In the previous section we saw solved examples on right triangles with special angles: 30o, 60o and 45o. In this section, we will see two more special angles: 0o and 90o.
Consider the circle in fig.1 below.
1. It's center 'O' is at the intersection of two axes: the X axis and the Y axis. These two axes are perpendicular to each other.
2. Consider the radius OB1.
• It makes an angle θ1 with the horizontal axis.
3. From B1, a perpendicular B1A1 is dropped onto the horizontal axis.
• So we get a right triangle: ⊿OA1B1
4. Now, if we increase the value of θ, with O as the pivot point, the radius OB1 will rotate about O.
• Let the new value of θ be θ2.
• B1 will move to B2 and A1 will move to A2.
• Here also we get a right triangle: ⊿OA2B2
5. Let us take trigonometric ratios in both the triangles:
• In ⊿OA1B1, sin θ1 = A1B1⁄OB1
• In ⊿OA2B2, sin θ2 = A2B2⁄OB2
6. But denominators are the same because OB1 = OB2 (since radii of the same circle)
• Also A1B1 < A2B2 (since A2B2 is nearer to the Y axis)
7. Since denominators are the same, from (5), we can easily see that sin θ1 is less than sin θ2
■ In general we can say that, when the angle increases, the sine value increases
Now we will see the extreme values.
• We have seen that when the angle increases, the sine value increases. Then what is the maximum value?
To find the answer, consider the fig.2 below:
1. The angle θ is now increased to θ3. This θ3 is very nearly equal to 90o.
2. Consider the right triangle ⊿OA3B3
• We can see that the opposite side A3B3 is very nearly equal to the hypotenuse OB3
• When θ3 is exactly equal to 90, A3B3 will become exactly equal to OB3
• So when θ3 is very close to 90, we can consider A3B3 to be very close to OB3.
♦ In such a condition, the difference in lengths between A3B3 and OB3 will be very small like 0.000001 cm
♦ Even when the difference in lengths is such a low value, the ⊿OA3B3 will exist
♦ But if the difference become zero, the ⊿OA3B3 will disappear because, A3B3 will coincide with OB3.
•For practical purposes, when the difference between A3B3 and OB3 is very small, we can take:
θ3 = 90o and A3B3 = OB3.
3. we will get:
sin θ3 = sin 90 = A3B3⁄OB3 = 1 (since numerator = denominator)
■ That is., we can take sin 90 = 1
Now we will see the other extreme:
• We have seen that when the angle increases, the sine value increases. It increases upto a maximum of 1.
• That means, when angle decreases, the sine value decreases.
• Then what is the minimum value?
To find the answer, consider the fig.3 below:
1. θ is now decreased to θ0. This θ0 is very nearly equal to 0o
2. Consider the right triangle: ⊿OA0B0
• We can see that the opposite side A0B0 is now very small
• When θ0 is exactly equal to 0, the length of the opposite side A0B0 will become exactly equal to 0 cm
• So when θ0 is very close to 0, we can consider A0B0 to be very close to 0 cm
♦ In such a condition, the length of A0B0 will be very small like 0.000001 cm
♦ Even when the length is such a low value, the ⊿OA0B0 will exist
♦ But if the length become zero, the ⊿OA0B0 will disappear because, A0B0 will coincide with OB0.
•For practical purposes, when the length of A0B0 is very small, we can take:
θ0 = 0o and A0B0 = 0 cm
3. we will get:
sin θ0 = sin 0 = A0B0⁄OB0 = 0 (since numerator = 0)
■ That is., we can take sin 0 = 0
So we got three information:
• When the angle increases, sine value increases
• The minimum value is: sin 0 = 0
• The maximum value is: sin 90 = 1
Next we will consider cosine. The same three figs. can be used for this purpose also
1. Consider fig.1 above. For the angle θ, the opposite sides are A1B1, A2B2 … so on
• As θ increases, these opposite sides increases
2. But the adjacent sides are OA1, OA2 . . . so on
• As θ increases, these adjacent sides decreases.
3. Let us see the effect of this decrease when we take the cosine ratios:
• In ⊿OA1B1, cos θ1 = OA1⁄OB1
• In ⊿OA2B2, cos θ2 = OA2⁄OB2
4. But denominators are the same because OB1 = OB2 (since radii of the same circle)
• Also OA2 < OA1 (since A2 is nearer to the origin than A1)
5. Since denominators are the same, from (3), we can easily see that cos θ2 is less than cos θ1
■ In general, we can say that, when the angle increases, the cosine value decreases
Now we will see the extreme values for cosine.
• We have seen that when the angle increases, the cosine value decreases. Then what is the minimum value?
