Trigonometry Class 10 Notes

Introduction

The word trigonometry consists of two Greek words: ‘trigonon’ which means a triangle and ‘metron’ which means a measure. Hence the word trigonometry means measurement of triangles. Thus trigonometry is a branch of mathematics which deals with measurement of triangles. We know that there are six component of a triangle: three sides and three angles. Out of these six component if any three components (including at least one side) are given, we can determine the remaining components uniquely. But the actual method of determining their components is not discussed in plane geometry. It is only trigonometry which helps us to determine all the components of a triangle. Now – a – days trigonometry has much wider applications. Trigonometry is so important that higher studies of mathematics and science are not possible without its knowledge.

Angles in Plane Geometry and Trigonometry

In plane  geometry an angle is defined as a figure which is formed by two rays originating from the same initial point. An angle in plane geometry is supposed to lie between 0° and 360° i.e. angles in plane geometry are supposed to be positive. But angles in trigonometry are more general in nature. These angles may be positive or negative and their magnitudes may exceed 360°.

In trigonometry an angle is supposed to be formed by the revolution of a ray about its end point. The original position of the ray is called the initial side and the final position is called the terminal side of the angle.

Figure 

The angle formed is said to be positive if the revolving ray \overrightarrow {OA}   rotates in anticlock wise direction from initial side to terminal side and negative if the revolving ray \overrightarrow {OA} rotates in the clock wise direction from the initial side to terminal side.

Trigonometric ratio of angles

Let the rays OX and OY meet at O forming an acute angle XOY. We denote the measure of this angle by the Greek letter θ (read as theta) in degrees and OX as initial line and OY as terminal line. If  we take a point P on OY and draw PQ⊥OX,  then ΔPOQ is a right angled triangle.

Figure 

The side adjacent to angle θ is called base, that opposite to angle θ is called perpendicular and the side opposite to the right angle is called hypotenuse. Thus in figure OQ is base, PQ is perpendicular and OP is hypotenuse of ΔOPQ. Let us suppose that lengths of the base, perpendicular and hypotenuse of ΔOPQ are b, p and h respectively.

We now define the following six trigonometric ratios of angle θ in terms of b, p and h.

Sine θ (abbreviated as sin θ) =\frac{{{\rm{perpendicular}}}}{{{\rm{hypotenuse}}}}  = \frac{p}{h}

Cosine θ (abbreviated as cos θ) = \frac{{{\rm{base}}}}{{{\rm{hypotenuse}}}} = \frac{b}{h}

Tangent θ (abbreviated as tan θ) = \frac{{{\rm{perpendicular}}}}{{{\rm{base}}}} = \frac{p}{b}

Cosecant θ (abbreviated as cosec θ) =\frac{{{\rm{hypotenuse}}}}{{{\rm{perpendicular}}}} = \frac{h}{p}

Secant θ (abbreviated as sec θ) =\frac{{{\rm{hypotenuse}}}}{{{\rm{base}}}}  = \frac{h}{b}

Cotangent θ (abbreviated as cot θ) =\frac{{{\rm{base}}}}{{{\rm{perpendicular}}}}  = \frac{b}{p}

Relations Between Different Trigonometric Ratios

The different trigonometric ratios are connected with each other. If one of them is known, the remaining five trigonometric ratios can be found out. We now establish such relations between the trigonometric ratios

(1)  tan θ =\frac{p}{b}  = \frac{{p/h}}{{b/h}} =\frac{{{\rm{sin \theta }}}}{{{\rm{cos \theta }}}} , cos θ≠ 0

(2)  cot θ =\frac{b}{p}  =\frac{{b/h}}{{p/h}}  =\frac{{\cos \,\theta }}{{\sin \,\theta }} , sin θ ≠ 0

(3)  sec θ =\frac{h}{b}  =\frac{1}{{b/h}} =\frac{1}{{\cos \,\theta }} , cos θ ≠ 0

(4)  cosec θ=\frac{h}{p}  =  \frac{1}{{p/h}}= \frac{1}{{\sin \,\theta }}, sin θ ≠ 0

(5)  cot θ =\frac{b}{p} = \frac{1}{{p/b}} =\frac{{\rm{1}}}{{{\rm{tan \theta }}}} , tan θ ≠ 0

(6)  sin θ= \frac{{\rm{1}}}{{{\rm{cosec \theta }}}}, cosecθ ≠0

(7)  tan θ =\frac{{\rm{1}}}{{{\rm{cot \theta }}}} , cot θ ≠ 0

(8)  cos θ =\frac{{\rm{1}}}{{{\rm{sec \theta }}}} , sec θ ≠ 0

Trigonometric ratios of some standard angles

Now we shall deduce the numerical values of the trigonometric ratios of some standard angles by using some results on triangles in plane geometry.

