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Reciprocal Trigonometry FunctionsPowerPoint Presentation

Reciprocal Trigonometry Functions

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Reciprocal Trigonometry Functions

Cosecant,

Secant

and Cotangent

Provided sin x 0, cos x 0 and tan x 0

Third letter rule

Example: Find (3 dps)

Answers:

(i) 1.035 (ii) -3.236 (iii) -0.176

Cosec x

x

Graphs of cosec, sec and cotThe graphs of the reciprocal functions can be found by taking the corresponding sine, cosine and tangent graph and calculating the reciprocals of each point on the graph.

S

C

T

ExamplesGiven that sin A =4/5, where A is obtuse, and cosB = 3/2, where B is acute, find the exact values of:

Answers

Examples

Prove that (1 – cos A)(1 + sec A) sin A tan A

L.H.S.

(1 – cos A)(1 + sec A) = 1 + sec A – cos A –Cos A sec A

= 1 + sec A – cos A - 1

= sec A – cos A

= sin A tan A

= R.H.S.

Solving equations

Solve 2 tan2 x – 7 sec x + 8 = 0 for 0 x 360

2 (sec2x – 1) – 7 sec x + 8 = 0

2 sec2x – 2 – 7 sec x + 8 = 0

2 sec2x – 7 sec x + 6 = 0

(2 sec x – 3)(sec x – 2)= 0

sec x = 3/2 or sec x = 2

cos x = 2/3 or cos x = ½

x = 48.2 or x = 60

or: x = 360 – 48.2 or x = 360 - 60

complete solution: x = 48.2 or 60 or 300 or 311.8

Solving equations

Solve 2 cos x = cot x for 0 x 360

2 cos x = cos x/ sin x

2 cos x sin x = cos x

2 cos x sin x – cos x= 0

cos x(2 sin x – 1)= 0

cos x = 0 or sin x = ½

cos x = 0 x = 90 or 270

sin x = ½ x = 30 or 330

complete solution: x = 30 or 90 or 270 or 30

Solving equations

Solve 3 cot2 x – 10 cot x + 3 = 0 for 0 x 2

(3 cot x - 1)(cot x – 3) = 0

cot x = 1/3 or cot x = 3

tan x = 3 or tan x = 1/3

tan x = 3 x = 1.24c or 4.39c

tan x = 1/3 x = 0.32c or 3.46c

complete solution: x = 0.32c or 1.24c or 3.46c or 4.39c

Solving equations

Solve 5 cot2 x – 2 cosec x + 2 = 0 for 0 x 2

5(cosec2 x – 1) – 2 cosec x + 2 = 0

5cosec2 x – 5 – 2 cosec x + 2 = 0

5cosec2 x – 2 cosec x - 3 = 0

sin x = -5/3 not possible or sin x = 1 x = /2

Additional formulae

sin (A + B) = sin A cos B +sin B cos A

sin (A - B) = sin A cos B -sin B cos A

cos (A + B) = cos A cos B- sin A sin B

cos (A - B) = cos A cos B+ sin A sin B

Examples

Find the exact value of sin 75

sin (A + B) = sin A cos B +sin B cos A

sin (30 + 45) = sin 30 cos 45 +sin 45 cos 30

Examples

Express cos (x + /3) in terms of cos x and sin x

cos (A + B) = cos A cos B- sin A sin B

cos (x + /3) = cos x cos /3- sin /3sin x

Double angle formulae

sin (A + B) = sin A cos B +sin B cos A

sin (A + A) = sin A cos A + sin A cos A

sin 2A = 2 sin A cos A

cos (A + B) = cos A cos B - sin A sin B

cos (A + A) = cos A cos A- sin A sin A

cos (A + A) = cos2A - sin2A

cos 2A = cos2A - sin2A

cos 2A = 2cos2A - 1

cos 2A = 1 – 2sin2A

1

A

15

ExamplesGiven that cos A = 2/3, find the exact value of cos 2A.

