Right angled triangle C is the hypotenuse (Always the longest side)

For angle θ (a) is the opposite and (b )is the

adjacent

For angle α (b) is the opposite and (a) is the

adjacent

ca

bθ

α

Trigonometry Functions

Sine = opposite hypotenuse

Cosine = adjacent hypotenuse

Tangent = opposite adjacent

This is aTrigonometric Identity

Also, if we divide Sine by Cosine we get:

Also, if we divide Sine by Cosine we get:

Sin θ = Opposite/hypotenuse Cos θ = Adjacent/hypotenuse

The hypotenuses cancel each other out]So we get Opposite/adjacent which is Tan (tangent) so

Tan θ = sin θ cos θ

More functions

Cosecant Function: csc(θ) = Hypotenuse / Opposite

Secant Function:

sec(θ) = Hypotenuse / Adjacent

Cotangent Function:

cot(θ) = Adjacent / Opposite

We can also divide "the other way around" (such as hypotenuse/opposite instead of Opposite/hypotenuse):

which will give us three more functions

So using the inverse of these we get:

sin(θ) = 1/cosec(θ)

cos(θ) = 1/sec(θ)

tan(θ) = 1/cot(θ)

More functions

Also the other way around:cosec(θ) = 1/sin(θ)

sec(θ) = 1/cos(θ)

cot(θ) = 1/tan(θ)

And we also have:

(cot(θ) = cos(θ)/sin(θ) (from tan = sin/cos)

Pythagoras

ca

b

a2 + b2 = c2

C is the hypotenuse (the longest side)

θ

TRIGONOMETRY

Pythagoras

a2 + b2 = c2

Can be written as a2 b2 c2

c2 c2 c2+ = = 1

proof

sin θ can be written as a/c and cos θ can be written as b/c (cos)

(a/c)2 is sin2θ and(b/c)2 is cos2θ (a/c)2 +(b/c)2 = 1

sosin2θ + cos2θ = 1

(sin2 θ) means to find the sine of θ, then square it. (sin θ2) means square θ, then find the sine

Rearranged versions

sin2θ = 1 − cos2θ

cos2θ = 1 − sin2θ

Rearranged versionssin2θ cos2θ 1

cos2θ cos2θ cos2θ

tan2θ + 1 = sec2θ

tan2θ = sec2θ − 1 Or

sec2θ = 1 + tan2θ

+ =

OR

Rearranged versions

sin2θ cos2θ 1 sin2θ sin2θ sin2θ

1 + cot2θ = cosec2θ

cot2θ + 1 = cosec2θor

cosec2θ = 1 + cot2θ

+ =

Circular motion to sine curve

0o 90o 180o 270o 360o

time

y = R.sinθ

In the triangle formed by the first part of the motion. The vertical line (y) is the opposite of the angle formed (θ) and the hypotenuse is the radius of the circle (R)

Sinθ = y/R so y = R.sin θ. At 90o sinθ = 1 so y = R

θ yR

More sine curves

y = Rsinθ

y =2Rsinθ

y =0.5Rsinθ

y = sin2θ

90o 180o 270o 360o

For y = sin2θ two waves fit in 360o

For y = sin3θ three waves fit in 360o and so on

For y =sin 0.5θ one wave would stretch over 720o

Cosine curves (cosine 90o = 0) (cosine 0o =1)

Graph of sin2θ

Has to be positive because we cannot have a minus squared number

Graph of cos2θ

C

AS

T

A = All positive

S = Only sine positive

T = Only tangentpositive

C = only cosine positive

0o

90o

180o

270o

360o

C.A.S.T

Finding sin, cos and tan of angles.

Sin 245o = sin(245o – 180o) = sin 65o = 0.906Sin 245o = - 0.906 (third quadrant = negative)

Sin 118o = sin (180o – 118o) = sin 62o = 0.883Sin 118o = + 0.883 (second quadrant positive)

Cos 162o = cos (180o – 162o) = cos 18o = 0.951Cos 162o = - 0.851 (second quadrant negative)

Cos 285o = cos(360o – 285o) = cos 75o = 0.259Cos 285o = + 0.259 (fourth quadrant positive)

Tan 196o = tan(196o – 180o) = tan 16o = 0.287Tan 196o = + 0.287 (third quadrant positive)

Tan 282o = tan(360o – 282o) = tan 78o = 4.705Tan 282o = - 4.705 (fourth quadrant negative)

Finding angles

First quadrant θ

2nd quadrant 180 – θ

3rd quadrant 180 + θ

4th quadrant 360 – θ

Finding angles

90o 180o 270o 360o0o

30o

30o 180o -30o = 150o

180o + 30o =210o

360o - 30o =330o

150o

210o 330o

C

AS

T

Finding angles

Find all the angles between 0o and 360o to satisfy the equation8sinθ -4 = 0 (rearrange)

