Trigonometry

GIKPKC7 94107

Trigonometry

Page 1

Introduction 3/6/98

Trigonometric Ratios: Sine Cosine Tangent

O H A cos   H O tan   A

(90  )

sin  

O = Opposite Hint: ‘O’ on top, ‘H’ below, ‘A’ where it fits.

H = Hypotenuse

A = Adjacent



Inverse Ratios: Cosecant Secant Cotangent

1 H  sin  O 1 H sec    cos  A 1 A cot    tan  O cos ec 

Complementary Angles: A  sin 90    H O cos 90     H A tan 90    O cos ec90    

H A

H O A  cot 90    O sec90    

Results from Complementary Angles and Inverse Ratios: sin  = cos (90  ) cos  = sin (90  ) tan  = cot (90  ) cosec  = sec (90  ) cot  = tan (90  ) sec  = cosec (90  )

Luke Cole

Page 1

GIKPKC7 94107

Trigonometry

Page 2

Exact Ratios: Hint: 1

0

30

45

60

90

0

1 2

1

3 2 1 2

1

3

E

2 = 0.707 … 3 2 sin cos

1

tan

0

2 1

3 2 1

2 1

0

3

Angle of Depression & Elevation: 1

1 = Depression 2 = Elevation 1 = 2

2

Bearings: Compass

N A  N 50 E B  S 65 W

50 A W

E 25 B S N 50 A

W

E 25 B

Luke S Cole

Page 2

GIKPKC7 94107

Trigonometry

True North

Page 3

N 0

A  50 B  245

50 A 270

90 25 B 180

Angles of Any Magnitude: y 90 2nd Quadrant 1st Quadrant S A (sin +ve) (All +ve) x 180 0 T C (tan +ve) (cos +ve) 3rd Quadrant 4th Quadrant

Hint: All Stations Too Central

270 Proof: Construct a unit circle [r = 1, c(0, 0)]: 1st Quadrant y P(x, y)





x

x

y y 1 x cos    x 1 y tan   x  All +ve sin  

2nd Quadrant y P(x, y)

y y 1 x  cos 180      x 1 y tan 180     x  sin +ve sin 180     

Luke Cole

Page 3

GIKPKC7 94107

Trigonometry

3rd Quadrant

4th Quadrant



x

Page 4



P(x, y)

x P(x, y)

y

y y sin 180      y 1 x  cos 180     x 1 y y  tan 180      x x  tan +ve

y  y 1 x  cos 360      x 1 y  tan 360    x  cos +ve

sin 360   

Trigonometric Identities

10/6/98

From the Unit Circle: ¼ of it:

1

y x

Equation:

tan  

Proof: From unit circle: sin  = y …(1) cos  = x …(2) y tan   …(3) x Sub (1) & (2) into (3): sin  tan    cos  Equation: Proof: Using Pythagoras: x2 + y2 = 12

sin  cos 

sin2  + cos2  = 1

Luke Cole

Page 4

GIKPKC7 94107 

Trigonometry

sin2  + cos2  = 1

Page 5

…[1]

Equation: cosec2   cot2  = 1 Proof: Divide [1] by sin2 : sin 2  cos 2  1   2 2 sin  sin  sin 2   cosec2   cot2  = 1 Equation: sec2   tan2  = 1 Proof: Divide [1] by cos2 : sin 2  cos 2  1   2 2 cos  cos  cos 2   sec2   tan2  = 1

Sine Rule: Equation: Proof:

sin A sin B  a b

h c h = c.sin B h sin C  b h = b.sin C So, (1) = (2): c.sin B = b.sin C sin B sin C   b c Here, sin B 

A

…(1) c

h

b

…(2) B

a

C

Cosine Rule: Equation:

a2 = b2 +c2 – 2.b.c.cosA

Proof:

Luke Cole

or

cos A 

b2  c 2  a 2 2.b .c

Page 5

GIKPKC7 94107 In ACD, In BCD, In ACD, So,

Trigonometry

x2 + p2 = b2 (c – x)2 + p2 = a2 c2 – 2.c.x + b2 = a2 x cos A  b x = b.cosA a2 = b2 + c2 – 2.b.c.cosA

Page 6

C b

A

x

p

a

D c x B c

Area of a Triangle Rule: Equation: A = ½.a.b.sinC Proof: Since, A = ½.b.h h And, sin C  a h = a.sin C  A = ½.b.a.sin C

B a

C

h

b

c

A

More Special Results: Equation: cos (x – y) = cos x.cos y + sin x.sin y Proof: From the unit circle: P

Q y

  A x O x B At point Q(x, y): (1) cos  = x (2) sin  = y  Q(cos , sin )

At point P( x, y): (1) cos(180  ) =  x  cos  =  x cos  = x (2) sin (180  ) = y sin  = y  P(cos , sin )

Now, using the distance formula: PQ2 = (cos   cos )2 + (sin   sin )2 PQ2 = cos2   2.cos  .cos  + cos2  + sin2   2.sin  .sin  + sin2  Luke Cole

