Trigonometry II – Mathematics Form 3 Notes and K.C.S.E Topical Q&A
Trigonometry II  Mathematics Form 3 Notes and K.C.S.E Topical Q&A
Trigonometry II – Mathematics Form 3 Notes and K.C.S.E Topical Q&A
Trigonometry II – Mathematics Form 3 Notes and K.C.S.E Topical
Trigonometry II – Mathematics Form 3 Notes and K.C.S.E Topical Q&A
The Unit Circle
It is circle of unit radius and centre O (0, 0).
From 0^{0} – 90^{0} is the first quadrant
From 90^{0} – 180^{0} is the second quadrant
From 180^{0} – 270^{0} is the third quadrant
From 270^{0} – 360^{0} is the forth quadrant
Trigonometry II – Mathematics Form 3 Notes and K.C.S.E Topical Q&A
An angle measured anticlockwise from positive direction of x – axis is positive. While an angle measured clockwise from negative direction of x – axis is negative.
In general, on a unit circle
cos θ^{0} = x co – ordinate of p.
sin θ^{0} = y co – ordinate of p.
tan θ^{0} = y coordinate of p. = sinθ
x coordinate of p cosθ
Trigonometry II – Mathematics Form 3 Notes and K.C.S.E Topical Q&A
Trigonometric Ratios of Negative Angles
In general
 sin(0^{0}) = – sinθ
 cos(0^{0}) = cosθ
 tan (0^{0}) = – tanθ
Use of Calculators
Example
Use a calculator to find Tan 30^{0}
Trigonometry II – Mathematics Form 3 Notes and K.C.S.E Topical Q&A
Solution
 Key in tan
 Key in 30
 Screen displays 0.5773502
 Therefore tan 300 = 0.5774
To find the inverse of sine cosine and tangent
 Key in shift
 Then either sine cosine or tangent
 Key in the number
Trigonometry II – Mathematics Form 3 Notes and K.C.S.E Topical Q&A
Note;
Always consult the manual for your calculator. Because calculators work differently
Radians
One radian is the measure of an angle subtended at the centre by an arc equal in length to the radius of the circle.
Because the circumference of a circle is 2πr, there are 2π radians in a full circle.
Degree measure and radian measure are therefore related by the equation 360° = 2π radians, or 1 80° = π radians.
The diagram shows equivalent radian and degree measures for special angles from 0° to 360° (0 radians to 2π radians).
You may find it helpful to memorize the equivalent degree and radian measures of special angles in the first quadrant. All other special angles are just multiples of these angles.
Trigonometry II – Mathematics Form 3 Notes and K.C.S.E Topical Q&A
Example
Convert 125^{0} into radians
Solution
If 1^{c} = 360^{0} = 57.29
2π
Therefore 125^{0}= 125 = 2.182 to 4 S.F
57.29
Trigonometry II – Mathematics Form 3 Notes and K.C.S.E Topical Q&A
Example
Convert the 60^{0} to radians, giving your answer in terms π.
Solution
3600 = 2π^{c}
Therefore
60^{0} = ( 2π x 60)^{c}
360
= (^{π}/_{3})^{c}
Example
What is the length of the arc that that subtends an angle of 0.6 radians at the centre of a circle of radius 20 cm.
Solution
1^{c} is subtended by 20 cm
therefore 0.6^{c} is subtended by 20×0.6 cm = 12 cm
Trigonometry II – Mathematics Form 3 Notes and K.C.S.E Topical
Simple Trigonometric Graphs
Graphs of y=sin x
The graphs can be drawn by choosing a suitable value of x and plotting the values of y against the corresponding values of x.
The black portion of the graph represents one period of the function and is called one cycle of the sine curve.
Trigonometry II – Mathematics Form 3 Notes and K.C.S.E Topical Q&A
Example
Sketch the graph of y = 2sin x on the interval [–π, 4π].
Solution:
Note that y = 2 sin x = 2
(sin x) indicates that the yvalues for the key points will have twice the magnitude of those on the graph of y = sin x.
x  ^{π}/_{2}  π  3π  2π 
Y=2sin x  2  0  2  0 
To get the values of y substitute the values of x in the equation y =2sin x as follows
y=2 sin (360) because 2 π is equal to 360^{0}
Note;
 You can change the radians into degrees to make work simpler.
 By connecting these key points with a smooth curve and extending the curve in both directions over the interval [–π, 4π], you obtain the graph shown in below.

Trigonometry II – Mathematics Form 3 Notes and K.C.S.E Topical Q&A
Graphs of y=cos x
Example
Sketch the graph of y = cos x for 0^{0} ≤ x ≤ 360^{0} using an interval of 30^{0}
Solution:
The values of x and the corresponding values of y are given in the table below
x  0^{0}  30^{0}  60^{0}  90^{0}  120^{0}  150^{0}  180^{0}  210^{0}  240^{0}  270^{0}  300^{0}  330^{0}  360^{0} 
Y=cos x  1  0.8660  0.5  0  0.5  0.8660  1  0.8660  0.5  0  0.5  0.8660  1 
Definition of Amplitude of Sine and Cosine Curves
The amplitude of y = a sin x and y = a cos x represents half the distance between the maximum and minimum values of the function and is given by Amplitude = a
Graph of Tangents
Note;
 As the value of x approaches 90^{0} and 270^{0} tan x becames very large
 Hence the graph of y =tan x approaches the lines x =90^{0} and 270^{0} without touching them.
