CHAPTER ONE
TRIGONOMETRY (CALCULAS)
1.1. Radians
So far we have given solutions to trigonometric equations in degrees. However,
using the fact that 180 = π radians.
Consider the circle with, centre O, radius r and an arc AB subtending an angle of x
at O.
(a) The length of an arc of a given circle is proportional to the angle it subtends
at the centre. But an angle of 360 is subtended by an arc of length 2πr,
therefore an angle of x is subtended by an arc of length × 2πr
∴The length of the arc AB is × .
(b) The area of a sector of a given circle is proportional to the angle at the centre.
But a sector containing an angle of 360 is the whole circle, which has an
area of πr , therefore a sector containing an angle of x has an area of
× πr .
∴ The area of the sector OAB is × .
Example 1
Convert the following to degrees
(a) (b)
Solution
(a) = = 90
×
(b) = = 120
Example 2
Convert to radians, leaving π in your answer:
(a) 300 (b) 450
Solution
SIMPLIFYING PURE MATHEMATICS 2: BY MUKOOBA JOEL 0774454151: 0706355884 Page 1
(a) 300 = × 300 =
(b) 450 = × 450 =
We can give the solutions of trigonometric equations in radians.
Example 3
Solve the equation 2 sin θ − cos θ = 0, for −π ≤ θ ≤ π
Solution
Rearranging gives
2 sin θ = cos θ
⟹ =1
⟹ tan θ =
⟹ θ = tan
⟹ θ = 0.46 rad.
The other solution in the required range is θ = −π + 0.46 = −2.68 rad.
The solution are θ = 0.46 rad and −2.68 rad for the range −π ≤ θ ≤ π
Example 4
Solve the equation 2 cos θ = sec θ, for −2π ≤ θ ≤ 2π. Give your answer in radians.
Solution
Since sec θ = , we have
2 cos θ =
⟹ 2cos θ = 1
⟹ cos θ = ±
√
⟹θ=±
The other solutions are θ = ± 2π − =± π
∴ The solutions are θ = ± , ± π for the range −2π ≤ θ ≤ 2π
Exercise 1.1
1. Convert the following to degrees
(a) (b) (c) 4π (d)
2. Convert the following to radians leaving π in your answer
(a) 150 (b) 15 (c) 270
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3. What is the length of an arc which subtends an angle of 0.8 rad at the centre of a
circle of radius 10 cm? Ans(8cm)
4. The area of a sector of a circle, diameter 7 cm, is 18.375 cm2. What is the length
of the sector?
5. In the triangle XYZ, x = 29, y = 21 and z = 20. Calculate;
(a) the area of the triangle,
(b) the length of the perpendicular from Z to XY.
6. Solve each of the following equations for 0 ≤ θ ≤ 2π rad, giving your answers in
radians.
(a) cos θ = 0.4 (b) sin θ = 0.7 (c) tan θ = 4
Ans{(a)1.16, 5.12 (b)0.78, 2.37 (c)1.33, 4.47}
7. Solve each of the following equations for −π ≤ θ ≤ π giving your answers in
radians correct to two decimal places.
(a) sin(θ − 0.1) = 0.2 (b) cos(θ + 0.2) = 0.6 (c) tan(θ − 2) = 3
Ans{(a)0.30, 3.04 (b) − 1.13, 0.73 (c) − 1.39, 1.75 }
8. Solve each of the following equations for 0 ≤ θ ≤ 2π, giving your answer in
radians in terms of π
(a) cosec2θ = 2 (b) tan 2θ − = 1 (c) cos 3θ + cos θ = 0
Ans (a) , , , (b) , , , (c) , , , , ,
1.2. Derivatives of Trigonometric Functions
Small angles
When θ is a very small angle then
(i) sin θ ≈ θ
(ii) cos θ ≈ 1
(iii) tan θ ≈ ≈ ≈θ
The above approximations are required when proving the derivatives of
trigonometric functions from first principles.
