In the previous section, we saw the domain and range of tangent and cotangent functions. In this section, we will derive some trigonometric identities.
First we will derive two simple identities. They can be derived in 21 steps:
1. In fig.3.34(a) below, a ray has rotated by xc in the +ve direction.
• Another ray has rotated by the same xc in the negative direction.
 |
| Fig.3.34 |
2. Since the angles are same, the points P, Q and M are on the same vertical line.
• The x axis bisects the base PQ of the triangle OPQ
• Thus we get: QM = PM = b
3. So the coordinates of Q are: (a, -b)
• But x coordinate is cosine and y coordinate is sine.
• So we get:
sin (-x) = -b and cos (x) = a
4. For the +ve angle xc, the coordinates of P are: (a, b)
• But x coordinate is cosine and y coordinate is sine.
• So we get:
sin x = b and cos x = a
5. comparing the results in (3) and (4), we get:
sin (-x) = -sin x and cos (-x) = cos x
6. In fig.3.34(b) above, a ray has rotated by xc in the +ve direction.
♦ x is greater than $\frac{\pi}{2}$
• Another ray has rotated by the same xc in the negative direction.
7. From the fig., it is clear that:
♦ ∠MOP = (π - x)
♦ ∠MOQ = (π - x)
• Since these two angles are same, the points P, Q and M are on the same vertical line.
• The x axis bisects the base PQ of the triangle OPQ
• Thus we get: QM = PM = b
8. So the coordinates of Q are: (-a, -b)
• But x coordinate is cosine and y coordinate is sine.
• So we get:
sin (-x) = -b and cos (-x) = -a
9. For the +ve angle xc, the coordinates of P are: (-a, b)
• But x coordinate is cosine and y coordinate is sine.
• So we get:
sin x = b and cos x = -a
10. comparing the results in (8) and (9), we get:
sin (-x) = -sin x and cos (-x) = cos x
11. In fig.3.35(a) below, a ray has rotated by xc in the +ve direction.
♦ x is greater than π
• Another ray has rotated by the same xc in the negative direction.
 |
| Fig.3.35 |
12. From the fig., it is clear that:
♦ ∠MOP = (x - π)
♦ ∠MOQ = (x - π)
• Since these two angles are same, the points P, Q and M are on the same vertical line.
• The x axis bisects the base PQ of the triangle OPQ
• Thus we get: QM = PM = b
13. So the coordinates of Q are: (-a, b)
• But x coordinate is cosine and y coordinate is sine.
• So we get:
sin (-x) = b and cos (-x) = -a
14. For the +ve angle xc, the coordinates of P are: (-a, -b)
• But x coordinate is cosine and y coordinate is sine.
• So we get:
sin x = -b and cos x = -a
15. comparing the results in (13) and (14), we get:
sin (-x) = -sin x and cos (-x) = cos x
16. In fig.3.35(b) above, a ray has rotated by xc in the +ve direction.
♦ x is greater than $\frac{3\pi}{2}$
• Another ray has rotated by the same xc in the negative direction.
17. From the fig., it is clear that:
♦ ∠MOP = (2π - x)
♦ ∠MOQ = (2π - x)
• Since these two angles are same, the points P, Q and M are on the same vertical line.
• The x axis bisects the base PQ of the triangle OPQ
• Thus we get: QM = PM = b
18. So the coordinates of Q are: (a, b)
• But x coordinate is cosine and y coordinate is sine.
• So we get:
sin (-x) = b and cos (-x) = a
19. For the +ve angle xc, the coordinates of P are: (a, -b)
• But x coordinate is cosine and y coordinate is sine.
• So we get:
sin x = -b and cos x = a
20. comparing the results in (18) and (19), we get:
sin (-x) = -sin x and cos (-x) = cos x
21. We have considered the position of the -ve angle in all the four quadrants.
• We get the same result in all the four steps (5), (10), (15) and (20)
• So we can write the first and second identities:
Identity 1: sin (-x) = -sin x
Identity 2: cos (-x) = cos x
Next we will derive the third identity. It is related to the sum of two angles. It can be derived in 7 steps:
1. In fig.3.36 below, the green circle is a unit circle.
 |
| Fig.3.36 |
• Beginning from the +ve side of the x axis, the red ray rotated by xc in the anticlockwise direction.
