TRIGONOMETRIC RATIOS AND IDENTITIES
WEEK 6
SUBJECT: FURTHER MATHEMATICS
CLASS: SS1
TOPIC: TRIGONOMETRIC RATIOS AND IDENTITIES:
CONTENT:
- GRAPHS OF TRIGONOMETRIC RATIOS
- TRIGONOMETRIC EQUATION
GRAPHS OF TRIGONOMETRIC RATIOS
Above are the graphs of the six trigonometric functions: sine, cosine, tangent, cosecant, secant and cotangent.
In the unit circle, the value of the hypotenuse is r = 1 so that sin = y and cos. In other words, as we progressed from geometrical figures to a situation in which there was just one input (one angle measure, instead of three sides and an angle) leading to one output (the value of the trig ratio). And this kind of relationship can be turned into a function.
Trigonometric functions like others have their graphical representation, which is of great importance to scientists. Thus, it is one of the basic knowledge required in mathematics. The sine, cosine and tangent can be represented graphically in either degrees or radians as units of measurements. Though degree is often used.
The methods applied in table of trigonometry curves are similar to that of quadratic graph.
Example 1: a. Copy and complete the table below for the relation y = 2 sin 3x – 1
X | 0° | 30° | 60° | 90° | 120° | 150° | 180° |
Y = 2sin3x -1 | -1 | 1 |
- Using a scale of 2cm to 30°on the x-axis and 1cm to 1 unit on the y- axis, draw the graph of y = 2sin3x – 1 for 0°≤ x ≤ 180°
- On the same axes draw the graph of y =
- Use your graph to find:
- Values of x for which 2 sin3x – 1 = 0
- Roots of the equation 2sins 3x
Solution
- Applying tabular format as in quadratic, we have
X | 0° | 30° | 60° | 90° | 120° | 150° | 180° |
2sin3x | 0 | 2 | 0 | -2 | 0 | 2 | 0 |
-1 | -1 | -1 | -1 | -1 | -1 | -1 | -1 |
Y | -1 | 1 | -1 | -3 | -1 | 1 | -1 |
- The graph is shown below
- The relation y = is a linear function since the highest degree of x is one. Thus preparing a three points table for the function; we have=
Table for
X | 0 | 90 | 180 |
0 | 0.5 | 1 | |
-2 | -2 | -2.0 | -2 |
Y | -2 | -1.5 | -1 |
- i. These are the points where the curve cut the x-axis 9, 50, 129 and 170
ii.These are the points of intersection between the linear line and the curve; traced to the x-axis 66, 117 and 180. Since the given equation is a combination of the two functions.
TRIGONOMETRIC EQUATION
An equation that contains trigonometric functions is called trigonometric equations. Trigonometric identities are equations involving the trigonometric functions that are true for every value of the variables involved. Trigonometric identities can be used along with algebraic methods to solve the trigonometric equations.
From the definition of trigonometric ratios
From i.
From ii.
Adding iii. And iv.
1 y
R
O
Since triangle OPR is a right-angled triangle
Equating v and vi
Dividing both side of vii by
Dividing vii through by
Example 1: Solve the following equation for values of between 0° and 360°:
Solution
But by trig identities 1 +
Substituting
becomes
Put sec
P = 2 or
Thus sec
But sec
Example 2: Find the value of
Solution
Recall that
Take square root of both sides
This is only possible in the 45° triangle for special trig ratio.
i.e. = 45°
Example 3:
Find all the solutions of the equation in the interval [ 0 , 2 π ) .
2 sin 2 x = 2 + cos x
The equation contains both sine and cosine functions.
We rewrite the equation so that it contains only cosine functions using the Pythagorean Identity sin 2 x = 1 − cos 2 x .
2 ( 1 − cos 2 x ) = 2 + cos x 2 − 2 cos 2 x = 2 + cos x − 2 cos 2 x − cos x = 0 2 cos 2 x + cos x =0
Factoring cos x we obtain, cos x ( 2 cos x + 1 ) = 0 .
By using zero product property , we will get cos x = 0 , and 2 cos x + 1 = 0 which yields cos x =− 1 2 .
In the interval [ 0 , 2 π ) , we know that cos x = 0 when x = π 2 and x = 3 π 2 . On the other hand, we also know that cos x = − 1 2 when x = 2 π 3 and x = 4 π 3 .