cos 2A = 2cos2A - 1

Given that sin A = ¼ , find the exact value of sin 2A.

sin 2A = 2 sin A cos A

Solving equations

Solve cos 2A + 3 + 4 cos A = 0 for 0 x 2

=2 cos2A - 1+ 3 + 4 cos A = 0

=2 cos2A + 4 cos A + 2= 0

= cos2A + 2 cos A + 1 = 0

= cos2A + 2 cos A + 1 = 0

= (cos A + 1)2 = 0

= cos A = - 1

A =

Solving equations

Solve sin 2A = sin A for - x

=2sin A cos A = sin A

=2 sin A cos A – sin A = 0

= sin A(2 cos A – 1) = 0

sin A = 0 or cos A = ½

sin A = 0 A = - or 0 or

cos A = ½ A = - /3 or /3

Complete solution: A = - or - /3 or 0 or /3 or

Solving equations

Solve tan 2A + 5 tan A = 0 for 0 x 2

tan A = 0 A = 0 or or 2

7 – 5tan2 A = 0 tan A = 7/5 A = 0.97 , 2.27, 4.01 or 5.41c

Complete solution: A= 0.97 , 2.27, 4.01, 5.41c 0, or 2

Harmonic form

If a and b are positive

a sin x + b cos x can be written in the form R sin( x + )

a sin x - b cos x can be written in the form R sin( x - )

a cos x + b sin x can be written in the form R cos( x - )

a cos x - b sin x can be written in the form R cos( x + )

Examples

Express 3 cos x + 4 sin x in the form R cos( x - )

R cos( x - ) = R cos x cos + R sin x sin

3 cos x + 4 sin x= R cos x cos + R sin x sin

R cos = 3 [1] R sin = 4 [2]

[1]2 + [2]2 : R2 sin2 x + R2 cos2 x = 32 + 42

R2(sin2 x + cos2 x ) = 32 + 42

R2= 32 + 42 = 25 R = 5

[2] [1]: tan = 4/3 = 53.1

3 cos x + 4 sin x = 5 cos( x + 53.1 )

Examples

Express 12 cos x + 5 sin x in the form R sin( x + )

R sin( x + ) = R sin x cos + R cos x sin

12 cos x + 5 sin x= R sin x cos + R cos x sin

R cos = 12 [1] R sin = 5 [2]

[1]2 + [2]2 : R2 cos2 x + R2 sin2 x = 122 + 52

R2(cos2 x + sin2 x ) = 122 + 52

R2= 122 + 52 = 169 R = 13

[2] [1]: tan = 5/12 = 22.6

12 cos x + 5 sin x = 13 sin( x + 22.6 )

Examples

Express cos x - 3 sin x in the form R cos( x + )

R cos( x + ) = R cos x cos - R sin x sin

cos x - 3 sin x = R cos x cos - R sin x sin

R cos = 1 [1] R sin = 3 [2]

[1]2 + [2]2 : R2 cos2 x + R2 sin2 x = 12 + (3 ) 2

R2(cos2 x + sin2 x ) = 12 + 3

R2= 1 + 3 = 4 R = 2

[2] [1]: tan = 3 = 60

cos x + 3 sin x = 2 cos( x + 60 )

Solving equations

Solve 7 sin x + 3 cos x = 6 for 0 x 2

R sin( x + ) = R sin x cos + R cos x sin

7 sin x + 3 cos x= R sin x cos + R cos x sin

R cos = 7 [1] R sin = 3 [2]

R2 = 72 + 32 R = 7.62

[2] [1]: tan = 3/7 = 0.405c (Radians)

7 sin x + 3 cos x = 7.62 sin( x + 0.405)

7.62 sin( x + 0.405 ) = 6 x + 0.405 = sin-1(6/7.62)

x + 0.405 = 0.907 or 2.235

x = 0.502c or 1.830c

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