• Beginning from the +ve side of the x axis, the magenta ray rotated by (x+y)c in the anticlockwise direction.
♦ So the angle between the red and magenta rays is yc
• Beginning from the +ve side of the x axis, the blue ray rotated by yc in the clockwise direction.
2. Important points and their coordinates:
• P1 is the tip of the red ray.
♦ It's coordinates will be: (cos x, sin x)
• P2 is the tip of the magenta ray.
♦ It's coordinates will be: (cos (x+y), sin (x+y))
• P3 is the tip of the blue ray.
♦ It's coordinates will be: (cos (-y), sin (-y))
• P4 is the point where the unit circle cuts the +ve side of the x axis.
♦ It's coordinates will be: (1, 0)
3. Using the five points O, P1, P2, P3 and P4, we can form two triangles:
P1OP3 and P2OP4
These two triangles are congruent. It can be proved in 5 steps:
(i) ∠ P2OP3 = (2π - x - y - y) = (2π - x - 2y)
(ii) In the triangle P1OP3,
∠ P1OP3 = [y + ∠ P2OP3] = [y + (2π - x - 2y)] = [2π - x - y]
(iii) In the triangle P2OP4,
∠ P2OP4 = [∠ P2OP3 + y] = [(2π - x - 2y) + y] = [2π - x - y]
(iv) So the same angle [2π - x - y] is present in both the triangles.
(v) In triangle P1OP3, this [2π - x - y] angle is included in between two radii.
• In triangle P2OP4 also, this [2π - x - y] angle is included in between two radii.
• So it is a SAS congruence.
4. Since the two triangles are congruent, we get: P1P3 = P2P4
5. Lengths of the these two sides can be calculated using distance formula.
| • The distance between two points A (x1, y1) and B (x2, y2) is given by: AB2 = [(x2 - x1)2 + (y2 - y1)2] • Details can be seen here. |
◼ Let us first calculate P1P3. It can be written in 3 steps:
(i) First we write the coordinates:
♦ Coordinates of P1 are: (cos x, sin x)
♦ Coordinates of P3 are: (cos (-y), sin (-y))
(ii) Now we apply the distance formula:
(P1P3)2 = [(cos (-y) - cos x)2 + (sin (-y) - sin x)2]
(iii) Using identities 1 and 2, we have: sin (-y) = -sin y and cos (-y) = cos y
• So the result in (ii) becomes:
(P1P3)2 = [(cos y - cos x)2 + (-sin y - sin x)2]
⇒ (P1P3)2 = cos2y - 2 cos x cos y + cos2x + (-sin y)2 - 2 (-sin y) (sin x) + sin2x
⇒ (P1P3)2 = cos2y - 2 cos x cos y + cos2x + sin2y + 2 sin y sin x + sin2x
⇒ (P1P3)2 = 2 - 2 cos x cos y + 2 sin y sin x
[∵ cos2y + sin2y = 1 and cos2x + sin2x = 1]
◼ Next let us calculate P2P4. It can be written in 2 steps:
(i) First we write the coordinates:
♦ Coordinates of P2 are: (cos (x+y), sin (x+y))
♦ Coordinates of P4 are: (1, 0)
(ii) Now we apply the distance formula:
(P2P4)2 = [(1 - cos (x+y))2 + (0 - sin (x+y))2]
⇒ (P2P4)2 = [1 - 2 cos (x+y) + cos2(x+y) + 02 - 2 0 sin (x+y) + sin2(x+y)]
⇒ (P2P4)2 = [1 - 2 cos (x+y) + cos2(x+y) + sin2(x+y)]
⇒ (P2P4)2 = [2 - 2 cos (x+y)]
[∵ cos2(x+y) + sin2(x+y) = 1]
