BC/AC   clearly: Sin A=Cos C  In the given triangle A+C=90° so we can write C=(90°-A). This gives us freedom to write Sin A=Cos (90°-A) similarly we can write these relationships            

                                                     tan A=Cot (90°-A)

                                                     CotA=tan (90°-A)

                                                     SecA=Cosec (90°-A)

                                                     Cosec A=Sec (90°-A)

                                                     Sin A=Cos (90°-A)

                                                     Cos A= Sin (90°-A)

<img src="complimentary angles.jpg" alt="complimentary angles">

These formulas give us the freedom to solve trigonometric problems based on complementary angles. IB Maths Tutors use the following concept to solve complementary angle problems

Concept-3

In a trigonometric problem, If we are given two angles which are complementary to each other, we will change only one angle out of the given two angles. We should keep the second angle unchanged.

Example 1: Find value of  cos 56° / sin 34°

Solution:    56°+34°=90° so given angles are complementary. In the case of complementary angles, we should change only one angle. here we are changing 34°

 

cos 56°  =  cos (90° – 56°)  =  sin 34°

cos 56° / sin 34°  =  sin 34° / sin 34°  =  1

Hence the value of cos 56° / sin 34° is 1.

 

Example 2: Find the value of (cos 65° sin 18° cos 58°) / (cos 72° sin 25° sin 32°)

Solution : 

65° and  25°

18° and 72°

32° and 58° all are pairs of

cos 65°  =  cos (90° – 25°)  =  sin 25°

sin 18°  =  sin (90° – 72°)  =  cos 72°

cos 58°  =  cos (90° – 32°)  =  sin 32°

(cos 65° sin 18° cos 58°) / (cos 72° sin 25° sin 32°) is

=  (sin 25° cos 72° sin 32°) / (cos 72° sin 25° sin 32°)

=  1

Hence the value of the given trigonometric expression is 1.

 

Example3: I the value of sin 3A = cos (A – 26°), where 3A is an acute angle, then calculate  A. 

Solution: We are given that sin 3A = cos (A – 26°)

this relationship is called a trigonometric equation. To solve a trigonometric equation, we should convert it such that it should have only one trigonometric ratio in it. so we should either convert sin into cos or cos into sin.
As                                                      sin 3A = cos (90° – 3A), so we can rewrite as
cos (90° – 3A) = cos (A – 26°)

If we compare both sides

90° – 3A = A – 26°

                                                                 -3A-A=-90-26
                                                                     -4A=-116
                                                                        A=116/4
                                                                         A=29

To get a better understanding of the concept you should check all my posts of Mathematics tutors series. you can check them by clicking on the links given below

First post- concept one

Second post- concept two

Third post -Concept three (current post)

Fourth post -Concept four

                                                      

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