Download Trigonometry: Revision of grade 11

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Trigonometry: Revision of grade 11
∗
Pinelands High School
This work is produced by OpenStax-CNX and licensed under the
†
Creative Commons Attribution License 3.0
1 Trigonometry Summary
1.1 Basic knowledge and skills required for Grade 12
You should be able to:
1) Write down the denitions of the three trigonometric ratios
Soh Cah Toa ratios in terms of side opposite reference angle, side adjacent to reference angle and
hypotenuse (always the right angle)
c
sin (θ) = opp
hyp = a
adj
cos (θ) = hyp
= ab
opp
tan (θ) = adj = cb
Figure 1
2) Sketch and use the special (standard) triangles:
◦
◦
◦
Triangle with angles 30 , 60 , 90
∗ Version
1.1: Nov 17, 2010 5:41 am -0600
† http://creativecommons.org/licenses/by/3.0/
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Figure 2
and
◦
◦
◦
Triangle with angles 45 , 45 , 90
Figure 3
3) Write down the denitions of the three trigonometric ratios and their reciprocals for angles of any size
in a Cartesian plane (Syr Cxr Tyx)
Shield your rear, 'Cause x-rays Tan your exterior ratios in terms of the radius and of the
co-ordinates of the points at the end of the radius.
sin (θ) = yr
cos (θ) = xr
tan (θ) = xy
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Figure 4
4) Draw the CAST diagram
Which ratios are positive in each quadrant?
Figure 5
5) Distinguish between positive angles (anti-clockwise) and negative angles (clockwise)
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Figure 6
6) Note that:
cos (−θ) = cos (θ)
sin (−θ) = −sin (θ)
tan (−θ) = −tan (θ)
◦
◦
7) Write down and apply the 180 degrees rule ( 180 ± θ, 360 ± θ, etc)
(RATIO does NOT change, SIGN may change based on CAST)
a) Identify quadrant
b) Use CAST diagram to determine SIGN
c) Reduce angle to reference angle
Example 1
◦
◦
◦
◦
cos 180 − θ = −cos (θ) (Second quadrant, CAST: cos negative) tan 210 = tan 180 + 30 =
◦
tan 30
(Third quadrant, CAST: tan positive)
Remember the following when using the 180 degree rule:
- We assume θ is acute First quadrant
◦
- 180 − θ - Second quadrant
◦
- 180 + θ - Third quadrant
◦
- 360 − θ - Fourth quadrant
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Figure 7
◦
8) Write down and apply the 90 degree rule ( 90 ± θ)
(RATIO changes to CO-ratio; SIGN base on CAST and original ratio)
sine and cosine
a) Identify the quadrant
b) Use CAST digrams to determine SIGN and write it down
c) Change ratio to COratio
d) Reduce angle to reference angle
Example 2
◦
sin 90 − θ = cos (θ) (First quadrant; CAST: sin positive)
◦
◦
◦
◦
cos 120 = cos 90 + 30 = −sin 30
(Second quadrant, CAST: cos negative)
9) Write down the Pythagorean identity (Square identity)
sin2 (θ) + cos2 (θ) = 1
(1)
10) Write down the Quotient identity
tan (θ) =
sin (θ)
cos (θ)
(2)
11) Write down the sine rule, cos rule and area rule (for use in triangles)
sinB
sinC
a
b
c
Sine rule: sinA
or sinA
= sinB
= sinC
a = b = c
b2 +c2 −a2
2
2
2
Cosine rule: a = b + c − 2bccosA or cosA =
2bc
Area rule: Area∆ABC = 21 absinC
1.2 Applying your knowledge
(Be aware of the broad categories of questions that may be asked)
You must be able to:
1) Use a sketch to nd x, y or r and then ratios and angles in the Cartesian Plane (all four quadrants)
using the theorem of Pythagoras ( x2 + y 2 = r2 ) and the CAST diagram.
◦
◦
◦
2) Determine the ratio of any positive angle, as well as the angles such as 0 , 90 , 180 , etc, and negative
angles in the Cartesian Plane.
◦
3) Determine angles for which ratios are undened (denominator zero), e.g. tan 90 , etc.
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4) Simplify trigonometric expressions using the theory described in the knowledge section above.
