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.
FigureÂ
The angle formed is said to be positive if the revolving ray \overrightarrow {OA} Â 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.
FigureÂ
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.
FigureÂ
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°.
FigureÂ
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                     Â
AD2 = AB2 – BD
⇒       h2 = a2 – \frac{{{a^2}}}{4}
⇒       h2 = \frac{{3{a^2}}}{4}          ∴ h = \frac{{\sqrt 3 }}{2}a
FigureÂ
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 = AB2Â + BC2Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â
AC2 = a2Â + 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 .
FigureÂ
Thus we can form following table .
Illustrations
Ex.1. Find the values of the following expressions:
(a)                                             (b) + 5 cos 90° – cot2 30°
Solution= Â \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 }}}}Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â (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 2Â + 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 2Â \times \sqrt 2 (\sqrt 2Â - 1)}}{{2 \times (4 - 2)}}
= \frac{{55}}{2}(\sqrt 2Â - 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°.
Solution∴                                  x + y = 90°                                      … (i)
and cos (x – y) = 1 = cos 0°
∴                                   x – y = 0°                                                    … (ii)
Adding (i) & (ii) we get 2x = 90°
∴                    x = 45°
∴ by (i)                     45° + y = 90°
∴                                         y = 45°
∴                                   x = y = 45°
Ex.3.  If cosec 4 θ= 2 sin 4 θ, where θ is acute and sin θ > 0. Find the value of θ.
Solution⇒    \frac{1}{{\sin 4\theta }} = 2 sin 4 θ
⇒    \frac{1}{2} = sin2 4 θ ⇒  sin 4 θ =\frac{1}{{\sqrt 2 }}
⇒      sin 4 θ = sin 45° ⇒ 4 θ = 45°
⇒     θ =\frac{{45}}{4} =
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.
Solution∴  sin A =\frac{5}{{13}} = \frac{{BC}}{{AC}}
∴  Let BC = 5x, AC = 13x
then by Pythagoras theorem we have
AC2 = AB2 + BC2 ⇒ (13x)2 = (12)2 + (5x)2
⇒      169x2 = 144 + 25x2
⇒      144x2 = 144 ⇒  x2 = 1
⇒       x = ± 1             x = 1  ( sides are positive)
∴        BC = 5x = 5 × 1 = 5 cm
AC = 13x = 13 × 1 = 13 cm
Now     tan C = \frac{{AB}}{{BC}} = \frac{{12}}{5}Â
FigureÂ
Ex.5.  Find the length of both diagonals of a rhombus of side 14 cm each of whose two opposite angles is 60°.
Solution∴      ∠AOB = 90°
Let    ∠BAD = ∠BCD = 60°
∴     a diagonal AC bisects both ∠A and∠C
∴     ∠BAC = 30° Now in rt ΔAOB we have
sin 30° = \frac{{OB}}{{AB}} = \frac{{OB}}{{14}}
⇒     \frac{1}{2} × 14 = OB ⇒  OB = 7 cm
Again  cos 30° =\frac{{OA}}{{AB}} =\frac{{OA}}{{14}}
⇒     \frac{{\sqrt 3 }}{2} = \frac{{OA}}{{14}} ⇒  OA =\frac{{\sqrt 3 }}{2} × 14 = 7\sqrt 3 cm
∴       AC = 2OA = 2 × 7\sqrt 3 = 14\sqrt 3 cm
BD = 2OB = 2 × 7 = 14 cm
Figure
Ex.6.  If θ is an acute angle and sin θ = cos θ, find the value of 2 tan2 θ + sin2 θ – 1.
Solution⇒        tan θ = 1
We know,  tan 45º = 1
⇒        θ = 45º
Now       2 tan2 θ + sin2 θ – 1 = 2 tan2 45º + sin2 45º – 1
= 2.12 +{\left( {\frac{1}{{\sqrt 2 }}} \right)^2}– 1 = 2 + \frac{1}{2}– 1 = \frac{3}{2}
Ex.7.  An equilateral triangle is inscribed in a circle of radius 4 cm. Find its side ?
Solution∴      OD ⊥ BC
∴      BD = DC and OB and OC are bisectors of ∠B and ∠C respectively
Now in ΔOBD we have
cos 30° =\frac{{BD}}{{OB}} ⇒\frac{{\sqrt 3 }}{2} = \frac{{BD}}{4}
BD =\frac{{\sqrt 3 }}{2} ×4 = 2\sqrt 3
Now   BC = 2BD = 2 ×2\sqrt 3  = 4\sqrt 3 cm
∴      AB = BC = CA = 4\sqrt 3 cm
Figure
Ex. 8. If angle B and ∠Q are acute angles such that sin B = sin Q then prove that ∠B = ∠Q.
FigureÂ
SolutionLet us consider two right angles triangles ABC andΔPQR where sin B = sin Q
We have
sin B = \frac{{AC}}{{AB}}
and       sin Q  =\frac{{PR}}{{PQ}}
Then      \frac{{AC}}{{AB}} = \frac{{PR}}{{PQ}}
Therefore   \frac{{AC}}{{PR}} = \frac{{AB}}{{PQ}}= R,  say        … (1)
Now using phthagoras theorem,
BC = \sqrt {A{B^2} - A{C^2}}  and QR = \sqrt {P{Q^2} - P{R^2}}
So     \frac{{BC}}{{QR}} =\frac{{\sqrt {A{B^2} - A{C^2}} }}{{P{Q^2} - P{R^2}}} =\frac{{\sqrt {{R^2}P{Q^2} - {R^2}P{R^2}} }}{{\sqrt {P{Q^2} - P{R^2}} }} = \frac{{R\sqrt {P{Q^2} - P{R^2}} }}{{\sqrt {P{Q^2} - P{R^2}} }} = R  … (2)
From (1) and (2) we have
\frac{{AC}}{{PR}}=\frac{{AB}}{{PQ}}Â = \frac{{BC}}{{QR}}Then by using SSS similarity criterion
ΔACB ~ ΔPRQ
⇒       ∠B =∠Q
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 : 3Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â
           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}}]
=Â \frac{{P{M^2}}}{{O{P^2}}}+ \frac{{O{M^2}}}{{O{P^2}}} =Â \frac{{P{M^2} + O{M^2}}}{{O{P^2}}}
= \frac{{O{P^2}}}{{O{P^2}}}Â Â Â Â Â Â Â Â Â [ By Pythagoras theorem, PM2Â + OM2Â = 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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