( ) ( )
We can use the formula f (x) = lim → to differentiate from first
principles or use the approach of increments.
Here we look at the derivative of trigonometric functions from first principles
(1) y = sin x then = cos x
Proof
As x increases by δx then y will increase by δy
⟹ y + δy = sin(x + δx)
⟹ δy = sin(x + δx) − sin x
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Applying the factors formula
⟹ δy = 2 cos sin
= 2 cos sin
= 2 cos x + sin
⟹ =
since δx is very small, then sin →
×
⟹ =
As δx → 0, →
∴ = cos x … … … … … … … … . . ∎
(2) y = tan x then = sec x
Proof
Let y = f(x)
( ) ( )
⟹ f (x) = lim →
( )
= lim →
( )
( )
= lim →
( ) ( )
= lim → ( )
( )
= lim → ( )
( )
= lim → ( )
Since h is very small then sin h → h
= lim → ( )
=
=
= sec x … … … … … … … … . ∎
(3) If y = secx then = sec x tan x
Proof
As x increases by δx then y will increase by δy
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⟹ y + δy = sec(x + δx)
⟹ δy = sec(x + δx) − sec x
1 1
⟹ δy = −
cos(x + δx) cos x
cos x − cos(x + δx)
⟹ δy =
cos(x + δx) cos x
−[cos(x + δx) − cos x]
⟹ δy =
cos(x + δx) cos x
− − 2sin x + sin
⟹ δy =
cos(x + δx) cos x
δx δx
Since δx is very small, sin →
2 2
δy 2sin x + .
⟹ =
δx δx cos(x + δx) cos x
δy dy
As δx → 0, →
δx dx
dy sin x 1 sin x
∴ = = = sec x tan x … … … … . ∎
dx cos x cos x cos x cos x
Differentiation of
To differentiate functions of the form sin cos , we use the chain rule.
For example, if = sin 4 , then let = 4 gives
= =4
⟹ = =4
By the chain rule,
= × = ×4
∴ =4 4
In practice, we write = 4 × (4 ) = 4 4
Example 5
Find for each of the following functions.
(a) = cos 3 (b) = sin( + 2) (c) = cos √
Solution
(a) When = cos 3 ; = − sin 3 × (3 ) = −3 sin 3
(b) When = sin( + 2) ; = cos( + 2) × ( + 2) = 2 cos( + 2)
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(c) When = cos √ ; = − sin √ × √ =− sin √ =− sin √
√
So since (sin ) = and (cos ) = −
We also have the following integrals:
∫ =− +
∫ = +
Note when integrating and we proceed as follows,
∫ = sin + and ∫ = − cos +
Proof
Let = , = , =
∫ =∫ = ∫ = sin + = sin +
Example 6
Find each of the following integrals
(a) ∫ 5 (b) ∫ cos( − 2) (c) ∫ sin cos
Solution
(a) ∫ 5 = − cos 5 +
(b) Let = − 2, =3 ,⇒ =
1 1
⇒ cos( − 2) = cos = sin + = sin( − 2) +
3 3 3
(c) ∫ sin cos =∫ ( 2 ) = − 2 + =− 2 +
Differentiation of and
To differentiate functions of the form and , we also use the chain rule. For
example , if = , then this can be written as =( ) . Letting = gives
= and =
∴ = 2 and =−
By chain rule;
= . = 2 (− )
∴ = −2
Example 7
Find for each of the following functions.
(a) = (b) =( + ) (c) = 2
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Solution
(a) When = ; =4 ×( ) =4
(b) When =( + ) ; = 5( + ) ×( + )
= 5( + ) ( − )
(c) When = 2 =( 2 ) ; = 3( 2 ) ×( 2 )
= 3( 2 ) (−2 2 )
∴ = −6 2 2
Example 8
Find for each of the following functions
(a) = (b) = 2 (c) =
Solution
(a) =
Using the product rule, we have
= ( ) + ( )
= +
(b) = 2
Using the product, we have;
= (−2 2 )+ 2 (2 )
= −2 2 + 2 2
∴ = 2 ( 2 −2 )
(c) =
Using the quotient rule, we have;
( ) − ( )
=
( )− (− )
=
+
=
1
=
∴ =
Example 9
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Find each of the following integrals.