6. From (4), the two lengths P1P3 and P2P4 are equal. So their squares must also be equal.
• Equating the squares, we get:
2 - 2 cos x cos y + 2 sin y sin x = 2 - 2 cos (x+y)
⇒ -2 cos x cos y + 2 sin y sin x = -2 cos (x+y)
⇒ -cos x cos y + sin y sin x = -cos (x+y)
⇒ cos x cos y - sin y sin x = cos (x+y)
7. Thus we get the third identity:
Identity 3: cos (x+y) = cos x cos y - sin x sin y
Next we will derive the fourth identity. It is related to the difference of two angles. It can be derived in 3 steps:
1. We have the Identity 3: cos (x+y) = cos x cos y - sin x sin y
2. Put (-y) in place of y. We get:
cos (x+(-y)) = cos x cos (-y) - sin x sin (-y)
⇒ cos (x-y) = cos x cos y - sin x × -sin y
[∵ From first and second identities, cos (-y) = cos y and sin (-y) = -sin y]
3. Thus we get the fourth identity:
Identity 4: cos (x-y) = cos x cos y + sin x sin y
• Identities 3 and 4 follow a similar pattern. The only difference is in the sign.
• So we can write them together as:
Identities 3 & 4: cos (x±y) = cos x cos y ∓ sin x sin y.
• On the left side, we have ± and on the right side, we have ∓.
♦ If on the left side we apply the top sign '+', on the right side also, we must apply the top sign '-'
♦ If on the left side we apply the bottom sign '-', on the right side also, we must apply the bottom sign '+'.
Next we will derive the fifth identity. It can be derived in 3 steps:
1. We have the Identity 4: cos (x-y) = cos x cos y + sin x sin y
2. put x = $\frac{\pi}{2}$ and y = x. We get:
$\cos (\frac{\pi}{2} - x)=\cos \frac{\pi}{2} \sin x + \sin \frac{\pi}{2} \sin x$
$\cos (\frac{\pi}{2} - x)=0 \times \sin x + 1 \times \sin x$
3. Thus we get the fifth identity:
Identity 5: $\cos (\frac{\pi}{2} - x)= \sin x$
Next we will derive the sixth identity. It can be derived in 3 steps:
1. Using the identity 5, we can write:
$\cos \left[\frac{\pi}{2}-\left(\frac{\pi}{2} - x\right)\right]= \sin \left(\frac{\pi}{2} - x\right)$
2. But $\cos \left[\frac{\pi}{2}-\left(\frac{\pi}{2} - x\right)\right]$ is cos x
So the result in (1) becomes: $\cos x = \sin \left(\frac{\pi}{2} - x\right)$
3. Thus we get the sixth identity:
Identity 6: $ \sin \left(\frac{\pi}{2} - x\right)=\cos x$
Next we will derive the seventh identity. It can be derived in 4 steps:
1. Using the identity 5, we can write:
$\cos \left[\frac{\pi}{2}-(x+y)\right]= \sin (x+y)$
2. The left side can be rearranged. We get:
$\cos \left[\left(\frac{\pi}{2}-x\right)-y \right]= \sin (x+y)$
3. Using identity 4, we can expand the left side. We get:
$\cos \left(\frac{\pi}{2}-x\right) \, \cos y + \sin \left(\frac{\pi}{2}-x\right) \, \sin y = \sin (x+y)$
⇒ $\sin x \, \cos y + \cos x \sin y = \sin (x+y)$
4. Thus we get the seventh identity:
Identity 7: $\sin (x+y)=\sin x \, \cos y + \cos x \sin y$
Next we will derive the eighth identity. It can be derived in 2 steps:
1. In identity 7, put -y in place of y. We get:
$\sin (x+(-y))=\sin x \, \cos (-y) + \cos x \sin (-y) $
⇒ $\sin (x-y)=\sin x \, \cos y + \cos x \times -\sin y $
⇒ $\sin (x-y)=\sin x \, \cos y - \cos x \sin y $
2. Thus we get the eithth identity:
Identity 8: $\sin (x-y)=\sin x \, \cos y - \cos x \sin y $
• Identities 7 and 8 follow a similar pattern. The only difference is in the sign.
• So we can write them together as:
Identities 7 & 8: sin (x±y) = sin x cos y ± cos x sin y.
• On the left side, we have ± and on the right side, we have ±.
♦ If on the left side we apply the top sign '+', on the right side also, we must apply the top sign '+'
♦ If on the left side we apply the bottom sign '-', on the right side also, we must apply the bottom sign '-'.
