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# Introduction to trigonometry 

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• ### Introduction to trigonometry 

1. 1. 1. Introduction2.Trigonometric Ratios3.Trigonometric Ratios of Some Specific Angles4.Trigonometric Ratios of Complementary Angles5.Trigonometric Identities
2. 2. • In this chapter, we will study some ratios of thesides of a right triangle with respect to its acuteangles, called TRIGNOMETRIC RATIOS OF THEANGLE.• We will also define the trigonometric ratios forthe angles of measures 0° and 90°. We willcalculate trigonometric ratios for some specificangles and establish some identities involvingthese ratios, called TRIGNOMETRIC IDENTITIES.
3. 3. • Trigonometry is a branch ofmathematics that studies triangles andthe relationship between their sidesand the angles between these sides.• The word ‘TRIGONOMETRY’ is derivedfrom the Greek words ‘TRI’ ( meaningthree ), ‘GON’ ( meaning sides ) and‘METRON’ ( meaning measure ).
4. 4. • Let us consider a right triangle:• Here angle A is an acute angle. Note the position of the sideBC with respect to angle A. we will call the side opposite toangle A. AC is the hypotenuse of the right angled ∆le and theside AB is the part of angle A. so, we will call it as the sideadjacent to angle A.• The trigonometric ratios of angle A in right angle ABC aredefined as follows :• Sine of angle A= side opposite to angle A=BChypotenuse AC• Cosine of angle A= side adjacent to angle A=ABhypotenuse AC• Tangent of angle A= side opposite to angle A= BCside adjacent to angle A AB
5. 5. • Cosecant of angle A= hypotenuse = ACside opposite to angle A BC• Secant of angle A= hypotenuse = ACside of adjacent to angle A AB• Cotangent osf angle A= side adjacent to angle A= ABside opposite to angle A 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 cosecA, sec A and cot A are respectively, the reciprocals of the ratiossin A, cos A, and tan A.• Also, observe that tan A= BC/AC = sin A and cot A = cos AAB/AC cos A sin Aso, the trigonometric ratios of an acute angle in a right ∆le expressthe relationship between the angle and the length of its sides.
6. 6. • LET US TRY AN EXAMPLE :• QUESTION : given tan A = 4/3, find the other trigonometricratios of the angle A.• SOLUTION : let us first draw a right ∆ ABCnow we know that tan A = BC = 4 = opp.AB 3 adj.therefore, to find hypotenuse we use Pythagoras theorem,AC²= AB² + BC² = (4)² + (5)² = (25)² = 5 cmnow, we can write all the trigonometric ratios using theirdefinitions.sin A = BC = 4AC 5cos A = AB = 3AC 5Therefore, cot A = 3, cosec A = 5, and sec A = 54 4 3
7. 7. • Trigonometric ratios of 45°• In• Using definitions : sin 45° = BC = a = 1AC a ∫͞₂ ∫͞₂• Cos 45° = AB = a = 1AC a∫͞₂ = ∫͞₂• tan 45° = BC = a = 1AB a• Also, cosec 45° = 1 = ∫͞₂, sec 45° = ∫͞₂, cot 45° = 1 = 1sin 45° tan 45°
8. 8. • Trigonometric ratios of 30° and 60°• Consider an equilateral ∆le ABC. Since each angle isequal to 60°.• So, AB = BC = AC = 60°.• Now, ∆ABC ‗̴ ∆ ACD ( AD is a median and mediandivides a ∆ into two equal parts).• Therefore, BD = DC and /͟ BAD = /͟ CAD ( CPCT )• Now, ∆ ABD is a right angled triangle with /͟ BAD =30° and /͟ ABD = 60°• As we know, for finding friends the trigonometricratios, we need to know the lengths of the side ofthe ∆. So, let us suppose that AB = 2a.
9. 9. • Trigonometric ratios of 30° and 60° cont…..• Then, BD =½ BC = aAnd AD²= AB² - BD² = (2a)² - (a)² = 3a²,Now we have : sin 30°= BD = a = , cos 30°= AD = a∫͞₃ = ∫͞₃ ,tan 30°= BD = a = 1AB 2a AB 2a 2AD a∫͞₃ ∫͞₃Also, cosec 30° = 1 = 2, sec 30° = 1 = 2 , cot 30° =1 = ∫͞₃sin 30° cos 30° ∫͞₃Similarly, sin 60°=AD=a∫͞₃= ∫͞₃, cos 60°= 1, tan 60°= ∫͞₃,cosec 60°= 2 ,AB 2a 2 2∫͞₃Sec 60°= 2 and cot 60°= 1 .∫͞₃
10. 10. • Trigonometric ratios of 0° and 90°• Let us see what happens to the trigonometricratios of angle A, if it is made smaller andsmaller in the right triangle ABC, till itbecomes zero. As ∠ A gets smaller andsmaller, the length of the side BC decreases.The point C gets closer to point B, and finallywhen ∠ A becomes very close to 0°, ACbecomes almost the same as AB.