(a) ∫ cos (b) ∫
√
Solution
(a) In this case cos is a derivative of sin
Let = , = , = cos
⟹ ∫ cos =∫ = + = +
(b) We notice that 2 + is a function and cos is the derivative
⟹Let = 2 + , = , =
⟹∫ =∫ = 2√2 + +
√ √
Exercise 1.2
1. Differentiate the following with respect to from first principles
(a) = (b) = √ (c) = sin
2. Find for each of the following
(a) = 3 (b) =8 (c) = cos ( + 3) (d) = 8 sin
3. Differentiate each of the following with respect to
(a) = (b) =3 (2 + 3) (c) = sin( −3 ) (d) = cos
4. Find ( ) for each of the following 2x
(a) ( ) = (b) ( ) = (c) ( ) = (d) ( ) = 2√ 4
5. Find for each of the following
(a) = (1 + ) (b) = (c) = √1 − 6 (d) = (1 + )
(e) = − (f) =( + 2 ) (g) =−
√
6. Differentiate each of the following with respect to .
(a) = (b) = (c) = (d) =3 2 (e) = (f)
= (g) =
7. Show that =
8. Given that = 3 + 3 , where A and B are constants, show that
+9 =0
9. Given that = +3 , show that + =1
10. Given that = 4 , show that = −16
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11. Given that = 2 , show that −2 =
12. Find each of the following integrals
(a) ∫ 2 2 (b) ∫ cos (2 − 1) (c) ∫ cos( )
√
(d) ∫ 2( − 2) cos( −4 ) (e) ∫ (f) ∫ sin
√
13. Find each of these integrals
(a) ∫ 4 cos (b) ∫ (c) ∫ ( )
(d) ∫
√
(e) ∫(1 − )( − ) (f) ∫ 2 4 √6 + 4
Differentiation of , , and
If = , ℎ =
If = , ℎ =−
If = , ℎ =
If = , ℎ =−
Example 10:
Find for each of these functions.
(a) = 3 (b) = sec (2 − 1) (c) =4
Solution
(a) When = 3 ; = 3 (3 ) = 3 3
(b) When = sec (2 − 1) ; = (2 − 1) (2 − 1) × (2 − 1)
=4 (2 − 1) (2 − 1)
(c) Let = , =4
=8 =−
Using the chain rule,
⟹ = × = (8 ) × (− ) = −8
Example 11
Find for each of the following functions
(a) = 3 (b) =
Solution
(a) Using the product rule, we have
=3 ( ) + (3 )
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= −3 +3
= 3( − )
(b) Using the quotient rule,
2 ( ) − ( 2 )
=
2
2 − (−2 2 2 )
=
2
2 (1 + 2 2 )
=
2
1+2 2
=
2
Note the following
(i) ∫ = +
(ii) ∫ = +
(iii) ∫ =− +
(iv) ∫ =− +
Therefore ∫ = +
Example 12
Find each of these integrals,
√
(a) ∫ 2 3 3 (b) ∫ (1 − ) (c) ∫
√
Solution
(a) ∫ 2 3 3 = sec 3 +
(b) ∫ (1 − ) = − tan(1 − )+
√
(c) ∫ = −2 cot √ +
√
Applications
Example 13
Find the equation of the tangent to the curve = + at the point where = .
Solution
We need when = , since = + , we have
= 1+
When = , =1+ = 1+ =1+ =3
The gradient of the tangent line is 3. Therefore the equation of the tangent is of the form
=3 +
When = , = + 1
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∴ The tangent passes through the point , +1
⟹ +1=3 + , =1−
4 4 2
The equation of the tangent is = 3 +1− or 2 − 6 = 2 −
Example 14
A curve is given by the equation = 2 =2 . Find the equation of the
normal to the curve at the point where = .
