TRIGONOMETRY
1. The Sine Rule
We can prove the sine rule like this. Let x and y be the perpendiculars from vertices A and B to the opposite sides a and b.
Considering the two right-angled triangles with x as one of the sides,
Comparing expressions for x,
Similarly, considering the two right-angled triangles with y as one of the sides,
Comparing expressions for y,
…and so we have the sine rule.
We now move onto using the sine rule. In order to use it we must have a combination of two sides and one angle, or one side and two angles.
Example 1 : Find the missing angles and sides in the triangle below.
The angles of the triangle add up to 180°.
To find b,
To find c, we use a rather than b since a is an integer and therefore more convenient.
Example 2 : solve the triangle with , b = 5cm, a = 6cm.
Construction : draw the base b and a line leaving A at an angle of 30°. Set compasses to 6cm and draw an arc from C. The point of intersection of the arc and the line leaving A is the vertex B.
Solving using the sine rule (we put the sines in the numerators to make the algebra easier),
A look at the graph of the sine function tells us that there are two possible solutions to this equation in between 0° and 180°. Solving the equation we have and
However B cannot be 155.4°, since A is already 30°, so B = 24·6° and .
Finally we calculate c.
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Example 3 : Now for an ambiguous case - solve the triangle with , b = 6cm, a = 4cm.
Construction : very similar to the previous example, although note that now the arc crosses the line leaving A at two different points. This means that there are two solutions to the triangle.
Both of these angles are possible, as they are less than , so we consider each case.
• If B = 48.6°, then C = 180° − 30° − 48.6° = 100.4°. To find c,
• If B = 131.4°, then C = 180° − 30° − 131.4° = 18.6°. To find c,
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2. The Cosine Rule
In a right angled triangle, . However, for the acute-angled triangle shown below, it is clear that . And for the obtuse-angled triangle, . Therefore, we need a modified version of Pythagoras’ theorem for non-right angled triangles.
In fact the formula is the cosine rule
Notice how if C is acute, then is positive, and . Similarly, if C is obtuse, then is negative, and . In fact the formula appears in three forms…
To prove the cosine rule, drop a perpendicular of length h from C to the line AB. Let the base of the unshaded triangle be x. Notice that .
Using Pythagoras’ rule on the unshaded triangle,
Using Pythagoras’ rule on the shaded triangle,
Substituting from the first equation,
Because the formula contains three sides and one angle, we can use it if we know three sides, or two sides and the angle they enclose.
Example 1 : Solve the triangle with a = 8cm, b = 6cm and c = 4cm.
To find A,
We see from the cosine function that there is only one solution between 0° and 180°, and so we do not have to worry about any ambiguous cases. This is consistent with our experience that three sides uniquely define a triangle. It also means that the cosine rule may be less prone to human error than the sine rule!
To find B, we can use either the cosine rule or the sine rule. The cosine rule is marginally preferable as with the sine rule we would have to use the calculated value of A, rounding it off in the process. Choosing the cosine rule,
Finally we calculate C,
If this answer looks like it should be 28.9, then remember we must use the full calculator values for A and B, not the rounded off ones.
Example 2 : Solve the triangle with a = 7cm, b = 5cm and C = 119°.
To find c,
We can now use the sine or the cosine rule to calculate A or B. Choosing the sine rule for A,
We reject the solution A = 143.9° as there are not enough degrees in the triangle. Finally,
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3. Applications of the Sine and Cosine Rules
Example1:Find the angle between the hour hand and the minute hand of a clock at 12.15. If the hour hand is 8cm long and the minute hand is 10cm long, how far apart are the tips of the hands at this time?
The hour hand covers 360° ÷ 12 = 30° in one hour, and therefore 30° ÷ 4 = 7.5° in quarter of an hour. So the angle between the hands is 90° – 7.5° = 82.5°.
By the cosine rule,
Example2:A boat is sailing directly towards the foot of a cliff. The angle of elevation of a point on top of the cliff, and directly ahead of the boat, increases from 8° to 12° as the boat sails 100m. Find
a) the original distance from the boat to the point on top of the cliff
b) the height of the cliff.
a) We have A = 168°, B = 8°, c = 100m.
To find a,
The original distance from the boat to the top of the cliff is 298.05m.
b) To find the height of the cliff,
Example 3 : A ship leaves a port P and sails 5 nautical miles on a bearing of 052°, followed by 6 nautical miles on a bearing of 205° to arrive at the next port Q.
Find the distance and bearing of P from Q.
To find the angle X of the triangle, we use the property of ‘interior angles’.
Using the cosine rule,
Using the sine rule,
We reject the solution as it does not fit with our diagram. The bearing of P from Q is given by .
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4. The Area of a Triangle
Suppose we label the angles of a triangle A, B and C. Then we label the side opposite angle A as a, the side opposite angle B as b, and the side opposite angle C as c. We can find a formula for the area of a triangle using this notation.
Clearly we could also have these alternative expressions for the area : or .
Example 1 : Find the area of this triangle.
Example2:The area of a triangle is 36 cm². Two sides of the triangle are 8cm and 10cm. Find the two possible angles between these two sides.