To find the answer, consider the fig 2 above
1. The angle θ is now increased to θ3. This θ3 is very nearly equal to 90o.
2. Consider the right triangle ⊿OA3B3
• We can see that the adjacent side OA3 is now very small
• When θ3 is exactly equal to 90, the length of the adjacent side OA3 will become exactly equal to 0 cm
• So when θ3 is very close to 90, we can consider OA3 to be very close to 0 cm
♦ In such a condition, the length of OA3 will be very small like 0.000001 cm
♦ Even when the length is such a low value, the ⊿OA3B3 will exist
♦ But if the length become zero, the ⊿OA3B3 will disappear because, A3B3 will coincide with OB3.
•For practical purposes, when the length of OA3 is very small, we can take:
θ3 = 90o and OA3 = 0 cm
3. we will get:
cos θ3 = cos 90 = OA3⁄OB3 = 0 (since numerator = 0)
■ That is., we can take cos 90 = 0
Now we will see the other extreme for cosine:
• We have seen that when the angle increases, the cosine value decreases. It decreases upto a minimum of zero.
• That means, when angle decreases, the cosine value increases.
• Then what is the maximum value?
To find the answer, consider the fig.3 above.
1. θ is now decreased to θ0. This θ0 is very nearly equal to 0o
2. Consider the right triangle: ⊿OA0B0
• We can see that the adjacent side OA0 is very nearly equal to the hypotenuse OB0
• When θ3 is exactly equal to 0, OA0 will become exactly equal to OB0
• So when θ3 is very close to 0, we can consider OA0 to be very close to OB0.
♦ In such a condition, the difference in lengths between OA0 and OB0 will be very small like 0.000001 cm
♦ Even when the difference in lengths is such a low value, the ⊿OA0B0 will exist
♦ But if the difference become zero, the ⊿OA0B0 will disappear because, OA0 will coincide with OB0.
• For practical purposes, when the difference between OA0 and OB0 is very small, we can take:
θ3 = 0o and OA0 = OB0.
3. we will get:
cos θ3 = cos 0 = OA0⁄OB0 = 1 (since numerator = denominator)
■ That is., we can take cos 0 = 1
So we got three information:
• When the angle increases, cosine value decreases
• The minimum value is: cos 90 = 0
• The maximum value is: cos 0 = 1
Next we will consider tangent ratio. This does not need lengthy calculations because we know that:
tan θ = sin θ⁄cos θ
So we get:
• tan 0 = sin 0⁄cos 0 = 0⁄1 = 0
• tan 90 = sin 90⁄cos 90 = 1⁄0 . So tan 90 is not defined because it is a division by zero.
Consider the circle in fig.1 below.
1. It's center 'O' is at the intersection of two axes: the X axis and the Y axis. These two axes are perpendicular to each other.
 |
| Fig.1 |
• It makes an angle θ1 with the horizontal axis.
3. From B1, a perpendicular B1A1 is dropped onto the horizontal axis.
• So we get a right triangle: ⊿OA1B1
4. Now, if we increase the value of θ, with O as the pivot point, the radius OB1 will rotate about O.
• Let the new value of θ be θ2.
• B1 will move to B2 and A1 will move to A2.
• Here also we get a right triangle: ⊿OA2B2
5. Let us take trigonometric ratios in both the triangles:
• In ⊿OA1B1, sin θ1 = A1B1⁄OB1
• In ⊿OA2B2, sin θ2 = A2B2⁄OB2
6. But denominators are the same because OB1 = OB2 (since radii of the same circle)
• Also A1B1 < A2B2 (since A2B2 is nearer to the Y axis)
7. Since denominators are the same, from (5), we can easily see that sin θ1 is less than sin θ2
■ In general we can say that, when the angle increases, the sine value increases
Now we will see the extreme values.
• We have seen that when the angle increases, the sine value increases. Then what is the maximum value?
To find the answer, consider the fig.2 below:
 |
| Fig.2 |
2. Consider the right triangle ⊿OA3B3
• We can see that the opposite side A3B3 is very nearly equal to the hypotenuse OB3
• When θ3 is exactly equal to 90, A3B3 will become exactly equal to OB3
• So when θ3 is very close to 90, we can consider A3B3 to be very close to OB3.
♦ In such a condition, the difference in lengths between A3B3 and OB3 will be very small like 0.000001 cm
♦ Even when the difference in lengths is such a low value, the ⊿OA3B3 will exist
♦ But if the difference become zero, the ⊿OA3B3 will disappear because, A3B3 will coincide with OB3.
•For practical purposes, when the difference between A3B3 and OB3 is very small, we can take:
θ3 = 90o and A3B3 = OB3.
3. we will get:
sin θ3 = sin 90 = A3B3⁄OB3 = 1 (since numerator = denominator)
■ That is., we can take sin 90 = 1
Now we will see the other extreme:
• We have seen that when the angle increases, the sine value increases. It increases upto a maximum of 1.
