Chapter 3.4 - Trigonometric Ratios Using Unit Circle

In the previous section, we saw the relation between degree and radian measure. In this section, we will see the general form of trigonometric ratios.

• In earlier classes, we have seen the trigonometric ratios for acute angles. For example, we have proved that sin 30^o = $\frac{1}{2}$, tan 45^o = $\frac{1}{\sqrt{2}}$ etc., (Details can be seen here)
• In our present discussion, we will see the general case which includes both acute and obtuse angles. First we will see the general case of acute angles. It can be written in 31 steps:
1. Consider a unit circle (circle with radius 1 meter).
• It is drawn in such a way that, it’s center O is at the origin of the coordinate axes in the Cartesian plane. So the coordinates of center will be (0, 0).
• This is shown in fig.3.13 below:

Fig.3.13

2. The circle cuts the coordinate axes at A, B, C and D.
• Since it is an unit circle,
    ♦ The coordinates of A will be (1, 0)
    ♦ The coordinates of B will be (0, 1 )
    ♦ The coordinates of C will be (-1, 0)
    ♦ The coordinates of D will be (0, -1)
3. A ray (indicated by the red arrow) rotates about O. Length of the ray is 1 meter. So the end point of the ray moves along the circumference of the unit circle.
• Let P be any position of the end point of the ray.
• Let the coordinates of P be (a, b).
• Let OP make an angle x radians with the x-axis.
    ♦ Then the length of arc AB will also be x meter.  
• M is the foot of the perpendicular drawn from P on the x-axis.
4. Now we have a right triangle OMP.
• In this right triangle, we get:
    ♦ $\cos x = \frac{OM}{OP} = \frac{OM}{1} = OM$.
    ♦ $\sin x = \frac{PM}{OP} = \frac{PM}{1} = PM$.
• But OM is the x coordinate 'a' of P. So we can write:
    ♦ OM = cos x = a
• Also PM is the y coordinate 'b' of P. So we can write:
    ♦ PM = sin x = b
5. In the right triangle OMP, we have: OM² + PM² = OP²
• So we get two important results:
    ♦ a² + b² = 1² = 1
    ♦ cos²x + sin²x = 1² = 1
6. Consider the situation in which x^c is extremely small. That is., P is very close to A.
• In such a situation:
    ♦ OM will be nearly equal to OA.
    ♦ P will be very close to M. So PM will be very small.
7. An extremely small angle x^c is an angle like 0.000000001^c
• For practical purposes, such an angle can be taken to be equal to zero ^c.
• So when x^c is extremely small,
$\sin x = b = \sin 0 = \frac{PM}{OP}=\frac{\text{An extremely small length}}{1}$
• 'An extremely small length' can be considered as zero length.
• So we can write:
When x is extremely small,
$\sin x = b = \sin 0 = \frac{PM}{OP}=\frac{0}{1}=0$
8. Also when x is extremely small,
$\cos x = a = \cos 0 = \frac{OM}{OP}=\frac{\text{A length very close to OP}}{OP}$
• 'A length very close to OP' can be considered as OP.
• So we can write:
When x is extremely small,
$\cos x = a = \cos 0 = \frac{OM}{OP}=\frac{OP}{OP}=\frac{1}{1}=1$
9. Remember that, in sin 0 and cos 0, the angle x is not exactly zero. (If it is exactly zero, there will be no triangle. If there is no triangle, we cannot take any trigonometric ratios). The angle is so small that, for practical purposes, it can be taken to be equal to zero.
• So sin 0 is a condition in which we are taking the sine of an extremely small angle (like 0.000000001^c)
• Also cos 0 is a condition in which we are taking the cosine of an extremely small angle (like 0.000000001^c)
• Through the above steps, we proved that:
    ♦ sine of such an angle can be taken as zero for practical purposes.
    ♦ cosine of such an angle can be taken as one for practical purposes
10. Let us check the coordinates:
• When x is extremely small, P is very close to A
    ♦ In (7), we got the y coordinate b as 0
    ♦ In (8), we got the x coordinate a as 1
• So the coordinates of P are (1,0)
• The coordinates of A are also (1,0)
• The coordinates are same because, when x is extremely small, it is practically impossible to distinguish between points A and P
11. We will write a summary of the above results:
sin 0 = 0, cos 0 = 1.

