The corresponding acute angle
Let θ be an angle that terminates in any quadrant. Then by the corresponding acute angle, we mean the shortest angular distance to the x-axis.
In each quadrant, φ is the corresponding acute angle of θ. φ is the shortest angular distance to the x-axis.
Examples.
If θ = 120° (Second quadrant), then φ = 60°.
If θ = 190° (Third quadrant), then φ = 10°.
If θ = 340° (Fourth quadrant), then φ = 20°.
Problem 1. Name the corresponding acute angle.
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a) 110°
70°
b) 225°
45°
c) −30°
30°
How, then, do we evaluate a function of any angle? According to the following theorem:
A function of any angle will equal plus or minus that same function
of the corresponding acute angle.
The sign will depend on the quadrant.
(We will prove that below.)
Example 1. Evaluate tan 120°.
Answer. The corresponding acute angle of 120° is 60°. Therefore, according to the theorem,
tan 120° = ±tan 60° = ±
Now in the second quadrant, tan θ (y/x) is negative (Topic 16). Therefore,
tan 120° = −
Example 2. Evaluate cos 195°.
Answer. The corresponding acute angle of 195° (third quadrant) is 15°. Therefore,
cos 195° = ±cos 15° = ± .966, from the Table.
In the third quadrant, cos θ (x/r) is negative. Therefore,
cos 195° = − .966
Example 3. Evaluate sec (−45°).
Answer. The corresponding acute angle of −45° (fourth quadrant) is 45°. Therefore,
sec 45° = ±sec 45° = ±
In the fourth quadrant, sec θ (r/x) is positive. Therefore,
sec 45° =
Problem 2. Draw a figure that illustrates the following.
The sine of an obtuse angle is equal to the sine of its supplement.
The cosine of an obtuse angle is equal to the negative of the cosine
of its supplement.
The supplement, 180° − θ, is the corresponding acute angle!
Problem 3. Evaluate each of the following. No tables.
a) sin 150°
The corresponding acute angle is 30°. And in the second quadrant, sin θ is positive. Therefore, sin 150° = sin 30° = ½.
b) cos 135°
The corresponding acute angle is 45°. And in the second quadrant, cos θ is negative. Therefore, cos 135° = −cos 45° = −½
.
c) tan 240°
The corresponding acute angle is 60°. In the third quadrant, tan θ is positive. Therefore, tan 240° = tan 60° = 
.
d) csc (−30)°
The corresponding acute angle is 30°. In the fourth quadrant, csc θ (y/r) is negative. Therefore, csc (−30)° = −csc 30° = −2.
Problem 4. Use the Table to evaluate the following.
a) cos 170°
The corresponding acute angle is 10°. In the second quadrant, cos θ is negative. Therefore, cos 170° = −cos 10° = − .985.
b) cos (−20°)
The corresponding acute angle is 20°. In the fourth quadrant, cos θ is positive. Therefore, cos (−20°) = cos 20° = .940.
c) sin (−20°)
The corresponding acute angle is 20°. In the fourth quadrant, sin θ is negative. Therefore, sin (−20°) = −sin 20° = − .342.
In the previous problem, we see an extremely useful theorem:
| cos (−θ) = cos θ, but sin (−θ) = −sin θ.
|
No matter in which quadrant θ falls, −θ has the same corresponding acute angle. And both θ and −θ fall in the same left-or-right half of the
x-y plane. The sign of the cosine depends only on which half. Therefore,
cos (−θ) = cos θ.
On the other hand, θ and −θ fall in opposite top-or-bottom halves of the plane. The sign of the sine depends on which of those halves. Therefore, sin θ and sin (−θ) will have opposite signs:
sin (−θ) = −sin θ.
That is what we wanted to prove.
Problem 5. Use the previous theorem to evaluate the following. No tables.
a) cos (−30°)
= cos 30° = /2
b) cos (−60°)
= cos 60° = ½
c) cos (−45°)
= cos 45° = ½
d) sin (−30°)
= −sin 30° = −½.
e) sin (−60°)
= −sin 60° = −/2
c) sin (−45°)
= −sin 45° = −½
Example 4. cos (θ + π) = −cos θ. Explain why.
Answer. No matter in which quadrant θ falls, θ and θ + π will have the
same corresponding acute angle. And θ + π will fall in the opposite left-or-right half of the plane. Therefore cos θ and cos (θ + π) will have opposite signs:
cos (θ + π) = −cos θ.
Note: cos (π + θ) = cos (θ + π).
Problem 6. cos (θ + 5π) = −cos θ. Explain why.
θ plus any odd multiple of π will be coterminal with θ + π. For, 2π is one revolution. Therefore, (θ + π) + 2π -- which is θ + 3π -- is coterminal with θ + π. (θ + π) + 4π -- which is θ + 5π -- is coterminal with θ + π. And so on. Therefore,
cos (θ + 5π) = cos (θ + π) = −cos θ.
Note: cos (5π + θ) = cos (θ + 5π).
Polar coördinates
We can specify the position of a point P by giving its distance r from the origin and the angle θ that r makes with the x-axis. Those are called the polar coördinates of P. (r, θ).
But the polar coördinates are easily related to the rectangular coördinates (x, y). Since (Topic 17)
The corresponding acute angle of 135° is 45°. In the second quadrant, the cosine is negative and the sine is positive. Therefore, x = r cos θ = 1· cos 135° = −cos 45° = −½
.
y = r sin θ = 1· sin 135° = sin 45° = ½.
cos (−θ) = cos θ, but sin (−θ) = −sin θ.
Therefore, x = 2· cos (−60°) = 2· cos 60° = 2· ½ = 1.
y = 2· sin (−60°) = 2(−sin 60°) = 2(−½
) = −.
