Thursday, November 23, 2023
Latest:

### EteacherG

a educational group

# Trigonometry Trigonometric Identities

trigonometry trigonometric identities trigonometric functions : The distances or heights can be found by using some mathematical techniques, which come under a branch of mathematics called ‘trigonometry’. The word ‘trigonometry’ is derived from the Greek words ‘tri’ (meaning three), ‘gon’ (meaning sides) and ‘metron’ (meaning measure). In fact, trigonometry is the study of relationships between the sides and angles of a triangle. The earliest known work on trigonometry was recorded in Egypt and Babylon. Early astronomers used it to find out the distances of the stars and planets from the Earth. Even today, most of the technologically advanced methods used in Engineering and Physical Sciences are based on trigonometrical concepts.

# Trigonometry Trigonometric Identities

In this chapter, we will study some ratios of the sides of a right triangle with respect to its acute angles, called trigonometric ratios of the angle. We will restrict our discussion to acute angles only. However, these ratios can be extended to other angles also. We will also define the trigonometric ratios for angles of measure 0° and 90°. We will calculate trigonometric ratios for some specific angles and establish some identities involving these ratios, called trigonometric identities.

## Trigonometry Trigonometric Identities

The trigonometric ratios of the angle A in right triangle ABC (see Fig. a) are defined as follows :

$$\displaystyle \text{sine}\,\,\,\text{of}\,\,\,\angle A\,\,=\,\,\frac{{\text{side}\,\,\,\text{opposite}\,\,\,\text{to}\,\,\,\text{angle}\,\,\,\text{A}}}{{\text{hypotenuse}}}\,\,=\,\,\frac{{BC}}{{AC}}$$

$$\displaystyle \text{cosine}\,\,\,\text{of}\,\,\,\angle A\,\,=\,\,\frac{{\text{side}\,\,\,adjacent\,\,\,\text{to}\,\,\,\text{angle}\,\,\,\text{A}}}{{\text{hypotenuse}}}\,\,=\,\,\frac{{AB}}{{AC}}$$

$$\displaystyle \text{tangent}\,\,\,\text{of}\,\,\,\angle A\,\,=\,\,\frac{{\text{side}\,\,\,\text{opposite}\,\,\,\text{to}\,\,\,\text{angle}\,\,\,\text{A}}}{{\text{side}\,\,\,\text{adjacent}\,\,\,\text{to}\,\,\,\text{angle}\,\,\,A}}\,\,=\,\,\frac{{BC}}{{AB}}$$

$$\displaystyle \text{cosecant}\,\,\,\text{of}\,\,\,\angle A\,\,=\,\,\frac{1}{{\text{sine}\,\,\,\text{of}\,\,\,\angle A}}\,\,=\,\,\frac{{\text{hypotenuse}}}{{\text{side opposite to angle A}}}=\frac{{AC}}{{BC}}$$

$$\displaystyle \text{secant}\,\,\,\text{of}\,\,\,\angle A\,\,=\,\,\frac{1}{{\text{cosine of}\,\,\,\angle A}}\,\,=\,\,\frac{{\text{hypotenuse}}}{{\text{side adjacent to angle A}}}=\frac{{AC}}{{AB}}$$

$$\displaystyle \text{cotangent of}\,\,\,\angle A\,\,=\,\,\frac{1}{{\text{tangent of}\,\,\,\angle A}}\,\,=\,\,\frac{{\text{side adjacent to angle A}}}{{\text{side opposite to angle A}}}=\frac{{AB}}{{BC}}$$

The ratios defined above are abbreviated as sin A, cos A, tan A, cosec A, sec A and cot A respectively. Note that the ratios cosec A, sec A and cot A are respectively, the reciprocals of the ratios sin A, cos A and tan A.

$$\displaystyle \text{tan A}=\frac{{BC}}{{AB}}=\frac{{\frac{{BC}}{{AC}}}}{{\frac{{AB}}{{AC}}}}=\frac{{\text{sin A}}}{{\text{cos A}}}$$

and $$\displaystyle \text{cot A}=\frac{{\cos A}}{{\sin A}}$$

So, the trigonometric ratios of an acute angle in a right triangle express the relationship between the angle and the length of its sides

Example 1 : Given tan A = $$\displaystyle \frac{4}{3}$$ , find the other trigonometric ratios of the angle A.
Solution : Let us first draw a right Δ ABC
Now, we know that $$\displaystyle \text{tan A}=\frac{{BC}}{{AB}}=\frac{4}{3}$$
Therefore, if BC = 4k, then AB = 3k, where k is a positive number.
Now, by using the Pythagoras Theorem, we have
AC2 = AB2 + BC2 = (4K)2 + (3K)2 = 25K2
So, AC = 5K
Now, we can write all the trigonometric ratios using their definitions.
$$\displaystyle \text{sin A}=\frac{{BC}}{{AC}}=\frac{{4K}}{{5K}}=\frac{4}{5}$$
$$\displaystyle \text{cos A}=\frac{{AB}}{{AC}}=\frac{{3K}}{{5K}}=\frac{3}{5}$$
Therefore,
$$\displaystyle \text{cotA}=\frac{1}{{\tan A}}=\frac{3}{4},\text{cosec A}=\frac{1}{{\text{sin A}}}=\frac{5}{4}\,\,\,\text{and}\,\,\,\text{sec A=}\frac{1}{{\cos A}}=\frac{5}{3}$$

