10 Math Trigonometry

Finish Line & Beyond TRIGONOMETRY Trigonometry is the study of relationships between the sides and angles of a triangle. Trigonometric Ratios

A

Side Opposite to C

B

Side Adjacent to C

C

In the above figure ∆ABC is a right angled triangle. ∠ C is an acute angle. Side BC is the Side Adjacent to ∠ C and side AB is the Side Opposite to ∠ C . The trigonometric ratio of

∠ C is given as follows:

Side Opposite to ∠ C Hypotenuse BC b cosC = = AC h AB p tanC = = BC b h AC cosecC = = p AB AC h secC = = BC b b BC cotC = = p AB sinC =

=

AB p = AC h

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Finish Line & Beyond Note: The value of trigonometric ratios do not vary with the value of sides if angles remain the same. Example: Consider Δ ACB, right-angled at C, in which AB = 35 units, BC = 21 units and A ABC = θ. Determine the values of (i) cos² θ + sin² θ, (ii) cos² θ – sin² θ.

A

35 Units

C

21 Units

B

Solution: (i) cos²θ + sin²θ

b p + (p²= h²-b² = 1225-441 =784, So, p=28) h h 21 28 =( )²+( )² 35 35 441 784 = + 1225 1225 1225 = =1 1225 =

(ii) cos²θ - sin²θ =

441 − 784 − 343 = = -0.28 1225 1225

Trigonometric Ratios of Some Specific Angles Trigonometric Ratios of 45°

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Finish Line & Beyond In Δ ABC, right-angled at B, if one angle is 45°, then the other angle is also 45°, i.e., X A = X C = 45° So, BC = AB , because it is an Isosceles Right Angled Triangle

A

B

C

Now, Suppose BC = AB = a. Then by Pythagoras Theorem, AC² = AB² + BC² = a² + a² = 2a², and, therefore, AC = a 2 Using the definitions of the trigonometric ratios, we have : sin 45° =

a 1 AB = = AC a 2 2

As cosec is reciprocal to sin So, cosec 45°= 2

a 1 BC = = AC a 2 2 And sec 45°= 2 AB a Tan 45°= = =1 BC a cos 45°=

And cot 45°= 1 Trigonometric Ratios of 30° and 60° Following triangle ABC is an equilateral triangle and AD ⊥ BC. So ∠ ACD =60° And ∠ BAD =30°. As AD bisects BC, so AB=2BD Or in the right angle triangle ABD h= 2b So, p²= h²-b² = 4b²-b²=3b² Or, p =b

3

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Finish Line & Beyond BD b 1 = = AB 2b 2 AD b 3 3 cos 30°= = = AB 2b 2 b 1 BD tan 30°= = = AD b 3 3

Now sin 30°=

A

B

D

C

Now, similarly ratios for 60° can be calculated as follows:

AD b 3 3 = = AB 2b 2 BD b 1 cos 60°= = = AB 2b 2 AD b 3 tan 60°= = = 3 BD b sin 60°=

Trigonometric Ratios of 0° and 90° When the acute angle is reduced upto 0° then the perpendicular becomes zero and the hypotenuse lies on the base. The triangle does not remain there, rather it

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Finish Line & Beyond becomes a straight line. Similarly, when the acute angle is increased to become 90° then the base becomes zero and the hypotenuse lies on the perpendicular.

p 0 = =0 h h b b h cos 0°= = = =1 h b h p 0 tan 0°= = =0 b b

So, sin 0°=

p p h = = =1 h p h b 0 cos 90°= = =0 h h p p tan 90= = =Undefined, also depicted as ∞ b 0 sin 90°=

Angle Ratios:

∠A

sin A cos A tan A

0° 0

30°

45°

60°

1

3 2 1 2

1 2

1

3 2 1

0

2 1 2 1

90° 1 0

3

∞

2

1

3 cosec A

∞

2

2

3 sec A

1

2

cot A

∞

3 3

2 1

2

1

1

0

3 Trigonometric Ratios of Complementary Angles In the right angled triangle both the acute angles are complementary to each other as they add up to 90°.

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Finish Line & Beyond A

B

C

∠ A is complementary to ∠ C AB 1. Now, in ∆ABC, sin C = AC BC And sin A = AC BC Or, sin (90°-C) = AC BC The value of cos C = AC So,

So it can be written that sin(90°-C) = cos C Or, sin (90°-θ) = cos θ 2. cos A =

AB = cos (90°-C) = sin C AC

So, cos (90°-θ) = sin θ

AB BC BC cot C = AB

3. tan C =

tan (90°-C) =

BC = cot C AB

Or, tan (90°-θ) = cot θ And cot (90°-θ) = tan θ

Similarly following relationships can also be calculated: cos (90°-θ) = cosec θ cosec (90°-θ) = cos θ Trigonometric Identities

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Finish Line & Beyond An equation is called an identity when it is true for all values of the variables involved. Similarly, an equation involving trigonometric ratios of an angle is called a trigonometric identity, if it is true for all values of the angle(s) involved. Let us prove the following trigonometric identity: 1. cos2 A + sin2 A = 1 This is true for all A such that 0° ≤ A ≤ 90°. So, this is a trigonometric identity. Solution: cos² A + sin² A

b p )² + ( )² h h b² + p² = h² h² = = 1 proved h² =(

2. 1+tan² A = sec² A solution: b²+ p² = h² dividing the above equation by b² we get

b² p² h² + = b² b² b²

Or, 1 + tan² A = sec² A proved Similarly dividing the equation by p² in stead of b² will give the following trigonometric identity: cot² A + 1 = cosec² A

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