Therefore, the solutions of the given equation in the interval [ 0 , 2 π ) are
{ π 2 , 3 π 2 , 2 π 3 , 4 π 3 } .
CLASS ACTIVITY
- Copy and complete the following table of values for y = 3 sin 2 – Cos
0o | 30o | 60o | 90o | 120o | 150o | 180o | |
Y | -1.0 | 0 | 1.0 |
(b) Using a scale of 2cm to 30o on theaxis and 2cm to 1 unit on the y axis, draw the graph of y=3sin 2-cosfor 0o≤≤180o
(c) Use your graph to find the:
(i) solution of the equation 3 sin2 – cos=0, correct to the nearest degree;
(ii) maximum value of y, correct to one decimal place.
- Prove that
- Solve
PRACTICE QUESTIONS
- By eliminating, or otherwise, find the value of a for which the equations:
Are consistent. Hence find in the range -180 ° ≤
- Simplify WAEC
- Evaluate, without using tables, sin45°cos15° – cos45°sin15°
WAEC
- Solve
- (a) Copy and complete the table for the relation y =2 cos 2x – 1.
X | 0o | 30o | 60o | 90o | 120o | 150o | 180o |
Y=2cos 2x-1 | 1.0 | 0 | 1.0 |
ASSIGNMENT
- Find the values of in the range 0° ≤ ≤ 360° which satisfy:
- ( + 1)() = 0
- Using a scale of 2cm=30o on the x -axis and 2cm =1 unit on the y –axis draw the graph of y = 2 cos2x +1/2 = 0
(c) On the same axis draw the graph of y =1/180 (x – 360).
(d) Use your graphs to find the:
(i) values of x for which 2 cos2x+ ½ = 0
(ii) roots of equation 2 cos2x – x/180 + 1 = 0
- (a) Copy and complete the following table of values for y = 9cos x + 5 sin x to one decimal place.
x | 0o | 30o | 60o | 90o | 120o | 150o | 180o | 210o |
Y | 10.3 | -0.2 | -5.3 | 10.3 |
(b) Using a scale of 2cm to 30o on the x-axis and 2cm to 1 unit on the y-axis, draw the graph of y = 9cos x + 5 sin x, for 0 ≤ x ≤ 210o
(c) Use your graph to solve the equation;
(i) 9cosx + 5 sin x = 0
(ii) 9cosx + 5 sin x = 3.5, correct to the nearest degree
(d) Find the maximum value of y, correct to one decimal place.
- (a) Copy and complete the table of the relation y – 2sin x – cos2x
x | 0o | 30o | 60o | 90o | 120o | 150o | 180o |
y | 0.5 | -1.0 |
Using a scale of 2 cm to 30o on the x-axis and 2cm to 0.5 unit on the y-axis, draw the graph of y = 2 sin x – cos2x, for 0o ≤ x ≤ 180o.
(b) Using the same axes, draw the graph of y = 1.25
(c) Use your graphs to find the:
(i) values of x for which 2 sin x – cos2x = 0.
(ii) roots of the equations 2 sin x – cos2x = 1.25.
- (a) Copy and complete the table of values for y = 3 sin x + 2cosx for 0o ≤ x ≤ 360o.
x | 0o | 60o | 120o | 180o | 240o | 300o | 360o |
y | 2.00 | – | – | – | – | – | 2.00 |
(b) Using a scale of 2cm to 60o on x-axis and 2cm to 1 unit on y-axis, draw the graph of y = 3 sin x + 2cosx for 0o ≤ x ≤ 360o.
(c) Use your graph to solve the equation 3 sin x + 2cosx = 1.5.
(d) Find the range of values of x for which 3 sin x + 2cos x< – 1.
- a. Copy and complete the table of values for y = sin x + 2cos x, correct to one decimal place.
x | 0o | 30o | 60o | 90o | 120o | 150o | 180o | 210o | 240o |
y | 2.2 | -1.2 | -2.0 | -1.9 |
(b) Using a scale of 2cm to 30o on the x-axis and 2cm to 0.5 units on the y-axis, draw the graph of y = sin x + 2cos x for 0o ≤ x ≤ 240o
(c) Use your graph to solve the equation:
(i) sin x + 2cos x = 0;
(ii) sin x – 2.1 – 2cos x.
(d) From the graph, find y when x = 171o.