Download Trigonometry Review (Parts 1,2,3 and Assessment)

Additional Mathematics
5/2/05
Trigonometry I – Unit 1
Notes and Examples
opposite
hypotenuse
cos x =
adjacent
hypotenuse
tan x =
sin2 x + cos2 x = 1
tan x =
sin x
cos x
1
SIN +ve
ALL +ve
TAN +ve
COS +ve
sin x =
opposite
adjacent
2
4
3
For angles > 90°
Example 1
Triangle ABC is right angled at B
a) Find AB given that AC = 3.2cm and ∠ C = 35°
b) Find ∠ A given that BC = 2.6cm and AB = 3 cm
c) Find AC given that ∠A = 41° and AB = 3cm
Solutions
a)
AB
= sin 35!
AC
b)
tan A =
c)
Be careful when finding the hypotenuse!
AB
AB
= cos 41!
∴ AC =
AC
cos 41!
BC
AB
∴ AB = AC sin35°
i.e. AB = 1.84cm
∴ ∠A = 40.9°
i.e. AC = 3.98cm
Example 2
Find θ within the range 0 ≤ θ ≤ 360°
a) sin θ = − 0.4
b) tan θ = 6.3
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Additional Mathematics
5/2/05
Solutions
a) Because the sin is negative we want two reflex angles i.e. those in the third and
fourth quadrants of the diagram in the box.
The acute angle with a sin = +0.4 is 23.6°
Therefore the angles required are 180 + 23.6 and 360 – 23.6
i.e.
θ = 203.6° or 336.4°
b) We want the angles in the first and third quadrants because tan is +ve
The acute angle is 81° and the angle in the third quadrant is 180 + 81°
i.e.θ = 81° or 261°
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Additional Mathematics
5/2/05
Trigonometry I – Unit 2
Notes and Examples
opposite
hypotenuse
cos x =
adjacent
hypotenuse
tan x =
sin2 x + cos2 x = 1
tan x =
sin x
cos x
1
SIN +ve
ALL +ve
TAN +ve
COS +ve
sin x =
opposite
adjacent
2
4
3
For angles > 90°
Example 1
Solve the following equations for 0 ≤ θ ≤ 360°
a) 4 sin2 x = 3
b) 2 cos2 x = cos x
c) 3cos2 x + 5sin x – 1 = 0
d) 3cos x = 4 sin x
e) 3cos2 x = 2sinx cos x
Solutions
Don’t forget ±
a) 4 sin2 x = 3
⇒ sin2 x = 0.75
∴sin x = ±
i.e.
We need values in all four
quadrants because of the ± sign
0.75
x = 60 °, 120 °, 240°, 300°
2
b) 2 cos x = cos x
2 cos2 x − cos x = 0
We can’t divide by cos x
as it could be zero
Now factorise
cos x (2cos x – 1) = 0
cos x = 0 or cos x =
1
2
It is important that you realise that
you cannot divide an equation by
any quantity that could be zero. It
is not just that you may lose a
solution but that any quantity
divided by zero is undefined.
x = 90°, 270°, 60°, 300 °
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Additional Mathematics
c) 3cos2 x + 5sin x – 1 = 0
3(1 – sin2 x) + 5sin x – 1 = 0
5/2/05
We need the identity
Tidies to give
3sin2 x – 5sin x – 2 = 0
Remember quadratics
must be equal to zero
(3sin x + 1) (sin x – 2) = 0
sin x = −
1
3
i.e. x = 199.5° or 340.5°
d) 3cos x = 4sin x
But sin x ≠ 2
We can divide through by cos x here
as it cannot be zero otherwise sin x
would have to be zero too at the
same time and this is not possible
i.e. tan x = 0.75
x = 36.9° or 216.9°
We have a quadratic with cos x
as a common factor. We can’t
divide by cos x for reasons
given before. Make quadratic
equal to zero, then factorise.
2
e) 3cos x = 2sin x cos x
cos x (3cos x – 2sin x) = 0
cos x = 0 or tan x = 1.5
x = 90°, 270°, 56.3°, 236.3°
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Additional Mathematics
5/2/05
Trigonometry I – Unit 3
Notes and Examples
opposite
hypotenuse
cos x =
adjacent
hypotenuse
tan x =
sin2 x + cos2 x = 1
tan x =
sin x
cos x
1
SIN +ve
ALL +ve
TAN +ve
COS +ve
sin x =
opposite
adjacent
2
4
3
For angles > 90°
Applications including the use of the sine and cosine rules
Example 1
Find all of the angles in triangle ABC given that a = 8cm, b = 9 cm, and c = 10 cm.
Solution
Since any obtuse angle must lie opposite the longest side, it would be sensible to
calculate one of the acute ones first.
cos A =
b2 + c 2 − a 2
2bc
cos A =
(81 + 100 − 64)
180
A = 49.5°
Using the cosine rule
Similarly using the
cosine rule again
B = 58.8°
We could use the sine rule to find B but this would mean using the calculated value
for A and could be inaccurate because of rounding errors.
C = 180 – (A + B)
C = 71.8°
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Additional Mathematics
Chapter 7: Trigonometry I
Chapter Assessment
1
The figure shows the cross section of a roof of a building. The length of the roof AC is
7 metres and the angles at B and C are 520 and 380 respectively.
A
7
C
380
h
x
520
B
Find the height, h, above the base and the length of the roof, AB.
3
3
and tan x = − .
5
4
[5]
2
Find the value of x in the range 0 ≤ x ≤ 360 for which sin x =
3
4
3
You are given that sin y = − and cos y = − . Find the exact value of tanx.
5
5
[2]
4
Solve the equation sin2x = 0.9 in the range 0 ≤ x ≤ 360.
[5]
5
(i)
Show that cosx − 2sin2x = 2cos2x + cosx − 2.
(ii)
Hence, or otherwise, solve the equation cosx − 2sin2x + 2 = 0 in the range 0 ≤ x ≤ 360.
[5]
6
[2]
[2]
Solve the following equations in the range −1800 ≤ x ≤ 1800.
(i)
sinx = 5cosx
[2]
(ii)
3tan2x + 2tanx −5 = 0
[5]
(iii) 4cos2x + sinx − 3 = 0
[5]
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Additional Mathematics
7
In the triangle ABC shown, AC = 8, angle C = 380 and angle B = 520.
A
8
380
C
8
9
520
B
(i)
Find the length of the side AB.
[3]
(ii)
Calculate the area of the triangle.
[3]
In the triangle ABC, AB = 6, BC = 8 and CA = 9.
Calculate the size of the largest angle in this triangle.
[4]
In the quadrilateral ABCD shown below, AB = 8, BC = 5, angle B = 700, angle DAC = 400
and angle DCA = 500.
B
700
5
8
C
0
50
A
45
0
D
(i)
Calculate the length of the diagonal AC.
[4]
(ii)
Calculate the length of the side AD.
[4]
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Additional Mathematics
10
A coast line runs East-West and on the coast there are two harbours, A and B.
A boat sails from A on a bearing of 0400 for 5 miles to point C before altering course to 1350
to sail directly to B.
C
1350
N
5
0
040
B
A
Find
(i)
the distance AB,
[5]
(ii)
the distance sailed.
[4]
Total: 60
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