Trigonometric Ratio of 0°

Let AX be the initial ray and P be a point on the terminal side AY. Let PN ⊥ AX and ∠PAN = θ where θ is very small acute angle.

Figure 

If we now make PN smaller and smaller keeping AP and AN unaltered, then θ becomes smaller and smaller. Mathematically, we say that as PN tends to zero, θ also tends to zero.

sin 0° =\frac{{PN}}{{AP}}  = 0          cot 0° =\frac{{{\rm{AN}}}}{{{\rm{PN}}}}  = ∞

cos 0° =\frac{{AN}}{{AP}}  = 1         sec 0° =\frac{{AP}}{{AN}}  = 1

tan 0° = \frac{{PN}}{{AN}} = 0          cosec 0° =\frac{{AP}}{{PN}} = ∞

Trigonometric Ratio of 90°

Let AX be the initial ray and P be a point on the terminal side AY. Let PN ⊥ AX and ∠ PAN = θ where θ is acute but very close to 90°.

If we now make AN smaller and smaller keeping AP and PN unaltered, then θ becomes larger and larger but it is always less than 90°. Mathematically, we say that as AN tends to zero, θ tends to 90° starting from values less than 90°.

Figure 

sin 90° =\frac{{PN}}{{AP}}  = 1         cot 90° =\frac{{{\rm{AN}}}}{{{\rm{PN}}}}  = 0

cos 90° = \frac{{AN}}{{AP}} = 0         sec 90° = \frac{{AP}}{{AN}} =∞

tan 90° = \frac{{PN}}{{AN}} = ∞        cosec 90° =\frac{{AP}}{{PN}}  = 1

Trigonometric Ratio of 30° and 60°

Let ΔABC is an equilateral triangle of side of length a we draw AD ⊥ BC, then in ΔABC

BD = DC = \frac{a}{2}

Let AD = h in ΔABD, by Pythagoras theorem we have                                           

AD² = AB² – BD

⇒            h² = a² – \frac{{{a^2}}}{4}

⇒            h² =  \frac{{3{a^2}}}{4}            ∴ h = \frac{{\sqrt 3 }}{2}a

Figure 

Now in ΔABD

sin 30° =\frac{{BD}}{{AB}}  =\frac{{a/2}}{a}  =\frac{1}{2}

cos 30° =\frac{{AD}}{{AB}}  =\frac{{\frac{{\sqrt 3 }}{2}a}}{a}  = \frac{{\sqrt 3 }}{2}

tan 30° = \frac{{BD}}{{AD}} =\frac{{a/2}}{{\sqrt 3 a/2}}  = \frac{1}{{\sqrt 3 }}

Similarly cot 30° =\sqrt 3 , sec 30° =\frac{2}{{\sqrt 3 }} , cosec 30° = 2

Also,

sin 60° =\frac{{AD}}{{AB}} =\frac{{\frac{{\sqrt 3 }}{2}a}}{a}  = \frac{{\sqrt 3 }}{2}

cos 60° = \frac{{BD}}{{AB}} =\frac{{a/2}}{a}  = \frac{1}{2}

tan 60° = \frac{{AD}}{{BD}} =\frac{{\sqrt 3 a/2}}{{a/2}}  =\sqrt 3

Similarly cot 60° =\frac{1}{{\sqrt 3 }} , sec 60° = 2, cosec 60° =\frac{2}{{\sqrt 3 }}

Trigonometric Ratio of 45°

Let ΔABC be a right angled triangle with ∠ABC = 90°, ∠BAC = 45°

Let AB = BC = a, then by Pythagoras theorem we have

AC² = AB² + BC²                                                                                                                   

AC² = a² + a²

∴  AC² = 2a²  ⇒  AC = a\sqrt 2

Now sin 45° = \frac{{BC}}{{AC}} =\frac{a}{{a\sqrt 2 }}  =\frac{1}{{\sqrt 2 }}

cos 45° =\frac{{AB}}{{AC}}  = \frac{a}{{a\sqrt 2 }} = \frac{1}{{\sqrt 2 }}

tan 45° =\frac{{BC}}{{AB}}  = \frac{a}{a} = 1

Similarly cot 45° = 1, sec 45° =\sqrt 2 , cosec 45° =\sqrt 2 .

