Trigonometry 10th

8/8/2019 Trigonometry 10th

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TRIGONOMETRIC RATIOS

These ratios are called Trigonometricratios of angles

By using some ratios of the sides of atriangle with respect to its acute angles wecan find the remaining sides and angles ofa triangle.


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The six trigonometric ratios of the ABC are:

Sin = perpendicular = BChypotenuse AC

Cos = adjacent side = ABhypotenuse AC


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Tan = sin = opposite side = BC

cos adjacent side ABCosec = 1 = hypotenuse = AC

sin opposite side = BC

sec = 1 = hypotenuse = ACcos adjacent side AB

cot = 1 = cos = adjacent side = AB

tan sin opposite side BC


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Example :If tan A =4/3 find other trigonometricratios?

Let ABC is a right triangle right angled at B

Tan A= opposite side = BC = 4

adjacent side AB 3

let BC = 4k and AB = 3k

AC2 = AB2 + BC2

= [3k] 2 + [4k]2

= 9k2 + 16k2

AC 2 = 25k 2

AC = 25k2 = 5k


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neglecting AC = - 5k since length cannot be negative

sin A = opposite side = BC = 4k = 4

hypotenuse AC 5k 5cos A = adjacent side = AB = 3k = 3

hypotenuse AC 5k 5

tan A = opposite side = BC = 4k = 4

adjacent side AB 3k 3cosec A= hypotenuse = 5k = 5

opposite side 4k 4

sec A = hypotenuse = 5 k 5

adjacent side 3k 3


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Cot A = adjacent side = 3k = 3

opposite side 4k 4

sin A = 4/5 , cos A = 3/5 , tan A = 4/3,

cosec A= 5/4, sec A = 5/3 , cot A = 3/4


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TRIGNOMETRIC RATIOS OF 450

Let ABC is a right triangle , right angled at B

let A=450

Then C = 180 [90+45]

= 180 135

= 450

In ABC C = 45 , A = C

then AB = BC = x

AC = AB2

+ BC2

= x2

+ x2

=2x2

= 2x


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sin 45 = BC = x = 1 AC 2x 2

cos 45 = AB = x = 1 AC 2x 2

tan 45 = BC = x =1 AC x

cosec 45 = 1 = 1 = 2

sin 45 1/1 2sec 45 = 1 = 1 = 2cos 45 1/12

cot 45 = 1 = 1 =1tan 45 1


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Trignometric ratios of 300 and 600

ABC is an equilateral triangleAD is altitude to the side BClet AB =BC = AC = xABD is congruent to ACD by RHS congruency rule

by CPCT BD = DCBD = BC = x/2

by Pythagoras theoremAB2 = BD2 + AD2

AD2= x2 x2/4= 3x2/4

AD = 3x2/4 = 3x/2


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600

sin 60 = AD = 3x/2 = 3 AB x 2

cos 60= BD = x/2 = x X 1 = 1

AB x 2 x 2

tan 60 = sin 60 = 3/2 = 3 X 2 = 3cos 60 2 1cosec 60 = 1 = 1 = 2

sin 60 3/2 3sec 60 = 1 = 1 = 2

cos 60

cot 60 = 1 = 1

tan 60 3


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300sin 30= BD = x/2 = x X 1 = 1

AB x 2 x 2cos 30= AD = 3x/2 = 3x X 1 = 3

AB x 2 x 2tan 30 = sin 30 = = 1 X 2 = 1

cos 30 3/2 = 2 3 3cosec 30= 1 = 1 = 2

sin 30 sec 30 = 1 = 1 = 2

cos 30 3/2 3cot 30 = 1 = 1 = 3

tan 30 1/ 3


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Trignometric ratio of 00

ABC is a right triangle, right angled at B. when angle C getssmaller than AB decreases when the point A closes to B thenangle C closes to zero . AC almost same as BC then