When simplifying expressions:
a) Use reduction formulae to change all angles to acute angles
b) It is often a good idea to write the expression in terms of sine and cosine as far as possible.
c) Look out for double angles and co-ratios
5) Find domain specic solutions to trigonometric equations: e.g. Solve for x if 2sin (x) = 0, 43 and
◦
◦
−360 ≤ x ≤ 360
REMEMBER the basic steps for solving a trigonometric equation:
a) Isolate the ratio with the unknown angle
b) Ignore the +/- sign and determine the reference angle
c) Use the +/- sign to determine the quadrant(s) in which the angle(s) must lie
d) Determine your solution(s), using the reference angle.
Remember the following when working out your solution using the reference angle:
◦
◦
Positive angles ( 0 ≤ θ ≤ 360 )
First quadrant: θ = refangle
◦
Second quadrant: θ = 180 − refangle
◦
Third quadrant: θ = 180 + refangle
◦
Fourth quadrant: θ = 360 − refangle
Figure 8
◦
◦
( −360 ≤ θ ≤ 0 )
◦
First quadrant: θ = 360 + refangle
◦
Second quadrant: θ = −180 − refangle
◦
Third quadrant: θ = −180 + refangle
◦
Fourth quadrant: θ = 0 − refangle
Negative angles
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Figure 9
◦
6) Find the GENERAL Solution to trigonometric equations (at step d) above, remember to add 360 k (for
◦
sine and cosine) or 180 k (for tan) to your answer and to specify that k ∈ Z)
Example 3
◦
2 sin x + 20 + 1 = 43
◦
2 sin x + 20
= −1
4
◦
−1
sin x + 20
= 8
Ref angle: sin−1
3rd quadrant:
1
8
(3)
= 7, 18
◦
◦
◦
x + 20 = 180 + 7, 18 + 360 k
x
or
=
167,18
+
◦
360 k
(4)
4th quadrant:
◦
◦
◦
x + 20 = 360 − 7, 18 + 360 k
x
=
332,82
+
◦
360 k
(5)
( k ∈ Z)
7) Solve right-angled triangles (SohCahToa)
8) Use the sine rule, cos rule and area rule in practical applications, i.e. solving triangles, nding angles
and side lengths, etc.
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Figure 10
9) Solve problems in two dimensions
NB: Make sure that you know how to use your calculator to nd angles and ratios
(Solution on p. 11.)
Exercise 1
In which quadrant does θ lie if:
a) sin (θ) < 0 and tan (θ)
b) cos (θ) > 0 and sin (θ) < 0
c) tan (θ) > 0 and cos (θ) > 0
Exercise 2
◦
◦
(Solution on p. 11.)
If cos (θ) = − √213 and 180 ≤ θ ≤ 360 , use a sketch to determine the value of:
a) tan (θ)
b) sin (θ) cos (θ)
Exercise 3
If tan (θ) = t and θ is acute, determine sin (θ) in terms of t.
Exercise 4
◦
◦
(Solution on p. 11.)
(Solution on p. 11.)
1) If x = 87, 6 and y = 240, 2 , use a calculator to evaluate each of the following correct to two
decimal places:
a) cos (x + y)
b) sin (2x − y) + tan2 (x)
sin(y)
c) cos(x)
+ 3tan (2x)
d) cos(y)
2
2) Without
using
nd the value of:
a calculator,
◦
◦
a) tan 310 sin 60
◦
◦
b) cos2 45 + sin 30
◦
◦
c) cos 30 + sin 60
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(Solution on p. 11.)
Exercise 5
Reduce each
thefollowing to a trigonometric ratio of x:
of
◦
a) sin 180 + x
◦
b) tan 90 + x
◦
c) cos 360 − x
d)
“ ◦
”
cos 90 −x
◦
sin(360 −x)
◦
◦
e) sin (x) − cos 90 − x − tan 180 − x
(Solution on p. 11.)
Exercise 6
1) Evaluate
using a calculator:
without
◦
a) tan 120
◦
b) cos 630
◦
◦
◦
c) sin 150 + tan 330 cos 30
d)
“
”
“
”
◦
◦
tan 315
+cos 300
◦
)+tan(135
) ◦
◦
◦
2) Prove that sin 240 tan 300 + cos 330 =
sin(150
◦
1
2
3+
Exercise 7
1) Use basic trigonometric identities to simplify the following:
a) tan (y) cos (y)
b) tan2 (y) sin2 (y) + tan2 (y) cos2 (y)
2
(y)
c) 1−cos
sin(y)
2) Prove the following identities:
sin2 (x)
= cos (x)
a) 1 − 1+cos(x)
b)
cos(x)
1+sin(x)
+ tan (x) =
√ 3
(Solution on p. 12.)
