Right angled triangle
description
Transcript of Right angled triangle
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