sin 0 = AB = 0 = 0

AC ACcos 0= BC = 1

AC

tan 0 = sin 0 = 0 = 0

cos 0 1cosec 0 = 1 = 1 = ND

sin 0 0


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Sec 0 = 1 = 1 = 1

cos 0 1Cot 0 = 1 = 1 = ND

tan 0 0


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Trignometric ratio of 900

ABC is a right triangle, right angled at B . Angle A gets largerwhen the point c gets closer to B. if C coincide with B thenangle C = 900

sin 90 = AB = 1

AC

cos 90= BC = 0 = 0

AC AC

tan 90 = sin 90 = 1 = N.D

cos 90 0cosec 90 = 1 = 1 = 1

sin 90 1


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Sec 90 = 1 = 1 = N.D

cos 90 0

Cot 90 = cos 90 = 0 = 0

sin 90 = 1


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VALUES OF ALL TRIGNOMETRIC RATIOSOF 0 , 30 ,45 ,60 ,AND 90 .

0 1/2 1/2 3/2 1

1 3/2 1/2 1/2 0

0 1/3 1 3 N.D

N.D 3 1 1/3 01 2/3 2 2 N.D

N.D 2 2 2/3 1

00 300 450 600 900

Tan A

Cosec ASec A

Cot A

A

Sin A

Cos A


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TRIGNOMETRIC RATIOS OFCOMPLEMENTARY ANGLES

Let ABC be a right triangle, right angled at B

A + B + C= 1800

A+ C =180 90

A+ C=90C = 90 - A

sin A = BC

AC

cos A = ABAC


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tan A =BC , cosec A = 1 = AC

AB sin A BC

sec A = 1 = AC , cot A = 1 = AB

cos A AB tan A BC

sin C= sin [90 A] = cos A

cos C = cos[90 A] = sin Atan C = tan [ 90 A] = cot A

cosec C= cosec [ 90 - A ] = sec A

sec C= sec [90 A ] = cosec Acot C = cot [90 A ] = tan A


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From the above we can conclude that

sin[90 A] = cos Acos[90 A] = sin A

tan [90 A] = cot A

cosec[90 A] = sec Asec [90- A] = cosec A

cot [90 A] = tan A


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Trignometric identity

An equation involving trignometricratios of an angle [say] is said to

be a trignometric identity if it issatisfied for all values of forwhich the given trignometric ratiosare defined.


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There are three major identities. they are:

Sin2A+ cos2A = 1

1+ Tan 2A = sec2A

1+ cot2A= cosec2A By these three identities we can form

many identities like;

Cot2

A = cosec2

A 1 Cos 2 A =1- sin2A


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IDENTITY DERIVATION : sin 2 A + cos2 A = 1

ABC is a right triangle right angled at B , by Pythagoras theoremAB2 + BC2 = AC 2 ..[1]

divide eq [1] by AC 2 on both sidesAB2 + BC2 = AC2

AC 2 AC 2

AB2 + BC2 = 1AC2 AC2

AB 2 + BC 2 = 1 AC AC(cos A) 2 + (sin A) 2 =1sin 2 A + cos2 A = 1 .{1}


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1+ tan 2A = sec 2ADivide eq [1] by AB2 on both sides

AB2 + BC2 = AC2

AB2 AB2

AB 2 + BC 2 = AC2

AB AB AB1+ BC 2 = AC 2

AB AB

1+ tan2

A = sec2

A ..{2}


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1+ cot 2A= cosec2A

Divide eq [1] by BC2 on both sides

AB 2 + BC 2 = AC 2

BC BC BCAB 2 +1 = AC 2

BC BC

cot2A+ 1 = cosec 2A1+ cot2 A = cosec2A


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EXAMPLE : PROVE THAT SEC A[1- SIN A] [SEC A + TAN A] = 1

LHS = sec A [1- Sin A] [sec A + tan A ]= 1 [1 sin A] 1 + sin Acos A cos A cos A

= [1- sin A ] [1+ sin A ] = 1 sin2A

cos2 A cos2 A= cos2A

cos2A

= 1

=RHS

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Trigonometry 10th

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