Example 2 : If ∠B and ∠Q are acute angles such that sin B = sin Q, then prove that ∠B = ∠Q.
Solution : Let us consider two right triangles ABC and PQR where sin B = sin Q (see Fig. ).
We have $$\displaystyle \text{sin B =}\frac{{AC}}{{AB}}$$
and $$\displaystyle \text{sin Q =}\frac{{PR}}{{PQ}}$$
Then $$\displaystyle \frac{{AC}}{{AB}}=\frac{{PR}}{{PQ}}$$
Therefore, $$\displaystyle \frac{{AC}}{{PR}}=\frac{{AB}}{{PQ}}=k,\,\,\,\text{say}$$
Now, using Pythagoras theorem,
$$\displaystyle \text{BC =}\,\,\sqrt{{A{{B}^{2}}-A{{C}^{2}}}}$$
and $$\displaystyle \text{QR =}\,\,\sqrt{{P{{Q}^{2}}-P{{R}^{2}}}}$$
So, $$\displaystyle \frac{{BC}}{{QR}}=\frac{{\sqrt{{A{{B}^{2}}-A{{C}^{2}}}}}}{{\sqrt{{P{{Q}^{2}}-P{{R}^{2}}}}}}=\frac{{\sqrt{{{{k}^{2}}P{{Q}^{2}}-{{k}^{2}}P{{R}^{2}}}}}}{{\sqrt{{P{{Q}^{2}}-P{{R}^{2}}}}}}=\frac{{k\sqrt{{P{{Q}^{2}}-P{{R}^{2}}}}}}{{\sqrt{{P{{Q}^{2}}-P{{R}^{2}}}}}}=k$$
From (1) and (2), we have
$$\displaystyle \frac{{AC}}{{PR}}=\frac{{AB}}{{PQ}}=\frac{{BC}}{{QR}}$$
Then, by using Theorem 6.4, Δ ACB ~ Δ PRQ and therefore, ∠ B = ∠ Q.

Example 3 : Consider Δ ACB, right-angled at C, in which AB = 29 units, BC = 21 units and ∠ ABC = θ (see Fig. ). Determine the values of
(i) cos2θ + sin2θ,
(ii) cos2θ – sin2θ.
Solution :
In Δ ACB, we have

$$\displaystyle \begin{array}{l}AC=\sqrt{{A{{B}^{2}}-B{{C}^{2}}}}\\AC=\sqrt{{{{{\left( {29} \right)}}^{2}}-{{{\left( {21} \right)}}^{2}}}}\\AC=\sqrt{{\text{(29 }-\text{ 21) (29 + 21)}}}\\AC=\sqrt{{\left( 8 \right)\left( {50} \right)}}\\AC=\sqrt{{400}}\\AC=2\text{0}\,\,\,\text{units}\end{array}$$
So,
$$\displaystyle \begin{array}{l}\text{sin }\!\!\theta\!\!\text{ =}\frac{{AC}}{{AB}}=\frac{{20}}{{29}}\\\text{cos }\!\!\theta\!\!\text{ }=\frac{{BC}}{{AB}}=\frac{{21}}{{29}}\end{array}$$
Now, (i) cos2θ + sin2θ = $$\displaystyle {{\left( {\frac{{20}}{{29}}} \right)}^{2}}+{{\left( {\frac{{21}}{{20}}} \right)}^{2}}$$
$$\displaystyle \begin{array}{l}=\frac{{{{{20}}^{2}}+{{{21}}^{2}}}}{{{{{29}}^{2}}}}\\=\frac{{400+441}}{{841}}=1\end{array}$$

(ii) cos2θ − sin2θ = $$\displaystyle {{\left( {\frac{{21}}{{29}}} \right)}^{2}}-{{\left( {\frac{{20}}{{29}}} \right)}^{2}}$$
$$\displaystyle \begin{array}{l}=\frac{{\left( {21+20} \right)\left( {21-20} \right)}}{{{{{29}}^{2}}}}\\=\frac{{41}}{{841}}\end{array}$$

Example 4 : In a right triangle ABC, right-angled at B, if tan A = 1, then verify that
2 sin A cos A = 1.
Solution : In Δ ABC, $$\displaystyle \text{tan A =}\,\,\frac{{BC}}{{AB}}=1$$
i.e., BC = AB
Let AB = BC = k, where k is a positive number.
Now,
$$\displaystyle \begin{array}{l}\text{AC = }\sqrt{{A{{B}^{2}}+B{{C}^{2}}}}\\=\sqrt{{{{{\left( k \right)}}^{2}}+{{{\left( k \right)}}^{2}}}}\\=k\sqrt{2}\end{array}$$
Therefore,
$$\displaystyle \text{sin A = }\frac{{BC}}{{AC}}=\frac{1}{{\sqrt{2}}}$$
and $$\displaystyle \text{cos A = }\frac{{AB}}{{AC}}=\frac{1}{{\sqrt{2}}}$$
So, 2 sin A cos A = $$\displaystyle 2\left( {\frac{1}{{\sqrt{2}}}} \right)\left( {\frac{1}{{\sqrt{2}}}} \right)=1$$ , which is the required value.

error: Content is protected !!