Next we will derive the ninth identity. Here we will derive 8 simple identities from 9(a) to 9(h). They can be derived in steps:
1. Consider the Identity 3: cos (x+y) = cos x cos y - sin x sin y.
• We put $\frac{\pi}{2}$ in place of x and x in place of y. We get:
$\cos (\frac{\pi}{2}+x) = \cos \frac{\pi}{2}\, \cos x - \sin \, \frac{\pi}{2} \sin x$
⇒ $\cos (\frac{\pi}{2}+x) =0 \times \cos x - 1 \times \sin x$
⇒ $\cos (\frac{\pi}{2}+x) =- \sin x$
Identity 9(a): $\cos (\frac{\pi}{2}+x) =- \sin x$
2. Consider the Identity 7: $\sin (x+y)=\sin x \, \cos y + \cos x \sin y$
• We put $\frac{\pi}{2}$ in place of x and x in place of y. We get:
$\sin (\frac{\pi}{2}+x) = \sin \frac{\pi}{2}\, \cos x + \cos \, \frac{\pi}{2} \sin x$
⇒ $\sin (\frac{\pi}{2}+x) =1 \times \cos x + 0 \times \sin x$
⇒ $\sin (\frac{\pi}{2}+x) = \cos x$
Identity 9(b): $\sin (\frac{\pi}{2}+x) = \cos x$
3. Consider the Identity 4: cos (x-y) = cos x cos y + sin x sin y.
• We put π in place of x and x in place of y. We get:
cos (π - x) = cos π cos x + sin π sin x.
⇒ cos (π - x) = -1 × cos x + 0 × sin x.
⇒ cos (π - x) = - cos x
Identity 9(c): cos (π - x) = - cos x
4. Consider the Identity 8: $\sin (x-y)=\sin x \, \cos y - \cos x \sin y $
• We put π in place of x and x in place of y. We get:
sin (π - x) = sin π cos x - cos π sin x
⇒ sin (π - x) = 0 × cos x - (-1) × sin x
⇒ sin (π - x) = sin x
Identity 9(d): sin (π - x) = sin x
5. Consider the Identity 3: cos (x+y) = cos x cos y - sin x sin y.
• We put π in place of x and x in place of y. We get:
cos (π + x) = cos π cos x - sin π sin x
⇒ cos (π + x) = -1 × cos x - 0 × sin x
⇒ cos (π + x) = -cos x
Identity 9(e): cos (π + x) = -cos x
6. Consider the Identity 3: sin (x+y) = sin x cos y + cos x sin y.
• We put π in place of x and x in place of y. We get:
sin (π + x) = sin π cos x + cos π sin x.
⇒ sin (π + x) = 0 × cos x + (-1) × sin x
⇒ sin (π + x) = -sin x
Identity 9(f): sin (π + x) = -sin x
7. Consider the Identity 4: cos (x-y) = cos x cos y + sin x sin y.
• We put 2π in place of x and x in place of y. We get:
cos (2π - x) = cos 2π cos x + sin 2π sin x.
⇒ cos (2π - x) = +1 × cos x + 0 × sin x.
⇒ cos (2π - x) = cos x
Identity 9(g): cos (2π - x) = cos x.
8. Consider the Identity 8: $\sin (x-y)=\sin x \, \cos y - \cos x \sin y $
• We put 2π in place of x and x in place of y. We get:
sin (2π - x) = sin 2π cos x - cos 2π sin x
⇒ sin (2π - x) = 0 × cos x - 1 × sin x
⇒ sin (2π - x) = -sin x
Identity 9(h): sin (2π - x) = sin x
• The above identities from 9(a) to 9(h) are related to sin x and cos x. Using them, we can derive similar identities for tan x, cot x, sec x and csc x.
• Let us see an example:
$\tan (\frac{\pi}{2}+x)=\frac{\sin (\frac{\pi}{2}+x)}{\cos (\frac{\pi}{2}+x)}=\frac{\cos x}{\sin x}=\tan x$
• Thus we get the identity: $\tan (\frac{\pi}{2}+x)=\tan x$
In
the next
section, we will see a few more identities.
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