• That means, when angle decreases, the sine value decreases.
• Then what is the minimum value?
To find the answer, consider the fig.3 below:
 |
| Fig.3 |
2. Consider the right triangle: ⊿OA0B0
• We can see that the opposite side A0B0 is now very small
• When θ0 is exactly equal to 0, the length of the opposite side A0B0 will become exactly equal to 0 cm
• So when θ0 is very close to 0, we can consider A0B0 to be very close to 0 cm
♦ In such a condition, the length of A0B0 will be very small like 0.000001 cm
♦ Even when the length is such a low value, the ⊿OA0B0 will exist
♦ But if the length become zero, the ⊿OA0B0 will disappear because, A0B0 will coincide with OB0.
•For practical purposes, when the length of A0B0 is very small, we can take:
θ0 = 0o and A0B0 = 0 cm
3. we will get:
sin θ0 = sin 0 = A0B0⁄OB0 = 0 (since numerator = 0)
■ That is., we can take sin 0 = 0
So we got three information:
• When the angle increases, sine value increases
• The minimum value is: sin 0 = 0
• The maximum value is: sin 90 = 1
Next we will consider cosine. The same three figs. can be used for this purpose also
1. Consider fig.1 above. For the angle θ, the opposite sides are A1B1, A2B2 … so on
• As θ increases, these opposite sides increases
2. But the adjacent sides are OA1, OA2 . . . so on
• As θ increases, these adjacent sides decreases.
3. Let us see the effect of this decrease when we take the cosine ratios:
• In ⊿OA1B1, cos θ1 = OA1⁄OB1
• In ⊿OA2B2, cos θ2 = OA2⁄OB2
4. But denominators are the same because OB1 = OB2 (since radii of the same circle)
• Also OA2 < OA1 (since A2 is nearer to the origin than A1)
5. Since denominators are the same, from (3), we can easily see that cos θ2 is less than cos θ1
■ In general, we can say that, when the angle increases, the cosine value decreases
Now we will see the extreme values for cosine.
• We have seen that when the angle increases, the cosine value decreases. Then what is the minimum value?
To find the answer, consider the fig 2 above
1. The angle θ is now increased to θ3. This θ3 is very nearly equal to 90o.
2. Consider the right triangle ⊿OA3B3
• We can see that the adjacent side OA3 is now very small
• When θ3 is exactly equal to 90, the length of the adjacent side OA3 will become exactly equal to 0 cm
• So when θ3 is very close to 90, we can consider OA3 to be very close to 0 cm
♦ In such a condition, the length of OA3 will be very small like 0.000001 cm
♦ Even when the length is such a low value, the ⊿OA3B3 will exist
♦ But if the length become zero, the ⊿OA3B3 will disappear because, A3B3 will coincide with OB3.
•For practical purposes, when the length of OA3 is very small, we can take:
θ3 = 90o and OA3 = 0 cm
3. we will get:
cos θ3 = cos 90 = OA3⁄OB3 = 0 (since numerator = 0)
■ That is., we can take cos 90 = 0
Now we will see the other extreme for cosine:
• We have seen that when the angle increases, the cosine value decreases. It decreases upto a minimum of zero.
• That means, when angle decreases, the cosine value increases.
• Then what is the maximum value?
To find the answer, consider the fig.3 above.
1. θ is now decreased to θ0. This θ0 is very nearly equal to 0o
2. Consider the right triangle: ⊿OA0B0
• We can see that the adjacent side OA0 is very nearly equal to the hypotenuse OB0
• When θ3 is exactly equal to 0, OA0 will become exactly equal to OB0
• So when θ3 is very close to 0, we can consider OA0 to be very close to OB0.
♦ In such a condition, the difference in lengths between OA0 and OB0 will be very small like 0.000001 cm
♦ Even when the difference in lengths is such a low value, the ⊿OA0B0 will exist
♦ But if the difference become zero, the ⊿OA0B0 will disappear because, OA0 will coincide with OB0.
• For practical purposes, when the difference between OA0 and OB0 is very small, we can take:
θ3 = 0o and OA0 = OB0.
3. we will get:
cos θ3 = cos 0 = OA0⁄OB0 = 1 (since numerator = denominator)
■ That is., we can take cos 0 = 1
So we got three information:
• When the angle increases, cosine value decreases
• The minimum value is: cos 90 = 0
• The maximum value is: cos 0 = 1
Next we will consider tangent ratio. This does not need lengthy calculations because we know that:
tan θ = sin θ⁄cos θ
So we get:
• tan 0 = sin 0⁄cos 0 = 0⁄1 = 0
• tan 90 = sin 90⁄cos 90 = 1⁄0 . So tan 90 is not defined because it is a division by zero.
So we have completed this present discussion on the two special angles 0o and 90o.
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