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12. We have completed the discussion on the situation when P is very close to A.
• We will now consider the other extreme case when P is very close to B.
• The following steps from 13 to 18 will give the details.
13. Consider the situation in which x is very close to ${\frac{\pi}{2}}^c$ (90^o).
• PM will be nearly equal to OB.
• M will be very close to O. So OM will be very small.
• An angle x very close to 90^o, is an angle like 89.999999999^o
    ♦ For practical purposes, such an angle can be taken to be equal to 90^o (${\frac{\pi}{2}}^c$).
14. So when x is close to ${\frac{\pi}{2}}^c$,
$\sin x = b = \sin \frac{\pi}{2} = \frac{PM}{OP}=\frac{\text{A length very close to OP}}{OP}$
• 'A length very close to OP' can be considered as OP.
• So we can write:
When x is very close to ${\frac{\pi}{2}}^c$,
$\sin x = b = \sin \frac{\pi}{2} = \frac{PM}{OP}=\frac{OP}{OP}=\frac{1}{1}=1$
15. Also when x is close to ${\frac{\pi}{2}}^c$,
$\cos x = a = \cos \frac{\pi}{2} = \frac{OM}{OP}=\frac{\text{An extremely small length}}{1}$
• 'An extremely small length' can be considered as zero length.
• So we can write:
When x is very close to ${\frac{\pi}{2}}^c$,
$\cos x = a = \cos \frac{\pi}{2} = \frac{OM}{OP}=\frac{0}{1}=0$
16. Remember that, in $\sin \frac{\pi}{2}$ and $\cos \frac{\pi}{2}$, the angle x is not exactly ${\frac{\pi}{2}}^c$. (If it is exactly ${\frac{\pi}{2}}^c$, there will be no triangle. If there is no triangle, we cannot take any trigonometric ratios). The angle is so close to ${\frac{\pi}{2}}^c$ that, for practical purposes, it can be taken to be equal to ${\frac{\pi}{2}}^c$.
• So $\sin\frac{\pi}{2}$ is a condition in which we are taking the sine of an angle very close to ${\frac{\pi}{2}}^c$
• Also cos $\frac{\pi}{2}$ is a condition in which we are taking the cosine of an angle very close to ${\frac{\pi}{2}}^c$
• Through the above steps, we proved that:
    ♦ sine of such an angle can be taken as one for practical purposes.
    ♦ cosine of such an angle can be taken as zero for practical purposes.
17. Let us check the coordinates:
• When x is close to ${\frac{\pi}{2}}^c$, P is very close to B.
    ♦ In (18), we got the y coordinate b as 1
    ♦ In (19), we got the x coordinate a as 0
• So the coordinates of P are (0,1)
• The coordinates of B are also (0,1)
• The coordinates are same because, when x is close to ${\frac{\pi}{2}}^c$, it is practically impossible to distinguish between points B and P
18. So we can write a summary of the above results:
• sin 0 = 0, cos 0 = 1.
• sin $\frac{\pi}{2}$ = 1, cos $\frac{\pi}{2}$ = 0.

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19. We have completed the discussion on the situation when P is very close to A.
• We have also completed the discussion on the situation when P is very close to B.
• We will now consider the intermediate points. The important intermediate points are those when x = 30^o (${\frac{\pi}{6}}^c$), 60^o (${\frac{\pi}{3}}^c$), and 45^o (${\frac{\pi}{4}}^c$).
• The following steps from 20 to 36 will give the details.
20. Consider fig.3.13 that we saw at the beginning of this section.
• Suppose that, x = 60^o (${\frac{\pi}{3}}^c$).
• So in the right triangle OMP, we have angle POM = 60^o.
21. Now construct a mirror image of triangle OMP. The mirror line should be PM.
This is shown in fig.3.14 below:

Fig.3.14

• We see that, mirror image angle OAP is 60^o. So the angle OPA will also be 60^o.
• Thus triangle OAP is an equilateral triangle.
22. Since it is an equilateral triangle, the perpendicular PM will bisect PA
So we get: OM = (1/2 × OA) = (1/2 × 1) meter = 1/2 meter
23. Thus in the right triangle OMP, we have the lengths of two sides:
OP = 1 meter, OM = 1/2 meter
• So the third side, PM = $\sqrt{OP^2 - OM^2}=\sqrt{1^2 - (1/2)^2 }= \frac{\sqrt{3}}{2}$
• Now we have all the three sides of the right triangle OMP. We can take the trigonometric ratios.
• We get:
   ♦ $\sin 60 = \sin \frac{\pi}{3}= \frac{PM}{OP} = \frac{\frac{\sqrt{3}}{2}}{1}=\frac{\sqrt{3}}{2}$ 
   ♦ $\cos 60 = \cos \frac{\pi}{3}= \frac{OM}{OP} = \frac{\frac{1}{2}}{1}=\frac{1}{2}$
• Note that here also, the coordinates (a,b) are the sine and cosine of the angle.
   ♦ $\sin 60 = \sin \frac{\pi}{3}= \frac{PM}{OP} = \frac{PM}{1} = PM = b =\frac{\sqrt{3}}{2}$ 
   ♦ $\cos 60 = \cos \frac{\pi}{3}= \frac{OM}{OP} = \frac{OM}{1}  = OM = a = \frac{1}{2}$
24. Next we will see the case of 30^o (${\frac{\pi}{6}}^c$)
• Consider fig.3.13 that we saw at the beginning of this section.
• Suppose that, x = 30^o (${\frac{\pi}{6}}^c$).
• So in the right triangle OMP, we have angle POM = 30^o.
This is shown in fig.3.15 below:

