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 cot.

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Reciprocal trigonometry functions
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


Graphs of cosec sec and cot

Cosec x

x

Graphs of cosec, sec and cot

The 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.


Graphs of cosec sec and cot1

sec x

x

Graphs of cosec, sec and cot


Graphs of cosec sec and cot2

cot x

x

Graphs of cosec, sec and cot


Examples

A

S

C

T

Examples

Find the exact values of:

Answers


Examples1

A

S

C

T

Examples

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

Answers



Examples2
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.


Examples3
Examples

Prove that cot A + tan A  sec A cosec A

L.H.S.

= R.H.S.


Examples4
Examples

R.H.S.

= R.H.S.


Solving equations
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 equations1
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 equations2
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 equations3
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
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


Examples5
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


Examples6
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


Examples7
Examples

L.H.S.

= R.H.S.


Double angle formulae
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



Examples8

4

1

A

15

Examples

Given 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 equations4
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 equations5
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 equations6
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
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 +  )


Examples9
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 )


Examples10
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 )


Examples11
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 equations7
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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