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GRADE 12 TRIGONOMETRY
TRIGONOMETRIC IDENTITIES:
Grade 11 Revision: Worksheet handed out on 18/04
COMPOUND ANGLE IDENTITIES:
cos(A – B) = cosA.cosB + sinA.sinB
cos(A + B) = cosA.cosB - sinA.sinB
sin(A – B) = sinA.cosB – cosA.sinB
sin(A + B) = sinA.cosB + cosA.sinB
Examples: Use the appropriate compound angle formulae to verify that:
Example 1:
Example 2:
sin(90° + ) = cosθ
cos3θ.sinθ – cosθ.sin3θ = - sin2θ
Solution:
Solution:
LHS = sin(90° + )
= sin90°.cosθ + cos90°.sinθ
= 1.cosθ + 0
= cosθ
LHS = cos3θ.sinθ – cosθ.sin3θ
= sinθ.cos3θ – cosθ.sin3θ
= sin(θ - 3θ)
= sin(-2θ)
= - sin2θ
Examples: Evaluate without using a calculator:
Example 3:
Example 4:
sin 60°.cos30° - sin30°.cos60°
Solution:
cos79°.cos 311° + sin101°.sin49°
= sin(60° - 30°)
= sin 30°
= cos79°.cos 49° + sin79°.sin49°…reduction formula
= cos(79° - 49°)
=
= cos30°
=
Example 5: cos 75°
Example 6: sin 285°
= cos(30° + 45°)
= sin (225o + 60o)
= cos 30o.cos 45o - sin 30o.sin 45o
= sin 225o.cos 60o + sin 60o.cos 225o
 3  2   1  2 

   

 




 2  2   2  2 
= (– sin 45o)(cos 60o) + (sin 60o)( – cos 45o)
2

6
2

4
4

2  1   3 
2
   
 

  
 2   2  2 
2






6 2
4


2 ( 3  1)
4

DOUBLE-ANGLE IDENTITIES
sin2A = 2sinA.cosA
cos2A = cos2A– sin2A
cos2A = 1 – 2 sin2A
cos2A = 2cos2A – 1
Note
*
1
2
*
sin 4A = 2sin2A.cos2A
sin 2A = sinA.cosA
Additional Examples: Prove the following identities:
1.
tan A 
1.
1  cos 2A
sin 2A
Solution:
R.H.S.




(sin 2 A  cos 2 A)  (cos 2 A  sin 2 A)
2 sin Acos A
2
sin A  cos 2 A  cos 2 A  sin 2 A
2 sin Acos A
2
2 sin A
2 sin Acos A
sin A
cos A
= tan A
= L.H.S.
2
6

4
4
 2 (1  3 )
4
3
cos 2 x  sin 2 x cos x  sin x

(cos x  sin x)3
1  sin 2 x
2.
Solution:L.H.S. 
3.
(cos x  sin x)(cos x  sin x)
(cos x  sin x)(cos x  sin x) 2
cos 2 P
 cos P  sin P
cos P  sin P
L.H.S.
cos x  sin x
cos x  2 sin x cos x  sin 2 x
cos x  sin x

1  2 sin x cos x
cos x  sin x

1  sin 2 x

2

cos 2 P  sin 2 P
cos P  sin P

(cos P  sin P)(cos P  sin P)
cos P  sin P
 cos P  sin P
= R.H.S.
= R.H.S.
RESTRICTIONS ON IDENTITIES:



AN IDENTITY IS NOT VALID WHEN ANY DENOMINATOR IS ZERO.
AN IDENTITY IS NOT VALID WHEN THERE IS A NEGATIVE VALUE UNDERNEATH THE
Tanθ is undefined for θ = 90° + 180°.k (remember the graph of tan?)
SOLVING TRIGONOMETRIC EQUATIONS: GRADE 11 REVISION
Summary of general solutions:
When the interval is not restricted, trigonometric equations have infinite many solutions due to the cyclic nature of
Trigonometric Functions (Graphs). The general solution of a trigonometric equation is the solution that holds for any
interval given.
The solution:
 = a + k.360o
OR
 = a + k.180o
sin  & cos  graphs
tan  graph
where ‘a’ represents an angle and k  Z
(remember the period of the graph)
A
ratio (angle) = fraction
Example 1
sin  = 1
Solution
 = 45 + k.360o
or
 = 135o + k.360o
;
kZ
Example 2
Solution
cos  = 0,5
 = 60o + k.360o
or
 = 300o + k.360o
;
kZ
Example 3
Solution
tan  = – 1
 = 135o + k.180o
or
 = 315o + k.180o
;
kZ
2
o
4
Example 4
Solution
tan( θ2  15o )   32
PAA : 56,309932 …ْ
Q2
o
θ
 15 = 180 - 56,3..ْ + k.180 or
2
θ=
2
138,7 + k.180
 = 277,4 + k.360 ;
θ
2
θ
2
Q4
 15 = 360 - 56,3.. + k.180
o
= 318,7 + k.180
k Z
The general solution for sin and cos is +k.360o as the functions repeat themselves every 360o (period of sin
and cos is 360o)
The general solution for tan is + k.180 as tan repeats itself every 180 (period of tan is 180)
Example 5
sin(3 + 30) = 0,734
Solution
PAA : 47,222785 …
Q1
Q2
o
3 + 30 = 47,2.. + k.360
3 = 17,2..o + k.360
 = 5,7o + k.120
B
or
3 + 30 = 180 – 47,2..o + k.360
3 = 102,7..o + k.360
 = 34,3o + k.120 ; k  Z
Use of Identities and factorisation
Example 1
3 sin2 + 2 sin  – 1 = 0. Determine the general solution and hence solve for ,  ε[
Solution
Let sin  = k
 3k2 + 2k – 1 = 0
(3k – 1)(k + 1) = 0
k = 13
 sin  = 13
or
k=–1
or sin  = – 1
  = 19,5o + k.360o or  = 160,5o + k.360o or  = 270o + k.360o ; k  Z
SS:  ε {19,5o ; 160,5o}
Example 2
Solution
6cos2x + cos x – 2 = 0
Find
1.
The general solution
2.
x for x  [-90; 270]
(3cos x + 2)(2cos x – 1) = 0
cos x =  23
or
cos x = 12
PAA: 48,18 … PAA: 60
x = 131,8o + k.360 or x = 228,2 + k.360 or x = 60 + k.360 or x = 300 + k.360 ; k  Z
x = 131,8; 228,2; 60 or – 60
SS: x ε {-60 o ; 60 o ;131,8o ; 228,2o}
Example 3
Solve for   [0; 360]
sin2 – sin .cos  – 2 cos2 = 0
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(sin  – 2 cos )(sin  + cos ) = 0
sin  = 2 cos or
sin  = – cos 
tan  = 2
or
tan  = – 1
PAA: 63.434949 …
or
PAA: 45
 = 63,4  + k.180
or
 = 135 + k.180
 ε {63,4; 135; 243,4 or 315}
Solution
C
;
kZ
;
kZ
Use of co- ratios
Example 1
sin x = cos 20. Solve for x [-360; 360]
Solution
sin x = cos(90 – 70)
sin x = sin 70
PAA: 70
x = 70 + k.360
or
x = 110 + k.360
x = 70 or x = - 290
or
x = 110 or x = - 250
SS: x ε {- 290; - 250; 70; 110}
Example 2
sin 3x = cos 2x. Solve for x [-180; 180]
Solution
sin 3x = sin (90 – 2x)
PAA: 90 – 2x
3x = 90 – 2x + k.360 or
3x = 180 – (90 – 2x) + k.360
5x = 90 + k.360
or
3x = 90 + 2x + k.360
x = 18 + k.72
or
x = 90 + k.360
;
kZ
x = 18 or x = 90 or x = 162 or x = - 54 or x = -126
SS: x ε {- 126; - 54; 18; 90; 162}
D
General Solution of certain special angles

sin x = 1
x = 90 + k.360


; kZ

sin x = 0
x = k.180 ; k  Z
x = 0o + k.360o
or
x = 180o + k.360o

E

sin x = – 1
x = – 90 + k.360 ; k  Z
or x = 270 + k.360
; kZ
cos x = – 1
x = 180 + k.360
; kZ
cos x = 0
x = 90 + k.180
kZ

cos x = 1
x = k.360 ; k  Z
i.e. x = 0o + k.360o

tan x = 0
x = k.180 ; k  Z
General Solution of where some ratios are undefined
tan x
;
x = 90 + k.180 ;
kZ
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GR.12 TRIGONOMETRY WORKSHEET 1
1.
y
r
sinA =
hence
y = r sinA
y
and
P(x; y)
cosA =
hence
A
O
x
x
r
x = r cosA
The POLAR CO-ORDINATES of P are:
______________________________
2. sin2A + cos2A = ______________
A
3. The COSINE RULE :
In  ABC:
BC2 = __________________________________
B
C
4. (xcosA-xsinA)2 = ____________________________________________
5. The DISTANCE FORMULA:
B(cosP; sinP)
AB2 =_____________________________________
_____________________________________
_____________________________________
A(cosQ; sinQ)
7
6.
P
Q
A
B
O
X
Consider the circle of radius 1 unit, with centre the point O at the origin. Let P and Q be the
POˆ Q  A  B
points on the circumference with POˆ X  A and QOˆ X  B .
Complete:
P is the point:(________; _________)
Q is the point:(_________; ________)
By the distance formula: (1)
PQ2 = ______________________________________________________________
______________________________________________________________
______________________________________________________________
______________________________________________________________
______________________________________________________________
By the Cosine rule: (2)
PQ2 = ______________________________________________________________
______________________________________________________________
From (1) and (2) above make cos(A-B) the subject of the formula.
___________________________________________________________________
___________________________________________________________________
HENCE: cos(A-B) = _________________________________________
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DEDUCTIONS: Complete:
1. cos(A + B) = cos[A – (-B)]
___________________________________________________
___________________________________________________
2. sin(A – B) = cos[90° - (A –B)]
= cos[(90° - A) + B]
or
cos[(90° + B) - A]
___________________________________________________
___________________________________________________
___________________________________________________
3. sin(A + B) =___________________________________________________
___________________________________________________
___________________________________________________
___________________________________________________
Using ordinary addition we have:
Complete:
sin2A = sin(A + A)
= ___________________________________________________
= ___________________________________________________
cos2A = ___________________________________________________
= ___________________________________________________
= ___________________________________________________