 Such lines are called asymptotes
Solution of Triangles
Sine Rule
If a circle of radius R is circumscribed around the triangle ABC ,then ^{a}/_{sin A} = ^{b}/_{sin B }= ^{c}/_{sin C }= 2R.
The sine rule applies to both acute and obtuse –angled triangle.
Example
Solve triangle ABC, given that CAB = 42.90, c= 14.6 cm and a =11 .4 cm
Solution
To solve a triangle means to find the sides and angles not given
^{a}/_{sin A} = ^{c}/_{sin C}
^{11.4}/_{sin 42.9} = ^{14.6}/_{sin C}
Sin C =14.6 sin42.9 = 0.8720
11.4
Therefore C =60.69^{0}
Trigonometry II – Mathematics Form 3 Notes and K.C.S.E Topical Q&A
Note;
The sine rule is used when we know
 Two sides and a nonincluded angle of a triangle
 All sides and at least one angle
 All angles and at least one side.
Cosine Rule
a^{2} = b^{2} + c^{2 }– 2bccos A OR b^{2} = a^{2} + c^{2} – 2accos B
Example
Find AC in the figure below, if AB= 4 cm , BC = 6 cm and ABC =78^{0}
Solution
Using the cosine rule
b^{2} + c^{2} – 2 accos B
b^{2} = a^{2} + c^{2} – 2 accos B
b^{2} = 4^{2} + 6^{2} – 2 x 4 x 6cos 780
= 16 + 36 – 48 cos 780
= 52 – 9.979
= 42.02 cm
Trigonometry II – Mathematics Form 3 Notes and K.C.S.E Topical Q&A
Note;
The cosine rule is used when we know
 Two sides and an included angle
 All three sides of a triangle
Past KCSE Questions on the Topic.
 Solve the equation
Sin ^{5θ}/_{2} = –^{1}/_{2} for 0^{0 }≤ θ≤ 180^{0}  Given that sin θ = 2/3 and is an acute angle find:
 Tan θ giving your answer in surd form
 Sec^{2}θ
 Solve the equation 2 sin2(x30^{0}) = cos 60^{0} for – 180^{0} ≤ x ≤ 180^{0}
 Given that sin (x + 30)^{0} = cos 2x^{0} for 0^{0}≤ x ≤90^{0} find the value of x. Hence find the value of cos^{2}3x^{0}.
 Given that sin a =^{1}/_{√5} where a is an acute angle find, without using Mathematical tables
 Cos a in the form of a√b, where a and b are rational numbers
 Tan (90^{0} – a).
 Give that xo is an angle in the first quadrant such that 8sin^{2}x + 2 cos x 5=0
Find: Cos x
 tan x
 Given that Cos 2x^{0} = 0.8070, find x when 0^{0 }≤ x ≤ 360^{0}
 The figure below shows a quadrilateral ABCD in which AB = 8 cm, DC = 12 cm, ∠ BAD = 45^{0}, ∠ CBD = 90^{0} and ∠BCD = 30^{0}.
Find: The length of BD
 The size of the angle ADB
 The diagram below represents a school gate with double shutters. The shutters are such opened through an angle of 63^{0}. The edges of the gate, PQ and RS are each 1.8 m
Calculate the shortest distance QS, correct to 4 significant figures  The figure below represents a quadrilateral piece of land ABCD divided into three triangular plots. The lengths BE and CD are 100m and 80m respectively. Angle ABE = 30^{0 }∠ACE = 45^{0} and ∠ACD = 100^{0
}
 Find to four significant figures:
The length of AE
The length of AD
The perimeter of the piece of land  The plots are to be fenced with five strands of barbed wire leaving an entrance of 2.8 m wide to each plot. The type of barbed wire to be used is sold in rolls of lengths 480m. Calculate the number of rolls of barbed wire that must be bought to complete the fencing of the plots.
 Find to four significant figures:
 Given that x is an acute angle and cos x = ^{2√}^{5}/_{5}, find without using mathematical5 tables or a calculator, tan (90 – x)^{0}.
 In the figure below ∠A = 62^{0}, ∠B = 41^{0}, BC = 8.4 cm and CN is the bisector of ∠ACB.
Calculate the length of CN to 1 decimal place.  In the diagram below PA represents an electricity post of height 9.6 m. BB and RC represents two storey buildings of heights 15.4 m and 33.4 m respectively. The angle of depression of A from B is 5.50 While the angle of elevation of C from B is 30.5^{0} and BC = 35m.
 Calculate, to the nearest metre, the distance AB
 By scale drawing find,
 The distance AC in metres
 ∠ BCA and hence determine the angle of depression of A from C
 Calculate, to the nearest metre, the distance AB
Trigonometry II – Mathematics Form 3 Notes and K.C.S.E Topical Q&A
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