Download Reciprocal Trigonometry Functions

Reciprocal Trigonometry Functions
Cosecant,
Secant and Cotangent
1
1
S ec x 
Tanx 
Cosx
Cotx
Provided sin x  0, cos x  0 and tan x  0
1
Co sec x 
S in x
Third letter rule
1
Co sec x 
S in x
1
S ec x 
Cosx
1
Cot x 
Ta n x
Example: Find (3 dps)
( i )cos ec 75 ( ii )sec108 ( iii )cot 280
Answers:
(i) 1.035
(ii) -3.236 (iii) -0.176
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.
Cosec x
2.0
1.0
x
0.0
-360
-270
-180
-90
0
-1.0
-2.0
90
180
270
360
Graphs of cosec, sec and cot
sec x
2.0
1.0
x
0.0
-360
-270
-180
-90
0
-1.0
-2.0
90
180
270
360
Graphs of cosec, sec and cot
cot x
4.0
2.0
0.0
-360
-270
-180
-90
0
-2.0
-4.0
90
180
270
360
x
Examples
Find the exact values of:
( i ) sec 23 ( ii ) cos ec 6 ( iii ) cot 6
Answers
( i ) sec
2
3

1
cos120
1
 cos 60

( ii ) co sec 6  sin160 
( iii ) cot 6  tan160 
1
3
1
1
2

  11  2
2
2
3
3
S
T
A
C
Examples
Given that sin A =4/5, where A is obtuse, and cosB = 3/2,
where B is acute, find the exact values of:
( i ) sec A ( ii ) cos ecB ( iii ) cot A
Answers
( i ) sec A cos1 A   315  53
( ii ) co sec B  sin1 B 
1
1
2
2
( iii ) cot A tan1 A   413   34
S
T
A
C
Trigonometric Identities
sin x  cos x  1
2
2
sin x
 tan x
cos x
cos x
 cot x
sin x
1  tan x  sec x
2
2
1  cot x  cosec x
2
2
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
1

 cos A
cos A
1  cos 2 A

cos A
sin 2 A

cos A
sin A sin A

cos A
= sin A tan A
= R.H.S.
Examples
Prove that cot A + tan A  sec A cosec A
L.H.S.
cot A  tan A 
cos A sin A

sin A cos A
cos 2 A  sin2 A

sin Acos A

1
sin Acos A
 cos ecAsec A
= R.H.S.
Examples
cos ec 
Pr ovethat sec  
cos ec   sin 
2
R.H.S.
cos ec 

cos ec   sin  1
1
sin 
 sin 
sin 
1

1  sin 2 

1
cos 2 
 sec 2 
= 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
or: x = 360 – 48.2 or
x = 60
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
5
2
2


3

5

2
sin
x

3
sin
x0
2
sin x sin x
3 sin 2 x  2 sin x  5  0
( 3 sin x  5 )(sin x  1 )  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
tan A  tan B
tan( A  B ) 
1  tan Atan B
tan A  tan B
tan( A  B ) 
1  tan Atan 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
1
2
2
3
 


2 2
2
2
2
6
2 6



4
4
4
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 /3 sin x
1
3
 cos x 
sin x
2
2
Examples
sin( A  B )
Pr ove that tan A  tan B 
cos Acos B
L.H.S.
tan A  tan B 
sin A sin B

cos A cos B
sin Acos B  sin B cos A

cos Acos B

sin( A  B )
cos Acos B
= R.H.S.
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
Double angle formulae
tan A  tan B
tan( A  B ) 
1  tan Atan B
tan A  tan A
tan( A  A ) 
1  tan Atan A
2 tan A
tan 2 A 
1  tan 2 A
Pr ovethat tan 2 A  tan A  tan A sec 2 A
2 tan A
2 tan A  tan A( 1  tan 2 A ) tan A( 1  tan 2 A )
 tan A 

2
2
1  tan A
1  tan A
1  tan 2 A
tan A sec 2 A tan A

 tan A sec 2 A
2
1  tan A
cos 2 A
Examples
Given that cos A = 2/3, find the exact value of cos 2A.
cos 2A = 2cos2A - 1
2
2
 2    1
3

8
1
1  
9
9
Given that sin A = ¼ , find the exact value of sin 2A.
sin 2A = 2 sin A cos A
4
1
15
15
A
 2 

15
4
4
8
1
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
2 tan A
2

5
tan
A

2
tan
A

5
tan
A(
1

tan
A)  0
2
1  tan A
 tan A[ 2  5( 1  tan2 A )]  0
 tan A[ 2  5  5 tan2 A )  0
 tan A[ 7  5 tan2 A )  0
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 +  )
R  a b
2
2
R cos   a and R sin   b
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