Page 6

GIKPKC7 94107

Trigonometry

Page 7

PQ2 = 2 – 2(cos  .cos  + sin  .sin ) …(1) Using the cos Rule: PQ2 = 12 + 12 – 2.1.1.cos (  ) PQ2 = 2 – 2.cos (  ) …(2) So, (1) = (2): 2 – 2(cos  .cos  + sin  .sin ) = 2 – 2.cos (  )  cos (  ) = cos  .cos  + sin  .sin  Now let  = x &  = y: cos (x – y) = cos x.cos y + sin x.sin y …[1] Equation: cos (x + y) = cos x.cos y  sin x.sin y Proof: From [1] let y =  y: cos [x – ( y)] = cos x.cos ( y) + sin x.sin ( y)  cos (x + y) = cos x.cos y  sin x.sin y

Equation: sin (x + y) = sin x.cos y + cos x.sin y Proof: From [1] let x = 90 – x: cos [(90 – x) – y] = cos (90 – x).cos y + sin (90 – x).sin y cos [90 – (x + y)] = sin x.cos y + cos x.sin y  sin (x + y) = sin x.cos y + cos x.sin y …[2] Equation: sin (x – y) = sin x.cos y – cos x.sin y Proof: From [2] let y =  y: sin [x + ( y)] = sin x.cos ( y) + cos x. sin ( y)  sin (x – y) = sin x.cos y – cos x.sin y Equation:

tan  x  y  

tan x  tan y 1  tan x . tan y

Proof: Since:

sin x  y  cos  x  y  sin x. cos y  cos x . sin y = cos x .cos y  sin x . sin y Divide by cos x. cos y: Luke Cole tan x  y  

Page 7

GIKPKC7 94107

Trigonometry

Page 8

sin x .cos y  cos x. sin y cos x .cos y cos x .cos y  sin x . sin y cos x .cos y tan x  tan y tan x  y   1  tan x. tan y tan x  y  



tan  x  y  

Equation:

…[3]

tan x  tan y 1  tan x . tan y

Proof: From [3] let y =  y:

tan x  tan y  1  tan x .tan  y  tan x  tan y tan x  y   1  tan x. tan y tanx   y  



Ratios of Double Angles: Equation: sin (2.x) = 2.sin x.cos x Proof: Since, sin (x + y) = sin x.cos y + cos x.sin y sin (x + x) = sin x.cos x + cos x.sin x = 2.sin x.cos x  sin (2.x) = 2.sin x.cos x Equation:

cos (2.x) = cos2 x  sin2 x = 2.cos2 x  1 = 1 – 2.sin2 x

Proof: Since, cos (x + y) = cos x.cos y  sin x.sin y cos (x + x) = cos x.cos x  sin x.sin x = cos2 x  sin2 x  cos (2.x) = cos2 x  sin2 x Formula: Proof: Since, tan x  y  

tan 2.x  

2. tan x 1  tan 2 x

tan x  tan y 1  tan x. tan y Luke Cole

Page 8

GIKPKC7 94107

Trigonometry

Page 9

tan x  tan x 1  tan x. tan x 2.tan x = 1  tan 2 x 2.tan x tan 2.x   1  tan 2 x tan x  x  



Ratios in Terms of “T Results”: Equation:

sin x 

2.t 1 t

Proof: Since, sin (2.x) = 2.sin x.cos x x x sin x  2. sin .cos 2 2 t 1 . = 2. 2 2 t 1 t 1 2.t sin x   1 t2 Equation:

cos x 

where, t  tan

2

t2  1 x 2

1t2

t 1

where, t  tan

2

1t Proof: Still working with the above triangle Since, cos (2.x) = cos2 x  sin2 x x x cos x  cos 2  sin 2 2 2

Luke Cole

x 2

x 2

Page 9

GIKPKC7 94107

Trigonometry 2

 1   t     cos x     2  2 t  1 t  1     =



1

t2 1 1 t2 cos x  1 t2

Equation:



2

Page 10

Note: T-results will always solve any trigonometry equations except when  = 180 (Since, tan 90 = Err)

t2 t2  1

tan x 

2.t 1 t2

where, t  tan

x 2

Proof: 2.tan x 1  tan 2 x x 2.tan 2 tan x  x 1  tan 2 2 2.t tan x  1 t2

Since, tan 2.x  



General Solutions 31/7/98 Equation:

sin  = sin  is  = 180  n + ( 1)n   n = Integer

Proof by e.g.: Find all the solutions for sin  



3 , here the domain is not restricted. 2

 = 60, 120,…… +  = 60, (180 – 60), (360 + 60), (540 – 60), …  can also be negative   = – (180 + 60), – (360 – 60), – [360 – (180 – 60)], … = (– 180 – 60), (– 360 + 60), (– 540 – 60), … The general solution for sin  = sin 60 is:  = 180  n + (– 1)n  60

Equation:

cos  = cos  is  = 360  n   n = Integer

Equation:

tan  = tan  is  = 180  n +  n = Integer Luke Cole

Page 10

GIKPKC7 94107

Trigonometry

Page 11

Note: These above equations are for +ve angles, change the equation to suite you need if the angle is –ve.

Trigonometry Graphs 16/9/98

y = sin x: 1 

1 Period: Amplitude:

T = 2. rad yA = 1 unit

y = cos x:

Luke Cole

Page 11

GIKPKC7 94107

Trigonometry

Page 12

Luke Cole

Page 12

1 

1 Period: Amplitude:

T = 2. rad yA = 1 unit

y = tan x: 1 

1

Period: Asymptote:

T =  rad  x  rad 2