Figure 

Thus we can form following table .

Illustrations

Ex.1. Find the values of the following expressions:

(a)                                                 (b)  + 5 cos 90° – cot2 30°

Sol.:(a) Given

=   \frac{{5{{\left( {\frac{1}{2}} \right)}^2} + {{\left( {\frac{1}{{\sqrt 2 }}} \right)}^2} + 4 \times {{(\sqrt 3 )}^2}}}{{2 \times \frac{1}{2} \times \frac{1}{2} + \frac{1}{{\sqrt 2 }}}}                          (using trigonometric table)

= \frac{{\frac{5}{4} + \frac{1}{2} + 12}}{{\frac{1}{2} + \frac{1}{{\sqrt 2 }}}}\frac{{\frac{{5 + 2 + 48}}{4}}}{{\frac{{\sqrt 2  + 2}}{{2\sqrt 2 }}}}

= \frac{{55}}{4} ×\frac{{2\sqrt 2 }}{{\sqrt 2  + 2}} = \frac{{55\sqrt 2 }}{{2(2 + \sqrt 2 )}}

= \frac{{55\sqrt 2 (2 - \sqrt 2 )}}{{2(2 + \sqrt 2 )(2 - \sqrt 2 )}} = \frac{{55 \times \sqrt 2  \times \sqrt 2 (\sqrt 2  - 1)}}{{2 \times (4 - 2)}}

= \frac{{55}}{2}(\sqrt 2  - 1)

(b) Given\frac{{\sin 60^\circ }}{{{{\cos }^2}45^\circ }}  + 5 cos 90° – cot² 30°

= \frac{{\frac{{\sqrt 3 }}{2}}}{{{{\left( {\frac{1}{{\sqrt 2 }}} \right)}^2}}} + 5.0 – (\sqrt 3 )2

= \frac{{\sqrt 3 }}{2}    ×\frac{2}{1}  + 0 – 3

= \sqrt 3 – 3

Ex.2.   If sin (x + y) = 1 and cos (x – y) = 1. Find the values of x and y when 0 ≤ x, y ≤ 90°.

Ex.3.   If cosec 4 θ= 2 sin 4 θ, where θ is acute and sin θ > 0. Find the value of θ.

Ex.4.   In ΔABC, ∠ABC = 90°. If AB = 12 cm and sin A =\frac{5}{{13}} , find the hypotenuse and other sides of the triangle, also find tan C.

Ex.5.   Find the length of both diagonals of a rhombus of side 14 cm each of whose two opposite angles is 60°.

Ex.6.   If θ is an acute angle and sin θ = cos θ, find the value of 2 tan² θ + sin² θ – 1.

Ex.7.   An equilateral triangle is inscribed in a circle of radius 4 cm. Find its side ?

Ex. 8.  If angle B and ∠Q are acute angles such that sin B = sin Q then prove that ∠B = ∠Q.

Figure 

Identities

            An equation which is true for every value of unknown quantity is called an identity. An identity involving trigonometric function is called a trigonometric identity.

Some well-known trigonometric identities

            We shall prove the following important identities :

(i)   sin² θ + cos² θ = 1, when 0° ≤ θ ≤ 90°.

(ii)  sec² θ = 1 + tan² θ, when θ ≠ 90º

(iii) cosec² θ = 1 + cot² θ, when θ ≠ 0º

(i)   To prove that sin² θ + cos² θ = 1 for 0º ≤ θ ≤ 90º, we consider three cases :

Case : 1

Let            θ = 0º.

∴              sin² θ + cos² θ = sin² 0º + cos² 0º

                = 0 + 1 = 1

Case : 2

Let θ          = 90º

∴                  sin² θ + cos² θ = sin² 90º + cos² 90º

                   = 1 + 0 = 1

Case : 3                                                                                               

            Let 0º < θ < 90º, i.e., let θ be an acute angle. Let ∠XOY = θ (in figure).

            We take a point P on the terminal ray OY and draw PM ⊥OX,

            where OX is the initial ray.

Figure

            Then,

sin² θ + cos² θ = {\left( {\frac{{PM}}{{OP}}} \right)^2} + {\left( {\frac{{OM}}{{OP}}} \right)^2} [ sin θ = \frac{{PM}}{{OP}} and cos θ = \frac{{OM}}{{OP}}]

=  \frac{{P{M^2}}}{{O{P^2}}}+ \frac{{O{M^2}}}{{O{P^2}}} =  \frac{{P{M^2} + O{M^2}}}{{O{P^2}}}

= \frac{{O{P^2}}}{{O{P^2}}}           [ By Pythagoras theorem, PM² + OM² = OP²]

            = 1

            Thus, in all the three cases, we have

        sin² θ + cos² θ = 1.

