Download Trigonometry (EM)

Higher Maths
Get Started
Revision Notes
Trigonometry
Trigonometry
solve trigonometric equations in a given interval
know and apply the
addition formulae
apply trigonometric formulae in
the solution of geometric problems
know and apply the
double angle formulae
solve trigonometric equations involving
addition formulae
and double angle formulae
y = sin(x)
The graph shows the solution of sin x = 0·7, 0 ≤ x ≤ 2π.
1.5
1
0.5
One of the solutions comes directly from the calculator:
x = sin–1(0·7) = 0·775 radians ( to 3 d.p.)
y = 0·7
0
0
1.57
3.14
4.71
6.28
The other comes from the fact that the sine wave is
symmetrical about x = π/2
… if x is an answer then so is π – x:
-0.5
π – x = π – sin–1(0·7) = 2·37 radians ( to 3 d.p.)
-1
-1.5
y = cos(x)
The graph shows the solution of cos x = 0·7, 0 ≤ x ≤ 2π.
1.5
One of the solutions comes directly from the calculator:
x = cos–1(0·7) = 0·795 radians ( to 3 d.p.)
1
The other comes from the fact that the cos wave is
symmetrical about x = π
0.5
0
0
-0.5
1.57
3.14
4.71
6.28
… if x is an answer then so is 2π – x:
2π – x = 2π – cos–1(0·7) = 2·37 radians ( to 3 d.p.)
-1
-1.5
Test
Yourself?
Standard formulae will be given.
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
Half-square and half-equilateral triangle should be known
√2
π/
π/
1
2
π/
4
1
6
√3
3
1
Test
Yourself?
Standard formulae will be given.
sin 2A = 2 sin A cos A
cos 2A = cos2 A – sin2 A
= 2 cos2 A – 1
= 1 – 2 sin2 A
Appreciate that variations exist
sin 4A = 2 sin 2A cos 2A
cos 4A = cos2 2A – sin2 2A
= 2 cos2 2A – 1
= 1 – 2 sin2 2A
sin A = 2 sin A/2 cos A/2
cos A = cos2 A/2 – sin2 A/2
= 2 cos2 A/2 – 1
= 1 – 2 sin2 A/2
Test
Yourself?
To solve equations of the form p cos 2x + q sin x = 0
•
expand the term in 2x … choosing the form which suits the other term
: p [1 – 2sin2 x] + q sin x = 0
•
express as a quadratic equation:
p + q cos x – 2p sin2 x = 0
•
solve the quadratic for two values of cos x
•
propagate solutions for both values of cos x.
Test
Yourself?
Remember basic geometric facts:
•
The sum of the angles of a triangle is 180˚
So, if A, B, and C are the three angles of a triangle
C = 180 – (A + B)
sin C = sin(180 – (A + B)) = sin (A + B)
a
•
•
•
b
a/
If an acute angle P has a sine of b then
• a right angle can be drawn thus:
• the third side can be calculated by Pythagoras viz. √(b2 – a2)
• the cosine and tangent can be read from the triangle.
P
When working with parallel lines, remember
• alternate angles are equal: x = y so, for example, sin x = sin y
• corresponding angles are equal: : x = w so, for example, sin x = sin w
• co-interior angles are supplementary: z = 180 – y
so, for example, sin z = sin (180 – y) = sin y
x
z
y
w
Look out for compound angles.
P
B
T
A
C
Q
R
S
PQT = PQR – TQS
So sin PQT = sin (PQR – TQS)
D
Test
Yourself?
ADC = ADB + BDC
So sin ADC = sin (ADB + BDC)
The diagram shows the graph of y = a sin(bx) + c ; 0 ≤ x ≤ 2π
(a)
(b)
Find the values of a, b and c
Solve for x in the interval 0 ≤ x ≤ 2π when y = 1·5
y = 3 sin(2x) + 1
4
3
2
reveal
1
0
0
-1
-2
1.57
3.14
4.71
6.28
The diagram shows the graph of y = a sin(bx) + c ; 0 ≤ x ≤ 2π
(a)
(b)
Find the values of a, b and c
Solve for x in the interval 0 ≤ x ≤ 2π when y = 1·5
(a)
y = 3 sin(2x) + 1
a = 3 (amplitude)
4
b = 2 (waves per 2π)
3
c = 1 (vertical-translation)
2
1
(b)
0
0
1.57
3.14
4.71
-1
sin 2x = 0·5 ÷ 3 = 0·167 (3 s.f.)
6.28
So
2x = sin–1(0·167),
or π – sin–1(0·167),
-2
or sin–1(0·167) + 2π, (π – sin–1(0·167)) + 2π
1st Solution from calculator: sin–1(…)
2nd Solution from symmetry: π – sin–1(…)
So
2x = 0·167, π – 0·167,
0·167 + 2π, and (π – 0·167) + 2π, …
More solutions from periodicity of function:
If x is a solution then so is 2π + x
= 0·167, 2·97, 6·45, 9·26, …
So
x = 0·084, 1·49, 3·23, 4·63
PQRS is a parallelogram with a base of 11 cm.
Its altitude, ST, is 12 cm.
QR is extended to meet the altitude at T. RT = 5 cm.
P
S
12
Q
(a)
(b)
(c)
(d)
T
11
R
5
State the length of RS.
Find the exact value of sin PQR.
State the exact value of sin SQR.
Hence find the exact value of sin PQS.
reveal
PQRS is a parallelogram with a base of 11 cm.
Its altitude, ST, is 12 cm.
