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# Reciprocal Trigonometry Functions - PowerPoint PPT Presentation

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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Presentation Transcript

Cosecant,

Secant

and Cotangent

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

Third letter rule

Example: Find (3 dps)

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

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.

sec x

x

Graphs of cosec, sec and cot

cot x

x

Graphs of cosec, sec and cot

S

C

T

Examples

Find the exact values of:

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:

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.

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

L.H.S.

= R.H.S.

R.H.S.

= R.H.S.

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

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

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

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

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

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

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

L.H.S.

= R.H.S.

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

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

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 = 

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 

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

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 +  )

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 )

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 )

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 )

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