8sinθ = 4Sinθ = 4/8 = 0.5

Sin-10.5 = 30o

and180o – 30o

= 150o

Find all the angles between 0o and 360o to satisfy the equation

6cos2θ = 1.8 (rearrange)cos2θ = 1.8÷6 = 0.3

cosθ = √0.3= ± 0.548

Cos-1 +0.548 (1st and 4th quadrant positive) = 56.8o and 360o – 56.8o = 303.2o

Cos-1 - 0.548 (2nd and 3rd quadrant negative) 180o – 56.8o = 123.2o

and 180o + 56.8o =236.8o

Finding angles

90o 180o 270o 360o0o

56.8o

56.8o 123.2o

56.8o

236.8o 303.2o

Red 1st and 4th quadrant(positive cos)

Blue 2nd and 3rd quadrant(negative cos)

Finding angles

• Solve for all angle between 0° and 360° • 2Tan2 B + Tan B = 6• (let Tan B = x) so• 2x2 + x = 6 or 2x2 + x – 6 = 0

• then solving as a quadratic equation using formula:• x = -b +/- √(b2 - 4ac) / 2a • Where a = 2; b= 1; and c = - 6

Finding angles• x = -1+/- √(12 – 4x2x-6) / 4

• = -1+/- √(1 + 48) / 4= -1+/- √(49) / 4= -1+/- (7) / 4+6/4 or -8/4

Tan B = 1.5 or -21st and 3rd quadrant 56.3o or(180 + 56.3) = 236.3o

2nd quadrant (180 - 63.43) = 116.57o

• 4th quadrant (360 – 63.43) = 296.57o

•

Formulae for sin (A + B), cos (A + B), tan (A + B) (compound angles)

• sin (A + B) = sin A cos B + cos A sin B

• sin (A - B) = sin A cos B - cos A sin B

• cos (A + B) = cos A cos B - sin A sin B

• cos (A - B) = cos A cos B + sin A sin B

• These will come in handy later

a sin θ ± b cos θ•can be expressed in the form

•R sin(θ ± α),•R is the maximum value of the sine

wave•sin(θ ± α) must = 1 or -1

• (α is the reference angle for finding θ)

Finding α

Using sin(A + B) = sin A cos B + cos A sin B, (from before)

•we can expand R sin (θ + α) as follows:• R sin (θ + α)

•  ≡ R (sin θ cos α + cos θ sin α) •  ≡ R sin θ cos α + R cos θ sin α

Finding α• So

a sin θ + b cos θ ≡ R cos α sin θ + R sin α cos θ

• a = R cos α

• b = R sin α

Finding α

b ÷ a =

•R sin α ÷ R cos α =

tan αtan α = b/a

Using the equation• Now we square the equation

• a2 + b2

• =R2 cos2 α + R2 sin2 α

• = R2(cos2 α + sin2 α)

• = R2 • (because cos2 A + sin2 A = 1)

compound angle formulae:

•Hence•R = √a2 + b2

•R2 = a2 + b2

•(pythagoras)

The important bits

tan α = b/a

•R2 = a2 + b2

•(pythagoras)

For the minus case

• a sin θ − b cos θ • =

• R sin(θ − α)

• tan α = b/a

Cosine version• a sin θ + b cos θ ≡ R cos (θ − α)

• Therefore: • tanα= a/b   

• (Note the fraction is a/b    for the cosine  case, whereas it is b/a  for the sine  case.)

• We find R the same as before:• R=√a2 +b2   

• So the sum of a sine term and cosine term have been combined into a single cosine term: • a sin θ + b cos θ ≡ R cos(θ − α)

Minus cosine version• If we have a sin θ − b cos θ and we need to express it in terms of a single cosine function, the formula we need to

use is: • a sin θ − b cos θ ≡ −R cos (θ + α)

Graph of 4sinθ

Graph of 4sinθ and 3 cosθ

Resultant graph of 4sinθ + 3 cosθ

The radian

r

Length of arc (s)

The radian is the length of the arc divided by the radius of the circle, A full circle is 2π radians

That is 3600 = 2π radians

Circular motion to sine curve

π/2 π Radians3π/2 2π

time

y = R.sinθ

In the triangle formed by the first part of the motion. The vertical line (y) is the opposite of the angle formed (θ) and the hypotenuse is the radius of the circle (R)

Sinθ = y/R so y = R.sin θ. At π/2 sinθ = 1 so y = R

θ yR

0

Angular velocity (ω)

• Angular velocity is the rate of change of an angle in circular motion and has the symbol

ωω = radians ÷ time (secs)

Angles can be expressed by ωt

example

• For the equation 3.Sin ωt - 6.Cos ωt:

• i) Express in R.Sin (ωt - α) form

• ii) State the maximum value

• iii) Find value at which maximum occurs •

example

R=√32 +62  R =√9 +36

R =√45R = 6.7

Maximum value is 6.7 

example

Tan α = b/aTan α = 6/3

= 263.4o

or1.107 radians

63.4o x π ÷ 180

example• Maximum value occurs when Sin (ωt - 1.1071) = 1, or (ωt

- 1.1071) = π/2 radians

• Since π/2 radians = 1.57, then (ωt - 1.107) = 1.57 rad.

• Therefore ωt = 1.57 + 1.107 = 2.6781 radians

• Maximum value occurs at 2.678 radians

examplea) 3Sin ωt - 6Cos ωt = 6.7Sin (ωt - 1.107)

b) maximum = 6.7

c) Maximum value occurs at 2.6781 radians