(ii) To prove that sec² θ = 1 + tan² θ, when θ ≠ 90º, i.e., when 0º ≤ θ < 90º, we consider two cases :

Case : 1

Let                      θ = 0

Then,                sec² θ = sec² 0º = 1

and                1 + tan² θ    = 1 + tan² 0º = 1 + 0 = 1

∴                     sec² θ = 1 + tan² θ.

Case : 2

            Let 0º < θ < 90º, i.e. θ be acute.

            Then, from figure , we have

sec² θ = {\left( {\frac{{OP}}{{OM}}} \right)^2} = \frac{{O{P^2}}}{{O{M^2}}}   .....(i)

 and          1 + tan² θ = 1 +  {\left( {\frac{{PM}}{{OM}}} \right)^2}= 1 + \frac{{P{M^2}}}{{O{M^2}}}

                 = \frac{{O{M^2} + P{M^2}}}{{O{M^2}}} =  \frac{{O{P^2}}}{{O{M^2}}}   [ By Pythagoras theorem, OM² + PM² = OP²]  … (ii)

From (i) and (ii), we have

    sec² θ = 1 + tan² θ

Hence, in all cases when θ ≠ 90º,

               sec² θ = 1 + tan² θ.

(iii) To prove that cosec² θ = 1 + cot² θ, when θ ≠ 0º, i.e., when 0º < θ ≤ 90º, we consider two cases :

Case : 1

Let                 θ = 90º

Then,          cosec² θ = cosec² 90º = 1

and           1 + cot² θ  = 1 + cot² 90º = 1 + 0 = 1

Hence,         cosec² θ = 1 + cot² θ.

Case : 2

      Let 0º < θ < 90º, i.e., let θ be acute.

Then, from figure , we have

 cosec² θ ={\left( {\frac{{OP}}{{PM}}} \right)^2}  = \frac{{O{P^2}}}{{P{M^2}}}   … (iii)

and         1 + cot² θ  = 1 + {\left( {\frac{{OM}}{{PM}}} \right)^2} = 1 + \frac{{O{M^2}}}{{P{M^2}}}

= \frac{{P{M^2} + O{M^2}}}{{P{M^2}}} = \frac{{O{P^2}}}{{P{M^2}}}         [ By Pythagoras theorem, PM² + OM² = OP²] … (iv)

From (iii) and (iv), we have

            cosec² θ = 1 + cot² θ in all cases when θ ≠ 0º

Trigonometric ratios of complemenary angles

            We know that two angles are said to be complementary, if their sum is 90º. Thus, angles θ and 90º – θ are complementary. We shall now express the trigonometric ratios of the complementary angle 90º – θ of a given angle θ in terms of the trigonometric ratios of θ, where θ is an acute angle.

            We consider a right-angled triangle ABC, where ∠ABC = 90º, ∠ACB = θ, so that ∠BAC = 90º – θ (figure).

Figure

            Clearly, ∠BAC and ∠ACB are complementary angles. We shall now express the trigonometric ratios of 90º – θ in term of those of θ.

            Now, with reference to the angle θ, BC is the base, AB is the perpendicular and AC is the hypotenuse.

            Hence, we have

            sin θ = \frac{{AB}}{{AC}} , cos θ = \frac{{BC}}{{AC}} , tan θ = \frac{{AB}}{{BC}} , cot θ = \frac{{BC}}{{AB}}, sec θ = \frac{{AC}}{{BC}} and cosec θ = \frac{{AC}}{{AB}}.

            Again, with reference to angle BAC = 90º – θ, AB is the base, BC is the perpendicular and AC is the hypotenuse.

            Hence, sin (90º – θ) =  \frac{{BC}}{{AC}}= cos θ

                  cos (90º – θ) =\frac{{AB}}{{AC}}  = sin θ

                  tan (90º – θ) = \frac{{BC}}{{AB}} = cot θ

                  cot (90º – θ) = \frac{{AB}}{{BC}} = tan θ

                  sec (90º – θ) =\frac{{AC}}{{AB}}  = cosec θ

            and cosec (90º – θ) = \frac{{AC}}{{BC}} = sec θ

            The above formulae should be memorised thoroughly.

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