Download III. Trigonometry

TRIGONOMETRY REVIEW
Circles
 Angles measured using degrees, radians, or revolutions
 One full revolution around a circle is equal to 360 and 2 radians
 Degree or radian measurements are equivalent over full revolutions
o E.g. 90 is equivalent to 450, 90 plus 360k where k is an integer
 Arc length
o Arc length for whole circle (or circumference) is 2r
o Arc length is equal to r
o Example: Arc Length of circle of radius 2 at =/2

o
arc length = /2 (2) =
Area of a sector:
o area of whole circle is r2
o the area of a sector with angle  is

r2
2
Trigonometric Identities
 sin 2   cos 2   1
 csc 2   cot 2   1

 tan 2   1  sec 2 
 cosA  B  cos A cos B  sin A sin B

 sin( A  B)  sin A cosB  cosAsin B

tan A  tan B

 tan( A  B) 
1 tan A tan B

 cos(90  A)  sin A

 cos(2A)  2cos 2 A 1
 sin( 2A)  2sin A cos A

2tan A

 tan( 2A) 
1 tan 2 A

x 
cos x  1

 cos  
2 
2
x 
1 cos x

 sin   
2 
2
sin 

 tan  
cos
 The phrase SOH – CAH – TOA is a very useful when finding sin , cos , tan 

when given the lengths of sides of a triangle.
Sin=Opposite/Hypotenuse

Cos=Adjacent/Hypotenuse
Tan=Opposite/Adjacent




1
sin 
1
sec 
cos
1
cos
cot  

tan  sin 
csc 
 Proofs


sin(A+B)
sin( A  B) 
RS QP

OR OR
RS QP OQ



 sin A cos B
OR OR OR
RS

 sin A cos B
OR
RS RQ


 sin A cos B
RQ OR
 cos A sin B  sin A cos B
cos(A+B)



RT RS  ST RS ST



OR
OR
OR OR
(1 cos(A  B)) 2  sin 2 (A  B)  (cos A  cos B) 2  (sin B  sin A) 2
1 2 cos(A  B)  cos 2 (A  B) sin 2 (A  B)  cos 2 A  2cos 2 A cos B  cos 2 B sin 2 A  2sin Asin B  sin 2 B
Sorry Arup I was having trouble understanding what to type for the proof.


tan(A+B)
sin( A  B) sin A cos B  cos A sin B

cos(A  B) cos A cos B  sin A sin B
1
sin A cos B  cos B sin B cos A cos B


1
cos A cos B  sin A sin B
cos A cos B
sin A cos B  cos A sin B
cos A cos B

cos A cos B sin A sin B

cos A cos B cos A cos B
tan A  tan B

1 tan A tan B
sin2
sin 2  sin(    )
tan( A  B) 

 sin  cos  cos sin 




 2sin  cos
cos2 
cos 2  cos(   )
 cos cos  sin  sin 
 cos 2   sin 2 
 cos 2   (1 cos 2  )




 2 cos 2   1
tan2 
tan 2  tan(   )
tan   tan 
1 tan  tan 
2 tan 

1 tan 2 

Finding Ratios with out finding the angle.

When given a ratio, for example sin  
2
it is possible to find other ratios such as
5
cos  , tan  , sin2  , cos2   , etc.
 To find the other trig functions with a single  , just use the Pythagorean theorem.
2
Using the example sin  :
5





52-22=x2 or 25-4=21=x2
2
21
So cos  =
, tan  =
, etc.
5
21
To find trig functions with more than 1  , manipulate the trigonometric identities.
2
 o E.g. give
 that sin   to sin2  use sin2  =2sin  cos 
5



2  21 4 21
2sin  cos  = 2 

5  5  25 
 



The Nature of trigonometric graphs
 
 Sinx
o Hasperiod of 2
o Is always continuous
 Cosx
o Has period of 2
o Is always continuous
o Similar to graph of sinx but offset by /2 to the left
 Tanx
o Has period of 
o Is undefined where x=/2 + k
 K is an integer
Composite Function
f(x) = a sin(b(x+c)) + d




a changes the amplitude of the graph.
b changes the period of the graph, condensing or expanding it
c moves the graph left or right
d moves the graph up or down.
Inverse Functions

arcsin x: has a range of (-  ,  )
has a domain of [-1,1]

arccos x: has a range of (-  ,  )
has a domain of [-1,1]
 