1
cos(x)
(Solution on p. 12.)
Exercise 8
Find the general solution to the following equations. Give answers to two decimal places:
a) sin (θ) = 0, 515
b) 3 −tan (θ) =2, 4
◦
c) cos θ + 20 = −0, 242
◦
d) 2sin θ − 15 + 1 = 0
◦
e) cos (2θ) = tan 24
Exercise 9
(Solution on p. 12.)
On the same system of axes, drawsketch graphs
of:
◦
◦
◦
f (x) = sin (x) and g (x) = cos 90 + x for the interval −180 ≤ x ≤ 180 . Use the graphs
to answer the following questions:
a) Describe g(x) in terms of a reection of f(x).
b) Explain why f(x) + g(x) = 0 for all values of x.
Exercise 10
Refer to the diagram:
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(Solution on p. 12.)
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Figure 11
a) Calculate the measurement of AB (correct to two decimal places)
b) Calculate the area of the triangle.
(Solution on p. 13.)
Exercise 11
Refer to the diagram.
Figure 12
a) Calculate the length of BC
b) Calculate the size of angle B.
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Solutions to Exercises in this Module
Solution to Exercise (p. 8)
a) Fourth
b) Fourth
c) First
Solution to Exercise (p. 8)
Figure 13
a)
b)
3
2
−3 √
−2
√
13 13
=
6
13
Solution to Exercise (p. 8)
sin (θ) = √
Solution to Exercise (p. 8)
1a) 0,65
b) 568,36
c) -20,97
d) -0,25
√ 3
2a) √13
=
2
2
b) √12 + 12 = 1
√
√
√
c) 23 + 23 = 3
1
2
Solution to Exercise (p. 8)
a) −sin (x)
b) −tan (x)
c) cos (x)
sin(x)
d) −sin(x)
= −1
e) sin (x) − sin (x) + tan (x) = tan (x)
Solution to Exercise (p. 9)
◦
√
1a) −tan 60 = − 3
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t
t2 + 1
(6)
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◦
b) cos 270 = 0
◦
◦
◦
c) sin 30 − tan 30 cos 30 =
d)
“ ◦”
“ ◦”
−tan 45
+cos 60
1
2
−
√
3
√1
3 2
=0
−1+ 1
= 1 −12 = 1
◦
45 )
◦
2
◦ )−tan(
◦
+ cos 30 =
−tan 60
2) LHS = −sin 60
sin(30
◦
√
3
2
√
√
3+
3
2
=
1
2
3+
√ 3 = RHS
Solution to Exercise (p. 9)
sin(y)
cos (y) = sin (y)
1a) cos(y)
2
b) tan (y) sin2 (y) + cos2 (y) = tan2 (y)
c)
sin2 (y)
sin(y)
= sin (y)
1+cos(x)−1+cos2 (x)
= cos(x)(1+cos(x))
= cos (x) = RHS
1+cos(x)
1+cos(x)
sin(x)
2
cos(x)+ cos(x) (1+sin(x))
(x)+sin(x)+sin2 (x)
1
= cos cos(x)(1+sin(x))
= cos(x)
=
(1+sin(x))
2a) LHS =
b) LHS =
RHS
Solution to Exercise (p. 9)
◦
◦
a) 31 + 360k or 149 + 360k, k ∈ Z
◦
b) 31 + 180k, k ∈ Z
◦
◦
c) 84 + 360k or −124 + 360k, k ∈ Z
◦
◦
d) 15 + 360k or 225 + 360k, k ∈ Z
◦
e) 12 + 90k, k ∈ Z
Solution to Exercise (p. 9)
Figure 14
a) Reection in the x-axis
b) If g(x) is a reection of f(x) in the x-axis, then g(x) = -f(x) and hence f(x) + g(x) = 0
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Solution to Exercise (p. 9)
a) 7,92 cm
b) 41, 96cm2
Solution to Exercise (p. 10)
a) 6,56 cm
◦
b) 39, 26
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