A small sketch of the sine curve between 0° and 180° illustrates the last step in the calculation.
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5. Radian Measure
One radian is defined as the angle subtended at the centre of a circle, by an arc equal in length to its radius. In general,
Therefore, for one revolution,
.
So 2π radians = 360°, and one radian is approximately 57°.
It is important to be able to convert between degrees and radians fluently. The table below shows some common angles in both units.
degrees / 0 / 30 / 45 / 60 / 90 / 120 / 135 / 150 / 180 / 240 / 270 / 300 / 360radians / 0 / / / / / / / / / / / /
The symbol for radians is a small superscript c, and so 2 radians is written 2c or 2 rad. If units are omitted, then an angle is in radians, not degrees!
Example 1 : Convert the following angles into degrees.
a) rad b) 2 rad
a) b)
Example 2 : Convert the following angles into radians.
a) 24° b) 7°
a) b)
When angles are measured in radians, we have several convenient formulas concerning circles…
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6. Arc Length
Consider the arc PQ, of length s, subtending angle θ radians at the centre of the circle. Now
And so
So a length of arc is found by multiplying the radius by the angle (measured in radians) subtended by the arc.
In fact, two points P and Q on a circle define two arcs, a shorter one (known as the minor arc) and a longer one (known as the major arc).
Example 1 : Find the length of the minor arc PQ,
We first write 135° as radians.
Example 2 : Find the perimeter of the shape opposite.
The perimeter consists of two arcs and two straight lines.
We first write 60° as radians.
Example3:A road tunnel 60m long has the cross-section shown opposite, consisting of an arc centre O and a straight line. It is required to paint the inside of the tunnel (not including the road itself). What area is to be painted?
We first find the angle θ. Splitting the isosceles triangle into two right-angled triangles gives us
The arc subtends an angle of radians.
The internal area is therefore
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7. Area of a Sector
Similarly, we also have minor sectors and major sectors. Consider the sector OPQ, with area A. Now
And so
Example 1 : Find the area of the major sector OPQ,
The major sector subtends .
Example2:A shelf has the shape of a sector of a circle of radius 8cm, with a concentric sector of radius 5cm removed, as shown in the diagram below. If the shelf has an area of 50cm2, find the angle subtended by the sectors.
Example3:Find a formula for the area A of the segment below. Hence find the area of the segment if r = 6cm and θ = 1.5 rad.
Substituting,
Example 4 : This shape is made up of arcs centered on the corners of an equilateral triangle of side length r. Show that the area of the shape is given by .
Example5:The diagram shows a triangle, and an arc whose centre is the point P.
a) Show that
b) Find the perimeter of the region R.
c) Find the area of the region R.
a) Using the cosine rule,
We only get the answer in radians if our calculator is in radian mode!
b) The perimeter is an arc length added to three straight edges.
c) The area is the triangle subtract the sector.
Here it is essential to have our calculator is in radian mode.
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8. Sine, Cosine and Tangent of Common Angles
Sines, cosines and tangents are commonly used with right-angled triangles.
Activity1:Do the Sine, Cosine and Tangent of Common Angles worksheet. Encourage students to learn these values.
0° / 0 / 1 / 030° / / /
45° / / / 1
60° / / /
90° / 1 / 0 / –
9. The Unit Circle
However sines, cosines and tangents also have a close relationship with the unit circle.
The diagram shows a circle whose centre O is at (0, 0), and whose radius is 1 unit. Let P be a point on the unit circle.
If we want to tell someone exactly where P is, we can give the angle between the positive x-axis and the line OP. We shall call this angle θ, and we shall always measure it anti-clockwise, starting from the positive x-axis.
Another way to specify the position of P is to give its coordinates. We can find them by considering the right-angled triangle in the diagram below.
The x-coordinate is the distance OA. Now
The y-coordinate is the distance AP. Now
So the coordinates of P are . So we define sine and cosine as the y and x coordinates of the point P, rather than and .
The unit circle is divided by the x and y axes into four quadrants, which are numbered anticlockwise as shown here.
If the angle θ is greater than 90°, then P will be in the 2nd, 3rd or 4th quadrants. We can now give a meaning to sines and cosines of angles greater than 90°, by simply saying that
For example, when θ = 130°, we find from a calculator that cos θ = –0.643, and sin θ = 0.766. We can see why by looking at the unit circle : when θ = 130°, the coordinates of P are (–0.643, 0.766).
Activity2:Use the Unit Circle worksheet to find the coordinates of points on the unit circle at angles 50°, 120°, 255° and 305°. Use calculators to check that the coordinates are indeed (cos θ, sin θ). Alternatively, let students choose their own points within each quadrant.
As an extension, investigate the sine and cosine of 430° and –65°.
10. The Sine Function
The sine of θ is the y-coordinate of the point P on the unit circle. Because of this, it is clear that...
sin 0° = 0, sin 90° = 1, sin 180° = 0, sin 270° = –1, sin 360° = 0.
If we plot a graph of sin θ against θ, we find it looks like this.
But after 360°, P is back to the start of the circle (1, 0) again. So the graph repeats itself every 360°. We say it is periodic, with period 360°.
This graph is called the sine function.
11. The Cosine Function