11. 11. Trigonometric ratios of 0 and 90 cont..• When ∠ A is very close to 0°, BC gets very close to 0and so the value of sin A =BC/AC is very close to 0.Also, when ∠ A is very close to 0°, AC is nearly thesame as AB and so the value of cos A =AB/AC is veryclose to 1.• This helps us to see how we can define the values ofsin A and cos A when A = 0°. We define : sin 0° = 0and cos 0° = 1.• Sin 0° = 0 and cos 0° = 1.• tan 0° = sin 0°= 0, cot 0°= 1 , which is not defined.cos 0° tan 0°
12. 12. • Trigonometric ratios of 0° and 90° cont…..• Sec 0°= 1 = 1 and cosec 0°= 1 , which is againnot defined.cos 0° sin 0°• Now, let us see what happens to the trigonometricratios of ∠ A, when it is made larger and larger in ΔABC till it becomes 90°. As ∠ A gets larger andlarger, ∠ C gets smaller and smaller. Therefore, as inthe case above, the length of the side AB goes ondecreasing. The point A gets closer to point B.Finally when ∠ A is very close to 90°, ∠ C becomesvery close to 0° and the side AC almost coincideswith side BC.
13. 13. Trigonometric ratios of 0 and 90 cont…..• When ∠ C is very close to 0°, ∠ A is very closeto 90°, side AC is nearly the• same as side BC, and so sin A is very close to1. Also when ∠ A is very close to 90°,• ∠ C is very close to 0°, and the side AB isnearly zero, so cos A is very close to 0.• So, we define : sin 90°= 1 and cos 90°= 0
14. 14. /͟ A 0° 30° 45° 60° 90°Sin A 0 1.21 .∫͞₂∫͞₃ .21Cos A 1 ∫͞₃21 .∫͞₂120Tan A 0 1∫͞₃1 ∫͞₃ not definedCosec A Not defined 2 ∫͞₂ 2 .∫͞₃1Sec A 1 2 .∫͞₃∫͞₂ 2 not definedCot A Not defined ∫͞₃ 1 1 .∫͞͞₃0
15. 15. • As we know that in ∆ABC, right-angled at B, we can see2 complementary angles.• Since /͟ A + /͟ C= 90°, they form such a pair. We have–• sin A = BC, cos A = AB, tan A = BC,AC AC AB• cosec A = AC, sec A = AC, cot A = AB __ ①BC AB BC
16. 16. Now let us write the trigonometric ratios for ∠ C = 90° – ∠ A.For convenience, we shall write 90° – A instead of 90° – ∠ A.What would be the side opposite and the side adjacent to theangle 90° – A?You will find that AB is the side opposite and BC is the sideadjacent to the angle90° – A. Therefore,sin (90° – A) = AB, cos (90° – A) = BC, tan (90° – A) = ABAC AC BCCosec (90°-A ) = AC, sec (90°-A ) = AC, cot (90°-A ) = BC __②AB BC AB• Now, compare the ratios in (1) and (2). Observe that :• Sin (90°-A ) = AB = cos A and cos (90°-A ) = BC = sin AAC AC
17. 17. Trigonometric ratios of complementary angles cont.• Also, tan (90°-A ) = AB = cot A, cot (90°-A) = BC = tan ABC AB• sc (90°-A) = AC = cosec A, cosec (90°-A) = AC = sec ABC AB• So, sin (90° – A) = cos A, cos (90° – A) = sin A,• tan (90° – A) = cot A, cot (90° – A) = tan A,• sec (90° – A) = cosec A, cosec (90° – A) = sec A,• for all values of angle A lying between 0° and 90°. Checkwhether this holds for A = 0° or A = 90°.• Note : tan 0° = 0 = cot 90°, sec 0° = 1 = cosec 90° and sec90°, cosec 0°, tan 90° and cot 0° are not defined.
18. 18. • In Δ ABC, right-angled at B, we have: AB²+BC² = AC²(1)Dividing each term of (1) by AC², we getAB² + BC²= AC²AC² AC² AC²i.e., (AB/AC)²+(BC/AC)²=(AC²/AC²)i.e., (cos A)²+(sin A)²= 1i.e., cos²A+ sin²A= 1 (2)This is true for all A such that 0° ≤ A ≤ 90°. So, this is atrigonometric identity.Let us now divide (1) by AB²
19. 19. • AB²+BC²=AC²AB² AB² AB²or, (AB/AB)² +(BC/AB)²=(AC/AB)²i.e., 1+tan²A = sec²A (3)Is this equation true for A = 0°? Yes, it is. What about A = 90°?Well, tan A and sec A are not defined for A = 90°. So, (3) istrue for all A such that 0° ≤ A < 90°.Let us see what we geton dividing (1) by BC². We getAB²+BC²=AC²BC² BC² BC²i.e., (AB²/BC²)+(BC²/BC²)=(AC²/BC²)i.e., cot²A+1 =cosec²A (4)
20. 20. Note that cosec A and cot A are not defined for A = 0°. Therefore(4) is true for all A such that 0° < A ≤ 90°.Using these identities, we can express each trigonometric ratio interms of other trigonometric ratios, i.e., if any one of the ratiosis known, we can also determine the values of othertrigonometric ratios.Let us see how we can do this using these identities. Suppose weknow that• tanA= 1 .then cotA=∫͞₃∫͞₃Since, sec²A = 1 + tan²A = 1+1 =4,sec=2 and cosA=∫͞₃3 3 ∫͞₃ 2Again sinA=∫͞͞͞͞͞͞1-͞co͞s²͞A=∫͞͞͞1-͞3 =1 Therefore, cosec A = 2.∫ 4 2
21. 21. Done by:Gayathri.