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TRIGONOMETRY
UNIT CIRCLE AND
TRIGONOMETRIC
RATIOS
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NOTES/ UNIT CIRCLE & TRIG RATIOS
UNIT CIRCLE AND TRIGONOMETRIC VALUES OF
ANGLES IN ALL QUADRANTS:
Point P is (x,y). Then point Q will be (-x,y) and points R and S will be?
R( -x,-y) and S ( x,-y)
If it is a unit circle, and OP makes an angle θ with the (positive) x-axis, then cos θ = x and sin θ = y
Therefore, the point P can be given as (cosθ, sinθ), when θ is in the first quadrant.
When the angle is in the 2 nd quadrant:
Now Q is (-x,y)
Then we get the x coordinate of Q will be (– cos θ).
Q(-x,y)
P(x,y)
But if we consider the triangle Q OB with angle 180
– θ, then, x = cos (180 – θ)
Therefore, we get, Cos θ = - Cos (180 – θ)
[Cos is negative]
180 - θ
θ
B
θ
O
Similarly, sin θ will be equal to sin (180 – θ).
Sin θ = Sin ( 180 – θ)
[ Sine is positive]
R(-x,-y)
Now if the angle is in the 3rd quadrant ,
then the coordinates will be (-x,-y) and the small
angle which makes the triangle will be θ and the actual angle would be 180 + θ.
S(x,-y)
And hence we will have,
Cos θ = - Cos ( 180 + θ)
[Cos is negative]
Sin θ = - Sin (180 + θ)
[ Sine is negative]
Now, it is easy to conclude for angles in 4th quadrant :
Cos θ = Cos (360 – θ)
Sin θ = - Sin ( 360 – θ)

[ Cos is positive]
[ Sine is negative]
REMEMBER: Angles are always measured from the
positive x- axis, anticlockwise.
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NOTES/ UNIT CIRCLE & TRIG RATIOS
Collecting the details from the above results, we get,
All functions (Sin, Cos and hence Tan) are positive in the First Quadrant.
Sine is positive and Cos is negative (and hence Tan is also negative) in the Second Quadrant.
Tan is positive ( Since both Cos and Sin are negative) in the Third Quadrant.
Cos is positive and Sin is negative ( and hence Tan is also negative) in the Fourth Quadrant.
Hence we get the following picture:
You have to
remember this as
Sin
All
ASTC or All Students
Take Coffee...
The quadrants start
Tan
Cos
from A moves
anticlockwise
What are these useful for?
From the above results, we see that Cos θ/Sinθ/Tanθ all have the same values for two angles in the
range of 0 to 360 degrees.
That is, Cos θ = 0.5, then θ can be 600 or 360 – 60 0.
And if Cos θ = - 0.5, then θ can be 180 – 60 0 or 180 + 600.
θ
How do we find it?
Let us take the first case: , Cos θ = 0.5:
360 - θ
First we notice that Cos θ is positive. And so, the angle should be
in the first or 4th quadrant.
If it was in the first quadrant, the angle would be θ = 600
If it was in the fourth quadrant, the angle would be 360 – 60 = 3000
Hence solving the equation, Cos x = 0.5 gives us two values of x in the range of 0 to 360 degrees:
x1 = 600 or
𝜋
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or x2 = 3000 or
5𝜋
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NOTES/ UNIT CIRCLE & TRIG RATIOS
Similarly, if Cos θ = - 0.5, then the angles would be in the 2nd and 3rd quadrant, since Cos is negative in
these two only.
Hence the angles will be 180 – θ and 180 + θ,
where θ is the corresponding angle formed in the first quadrant
for Cos θ = + 0.5
In this case, θ = 600 in the first quadrant.
180 -θ
θ
0
This 60 value is
called the
180 +θ
primary angle
But the solution of the equation Cos x = - 0.5 will be:
x1 = 180 – 60 = 1200 or 2π/3
or
Note: 600 is NOT a
solution for this
equation.
x2 = 180 + 60 = 2400 or 4π/3
Let us solve some more similar equations:
1. Solve for x , for 0 ≤ x ≤ 2π: Sin x = 0.45
Solution:
Two things to note:
•
•
The angles should be in radians
Sin x is positive
Since Sin x is positive, the angles are in 1st and 2nd quadrant only.
The first quad angle can be found out by using calculator:
Remember to
Sin x = 0.45
change your calc
Hence x = Sin -1 (0.45) = 0. 46 radians
mode to Radians
The 2nd quad angle can be found by π – 0.46 = 2.68 radians
Therefore, the solutions are:
Not 180 !
x1 = 0.46 or x2 = 2.68
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NOTES/ UNIT CIRCLE & TRIG RATIOS
2. Solve, for 0 ≤ x ≤ 2π, Sin x = - 0.27
Solution:
Two things to note:
•
Angles in radians
Since Sin x is negative, the angles are in 3rd and 4th quadrants.
In order to find the primary angle, we have to take the positive value as below:
Sin x = + 0.27
Hence x = Sin -1 (0.27) = 0.27 radians [ again remember radians here]
Now, we have that the angle is NOT in the first quadrant but only in 3rd and 4th quadrants, which is
given by 180 - and 360 - .
Hence the solutions are:
x1 = π – 0.27 = 2.8 or
x2 = 2π – 0.27 = 6.03
Try these problems:
Solve the following trigonometric equations for angles between 0 and 2π.
1.
2.
3.
4.
5.
6.
Cos x = 0.98
Cos x = - 0.34
Sin x = 0.532
Sin x = - 0.879
Tan x = 2.32
Tan x = - 1.56
Now let us solve the following:
Cos 2x = 0.35
Solution:
Note: the difference is we have 2x instead of x. [We can have anything here, like x+2, 2x-3 etc.]
Take 2x = y and proceed as before.
We will get y = cos-1 (0.35) = 0.94
Or y = 2π – 0.94 = 5.34
That is 2x = 0.94 or 2x = 5.34
Dividing by 2 should be done only
after both the values of y are found
and not on the primary value in case
of negative values.
Hence
x1= 0.47 or x2 = 2.66
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NOTES/ UNIT CIRCLE & TRIG RATIOS
Let us solve this following problem to understand negative value problems:
Sin 2x = - 0.78
Let 2x = y
Primary value for y = Sin -1 (0.78) = 0.89
Do not
divide
0.89
by 2...
The actual values of y will be π – 0.89 and 2π – 0.89, which are 2.25 and 5.39
To find x, you should divide these answers by 2.
Hence the solutions will be 1.17 or 2.69
Try out these:
Solve for x between 0 and 2π:
a. Sin 3x = -0.61
b. Sin ( 2x – 3) = 0.38
c. Cos (2x +1) = - 0.92
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