, ]
2 2
has a domain of (-  ,  )
Rules and Formulas for Triangles
1
 A = bh
2

arctan x: has a range of [-

Hero’s Formula: A =

s(s  a)(s  b)(s  c) where s =
abc
2
Cosine Rule: c 2  a 2  b 2  2ab cos c
a
b
c


 The Sine Rule:
sin A sin B sin C
1
 Alternate Area: A =
absinC
2
Half Angle Identities:
2 tan A
 We know that tan 2 A 
. From this it can be determined that
1  tan 2 A
2 tan A
1  tan 2 A
sin 2 A 
and
.
cos
2
A

1  tan 2 A
1  tan 2 A

 If we replace 2A by  and use t to denote tan we have:
2
2t
o tan  
1 t2
2t
o sin  
1 t2
1 t2
o cos  
1 t2
More Identities










1
(1  cos 2 )
2
1
cos 2   (1  cos 2 )
2
PQ
P Q
sin P  sin Q  2sin
cos
2
2
PQ
P Q
sin P  sin Q  2 cos
sin
2
2
PQ
P Q
cos P  cos Q  2 cos
cos
2
2
PQ
P Q
cos P  cos Q  2sin
sin
2
2
2sin A cos B  sin( A  B)  sin( A  B)
2 cos A sin B  sin( A  B)  sin( A  B)
2 cos A cos B  cos( A  B)  cos( A  B)
2sin A sin B  cos( A  B)  cos( A  B)
sin 2  
Practice Problems
1) A chord AB divides a circle of radius 2m into two segments. If AB subtends an angle
of 60 at the center of the circle, find the area of the minor segment.
2) Find the sine, cosine, and tangent of 243 .
3) Find the general solution of tan 7  tan 2 .
4) Solve sec2  tan 2   6 for angles in the range 180    180 .
5) Without using tables or a calculator, evaluate:
(a) sin 75
(b) cos105
(c) tan(15 )
6) Prove that:
1  cos 2 A
 tan A
sin 2 A
1  sin 2
7) Express
in terms of tan  .
1  sin 2
Solutions
1.
60 

3
1 2
1
 2 2
r   (22 ) 
m  2.094m 2
2
2
3
3
1  sin 2 1  tan 


Area of  AOB =
1  sin 2 1  tan 
Area of sector AOB =
Area of Minor Segment (shaded in) = (2.094 – 1.732) m 2 = .362 m 2
2.     180  63
sin 243   sin 63  .8910
cos 243   cos 63  .4540
tan 243  tan 63  1.9626
3. tan 7  tan 2 because the tangent function has a period of  ,
7  n  2
5  n

n
5
4.
sec 2  tan 2   6
sec 2   1  tan 2 
1  tan 2   tan 2   6
2 tan 2   5
tan 2   
5
2
5
2
  1.007, 4.149 radians
tan   
  57.689 , 237.689
radians
5.
sin 75  sin(45  30 )  sin 45 cos 30  cos 45 sin 30
2
3
2 1
)( )  ( )( )
2
2
2 2
2
 ( 3  1)( )
4
(a)  (
cos105  cos(60  45 )  cos 60 cos 45  sin 60 sin 45
1
2
3
2
(b)  ( )( )  ( )( )
2 2
2
2
(1  3) 2

4
tan(15 )  tan(45  60 ) 
(c)

tan 45  tan 60
1  tan 45 tan 60
1 3
 32
1 3
6.
1  cos 2 A 1  (cos 2 A  sin 2 A)

sin 2 A
2sin A cos A
2
1  cos A  sin 2 A

2sin A cos A
sin 2 A  sin 2 A
2sin 2 A


2sin A cos A
2sin A cos A
sin A

 tan A
cos A
2t
7. Using sin 2 
where t  tan  gives
1 t2
2t
1  t 2  2t (1  t ) 2
1  sin 2  1 


1 t2
1 t2
1 t2
2t
1  t 2  2t (1  t ) 2
1  sin 2  1 


and
1 t2
1 t2
1 t2
(1  t ) 2
2
1  sin 2
1 t 2
Hence
 1 t 2  (
)
1  sin 2 (1  t )
1 t
1 t2
1  sin 2 1  tan 


1  sin 2 1  tan 