Solution
We first need to find by differentiation parametrically
When =2 , =6 cos
When =2 , =6 (− ) = −6
By the chain rule,
= ×
6 cos
= =−
−6
When = , =− =−
√
Therefore the gradient of the normal is √3. The normal has equation of the form
= √3 +
√
When = , =2 = and =2 =
√
The normal passes through the point ,
√
⟹ = √3 + , = −2
∴ the equation of the normal is = √3 − 2
Example 15
Find the area enclosed between the curve = cos 2 , the x-axis and the y-axis
Solution
= 2
0
4
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=∫ 2
=
( )
= −
Example 16
A curve is given by the equations = 2 and =2 . Find the equation of the
normal to the curve at the point where =
Solution
We first find by differentiating parametrically.
When =2 , =6 cos
When =2 , =6 (− sin ) = −6 sin
By the chain rule,
= ×
=− = − tan
When = , = − tan =−
√
Therefore the gradient of the normal is √3.
The normal has the equation of the form = √3 +
√
When = , =2 = and =2 =
√
The normal passes through the point ,
√
⟹ = √3 + , = −2
∴ The equation of the normal is = √3 − 2
Differentiation of inverse trigonometric function
The inverse trigonometric functions can be differentiated as follows,
1. Given that = sin
⟹ =
SIMPLIFYING PURE MATHEMATICS 2: BY MUKOOBA JOEL 0774454151: 0706355884 Page 12
⟹ =1
⟹ =
⟹ =
⟹ =
⟹ =
√
2. Given that = tan
⟹ tan =
⟹ =1
⟹ =
⟹ =
⟹ =
3. Given that = sec
⟹ =
⟹ =1
⟹ =
⟹ =
⟹ =
⟹ =
√
Exercise 1.3
1. Find ( ) for each of the following
(a) ( ) = (b) ( ) = (c) ( ) = (d) ( ) = − 5
2. Find for each of the following
(a) = (1 + ) (b) = (c) =
√ ( )
3. Differentiate each of the following
(a) (b) (c) 3 (d) (e) (f) (g)
4. Show that =
5. Show that =
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6. Given that = +2 , show that +3 =2
7. Given that = , show that (1 + ) + =1
8. Given that = , show that = 4 (1 + )
9. Find the equation of the tangent to the curve = + at the point where =
10. Find the equation of the tangent and the normal to the curve = at the point
where = .
11. The normals to the curve = 2 at the points , 0 and , 0 meet at the
point C. Find the coordinates of the point C, and the area of the triangle ABC.
, , .
12. Find the equation of the tangent and the normal to the curve = at the point
where =
13. Find the coordinates of the points on the curve = (2 + 1), in the range
− ≤ ≤ , where the gradient is − . − , −√3 , , √3
14. Find the equation of the tangent and the normal to the curve ¸ = 6− , at the
point 2, . (4 + 5 = + 10, 20 − 16 = 5 − 32)
15. Given that = 2 , show that = 4 (2 − 1)
16. Find each of the following integrals,
(a) ∫ 2 (b) ∫ 3 6 6 (c) ∫ 8 tan (d) ∫ 4 8
17. Find each of the following integrals,
(a) ∫ 5 (b) ∫ (3 + ) (c) ∫ ( )
(d) ∫ 5 √2 + cot 5 (e)∫ tan (f) ∫ √cot
( )
(g) ∫ (ℎ) ∫ (i) ∫ √ +1 (j) ∫ ( ) ( )
√
(k) ∫ ( + 5) (l) ∫
√
18. Differentiate the following with respect to x
(a) cos (b) cot (c)
19. Find in the simplest form,
√
(a) (b) (c) (d) sin
√
20. Given that = − , show that −2 1− =0
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