QR is extended to meet the altitude at T. RT = 5 cm.
P
S
(a) By Pythagoras, SR = 13 and QS = 25.
12
Q
(a)
(b)
(c)
(d)
T
11
R
5
State the length of RS and of QS.
Find the exact value of sin PQR.
State the exact value of sin SQR.
Hence find the exact value of sin PQS.
(b) PQR = SRT (corresponding angles)
So sin PQR = sin SRT = 12/13
(c) SQR = SQT (corresponding angles)
So sin SQR = sin SQT = 12/25 = 3/4
(d) Sin PQS = sin(PQR – SQR)
= sin PQR.cos SQR – cosPQR.sin SQR
=
12/
=
132/
13
. 16/ 25 – 5/13 . 12/25
325
Find the exact value of
(a) sin 15˚
(b)cos 75˚
(c) tan 105˚
reveal
Find the exact value of
(a) sin 15˚
(a) sin 15˚ = sin (60 – 45)˚
= sin 60˚cos 45˚ – cos 60˚sin 45˚
(b)cos 75˚
(c) tan 105˚
√2
1
3 1 1 1
.
- .
2
2 2 2
=
3 -1
2 2
(b) cos 75˚ = cos (45 + 30)˚
= cos 45˚ cos 30˚ – sin45˚ sin30˚
1
2
45˚
=
60˚
1
30˚
√3
=
1
3 1 1
.
.
2 2
2 2
=
3 -1
2 2
(c) tan 105˚ = tan (60 + 45)˚
=
sin ( 60 + 45)˚ sin 60˚cos 45˚+cos60˚sin 45˚
=
cos ( 60 + 45)˚ cos60˚cos 45˚-sin 60˚sin 45˚
=
. 1 2 + 12. 1
1 . 1
- 32 . 1
2
2
3
2
2
2
=
3 +1
1- 3
A
x˚
B
C
Triangle ABC is isosceles with AB = AC
Angle ABC = x˚.
sin x˚ = 0·8 exactly.
Calculate the exact value of
(a)cos ACB
(b)cos BAC
reveal
A
x˚
B
C
Triangle ABC is isosceles with AB = AC
Angle ABC = x˚.
sin x˚ = 0·8 exactly.
Calculate the exact value of
(a)cos ACB
(b)cos BAC
(a) Angle ACB = angle ABC = x˚
We know sin x˚ = 0·8.
We can draw a right angle triangle,
hypotenuse 1 and opposite side 0·8
and use Pythagoras’ Theorem to deduce
that the third side is 0·6.
1
0·8
x˚
0·6
cos ACB = cos x˚ = 0·6.[from triangle]
(b) cos BAC = cos (180 – 2x)
= – cos 2x˚
= – [cos2 x˚ – sin2 x˚]
= – [0·62 – 0·82]
= 0·28 exactly
1
a Expand sin(30 + x)˚
b Hence solve the equation cos x˚ + √3 sin x˚ = 1 , 0 ≤ x ≤ 360
2
Solve sin 2x – cos x = 0, 0 ≤ x ≤ 2π
2 sin x cos x – cos x = 0
 cos x(2 sin x – 1) = 0
 cos x = 0 or sin x = ½
 x = π/2 or 2π – π/2 or π/6 or 2π – π/6
 x = π/2 , 3π/2 , π/6 , 11π/6
3
Solve 3 cos 2x˚ – 14 cos x˚ + 7 = 0 , 0 ≤ x ≤ 360
3[2cos2 x˚ – 1] – 14 cos x˚ + 7 = 0
 6cos2 x˚ – 14 cos x˚ + 4 = 0
 cos x˚ = [14 ± √(142 – 4.6.4)] ÷ 12 = 24/12 or 4/12
 cos x˚ = 2 (no solutions) or cos x˚ = 1/3
 x = 70·5 or 360 – 70·5 = 289·5
(to 1 d.p.)
reveal
1
a Expand sin(30 + x)˚
b Hence solve the equation cos x˚ + √3 sin x˚ = 1 , 0 ≤ x ≤ 360
a
sin(30 + x)˚ = sin 30˚ cos x˚ + cos 30˚ sin x˚ = 1/2 cos x˚ + √3/2 cos x˚
b
Given cos x˚ + √3 sin x˚ = 1 then 1/2 cos x˚ + √3/2 cos x˚ = 1/2
Þ sin(30 + x)˚ = 1/2
Þ 30 + x = 30 or 180 – 30 = 150 or 360 + 30 or 360 + 150 or
…
In the desired domain, x = 30 or 150.
2
Solve sin 2x – cos x = 0, 0 ≤ x ≤ 2π
2 sin x cos x – cos x = 0
cos x(2 sin x – 1) = 0
cos x = 0 or sin x = ½
x = π/2 or 2π – π/2 or π/6 or 2π – π/6
x = π/2 , 3π/2 , π/6 , 11π/6
Þ
Þ
Þ
Þ
3
Solve 3 cos 2x˚ – 14 cos x˚ + 7 = 0 , 0 ≤ x ≤ 360
3[2cos2 x˚ – 1] – 14 cos x˚ + 7 = 0
Þ 6cos2 x˚ – 14 cos x˚ + 4 = 0
Þ cos x˚ = [14 ± √(142 – 4.6.4)] ÷ 12 = 24/12 or 4/12
Þ cos x˚ = 2 (no solutions) or cos x˚ = 1/3
Þ x = 70·5 or 360 – 70·5 = 289·5 (to 1 d.p.)