Fig.3.15

25. In step 23, we have calculated sin 60 and cos 60.
• In our present triangle OMP, we have:
$\sin 60 = \frac{\sqrt{3}}{2} = \frac{OM}{OP} = \frac{OM}{1} = OM$
    ♦ That is., OM = $\frac{\sqrt{3}}{2}$
• Also we have:
$\cos 60 = \frac{1}{2} = \frac{PM}{OP} = \frac{PM}{1} = PM$
    ♦ That is., PM = $\frac{1}{2}$
26. Now we have all the three sides of the right triangle OMP. We can take the trigonometric ratios.
• We get:
   ♦ $\sin 30 = \sin \frac{\pi}{6}= \frac{PM}{OP} = \frac{\frac{1}{2}}{1}=\frac{1}{2}$ 
   ♦ $\cos 30 = \cos \frac{\pi}{6}= \frac{OM}{OP} = \frac{\frac{\sqrt{3}}{2}}{1}=\frac{\sqrt{3}}{2}$
• Note that here also, the coordinates (a,b) are the sine and cosine of the angle.
   ♦ $\sin 30 = \sin \frac{\pi}{6}= \frac{PM}{OP} = \frac{PM}{1} = PM = b =\frac{1}{2}$ 
   ♦ $\cos 30 = \cos \frac{\pi}{6}= \frac{OM}{OP} = \frac{OM}{1} = OM = a =\frac{\sqrt{3}}{2}$
27. Next we will see the case of 45^o (${\frac{\pi}{4}}^c$)
• Consider fig.3.13 that we saw at the beginning of this section.
• Suppose that, x = 45^o (${\frac{\pi}{4}}^c$).
• So in the right triangle OMP, we have angle POM = 45^o.
28. Now construct a mirror image of triangle OMP. The mirror line should be PM.
This is shown in fig.3.16 below:

Fig.3.16

• The resulting larger triangle is OQP
• We see that, angle OQP is 45^o. So angle OPQ will be 90^o.
• Thus triangle OAP is an isosceles triangle.
29. Since it is an isosceles triangle, the perpendicular PM will bisect angle OPQ.
• So we get:
angle OPM = angle QPM = 90/2 = 45^o
30. Now consider the triangle OPM. Two of it's angles are same (45^o). So it is an isosceles triangle. The equal sides are MO and MP
Let MO = MP = k
We get: MO² + MP² = OP²
⇒ k² + k² = 1²
⇒ 2k² = 1
⇒ k = $\frac{1}{\sqrt{2}}$
31. Now we have all the three sides of the right triangle OMP. We can take the trigonometric ratios.
• We get:
   ♦ $\sin 45 = \sin \frac{\pi}{4}= \frac{PM}{OP} = \frac{\frac{1}{\sqrt{2}}}{1}=\frac{1}{\sqrt{2}}$ 
   ♦ $\cos 45 = \cos \frac{\pi}{4}= \frac{OM}{OP} = \frac{\frac{1}{\sqrt{2}}}{1}=\frac{1}{\sqrt{2}}$
• Note that here also, the coordinates (a,b) are the sine and cosine of the angle.
   ♦ $\sin 45 = \sin \frac{\pi}{4}= \frac{PM}{OP} = \frac{PM}{1} = PM = b =\frac{1}{\sqrt{2}}$ 
   ♦ $\cos 45 = \cos \frac{\pi}{4}= \frac{OM}{OP} = \frac{OM}{1} = OM = a =\frac{1}{\sqrt{2}}$

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• Based on the above 31 steps, we can write the following three points:
(i) We want the sine and cosine of a given angle.
   ♦ For that, we start to rotate the red arrow beginning from the x-axis.
   ♦ The tip of the arrow moves along the circumference of the unit circle.
(ii) We stop the rotation when the given angle is reached.
   ♦ We mark the final position of the tip of the arrow as ‘P’.
   ♦ We write the x and y coordinates of P.
(iii) The x coordinate is the cosine and y coordinate is the sine.
• This is a new method for finding sine and cosine of a given angle.

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In the next section, we will see trigonometric ratios of 180^o.

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