Trigonometry

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16

TRIGONOMETRIC FUNCTIONS OF ANY ANGLE

The corresponding acute angle

The theorem

cos (−θ) and sin (−θ)

Polar coördinates

Proof of the theorem

HOW SHALL WE EVALUATE tan 120°, for example?  We will see that we will be referred back to an acute angle.

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.

The corresponding acute angle

In each quadrant, φ is the corresponding acute angle of θ.  φ is the shortest angular distance to the x-axis.

(The corresponding acute angle is often called the reference angle.)

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.

To see the answer, pass your mouse over the colored area.
To cover the answer again, click "Refresh" ("Reload").

  a)   150°  30°   b)   135°  45°   c)   210°  30°   d)   315°  45°
 
  e)   225°  45°   f)   240°  60°   g)   300°  60°   h)   −30°  30°

The theorem

How do we evaluate a function of any angle? The following theorem is the answer.

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° = ±square root of 3

In the second quadrant, tan θ  (y/x)  is negative.  Therefore,

tan 120° = −square root of 3.

Example 2.   Evaluate sec (−45°).

Answer.  The corresponding acute angle of −45° (fourth quadrant) is 45°.  Therefore,

sec 45° = ±sec 45° = ±square root of 2

In the fourth quadrant, sec θ  (r/x)  is positive.  Therefore,

sec 45° = square root of 2

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.

An obtuse angle

Let θ be an obtuse angle. Then the corresponding acute angle is its supplement, 180° θ.

sin θ = sin (180° θ).

  cos θ = −cos (180° θ).

Problem 3.   Evaluate each of the following.  No tables.

a)  sin 150°  

The corresponding acute angle is 30°.  In the second quadrant, sin θ is positive.  Therefore, sin 150° = sin 30° = ½.

b)  cos 135°  

The corresponding acute angle is 45°.  In the second quadrant, cos θ is negative.  Therefore, cos 135° = −cos 45° = −½square root of 2.

c)  tan 240°  

The corresponding acute angle is 60°.  In the third quadrant, tan θ is positive.  Therefore, tan 240° = tan 60°square root of 3.

d)  csc (−30)°  

The corresponding acute angle is 30°.  In the fourth quadrant, csc θ  (r/y)  is negative.  Therefore, csc (−30)° = −csc 30° = −2.

  e)  tan  3π
 4
The corresponding acute angle is  π
4
  See here

In the 2nd quadrant the tangent is negative. Therefore,

tan  3π
 4
 = −1.
  f)  cot  7π
 6
7π
 6
 = 7 · 30° = 210°.

The corresponding acute angle is 30°. In the 3rd quadrant the costangent is positive. Therefore,

cot  7π
 6
 = cot 30° sq-3
  g)  cos  5π
 3
5π
 3
 = 5 · 60° = 300°.

The corresponding acute angle is 60°. In the 4th quadrant the cosine is positive. Therefore,

cos  5π
 3
 = ½.

Problem 4.   What radian angle less than 2π is x?

a)  sin x = ½.  x = π/6 or 5π/6.

b)  tan x = 1.  x = π/4 or 5π/4.

c)  cos x = −½square root of 2.  x = 3π/4 or 5π/4.

We come now to the following theorem:

cos (−θ)  =  cos θ.   sin (−θ)  =  −sin θ.

cos minus x, sin minus x

Let (a, b) and (a, −b) be coördinates on a unit circle. Then according to the definitions,

cos θ = a.   cos (−θ) = a.

That is,

cos (−θ) = cos θ.

On the other hand,

sin θ = b.   sin (−θ) = −b.

That is,

sin (−θ) = −sin θ.

That is what we wanted to prove.

In the language of functions, cos θ is an even function: cos (−θ) = cos θ; while sin θ is an odd function: sin (−θ) = −sin θ.

It follows, then, that the graph of cos x is symmetrical with respect to the y-axis; the graph of sin x, with respect to the origin.

Problem 5.   Use the previous theorem to evaluate the following.

a)   cos (−30°)  = cos 30° = ½square root of 3

b)   cos (−60°)  = cos 60° = ½

c)   cos (−45°)  = cos 45° = ½square root of 2

d)   sin (−30°)  = −sin 30° = −½.

e)   sin (−60°)  = −sin 60° = −square root of 3/2

f)   sin (−45°)  = −sin 45° = −½square root of 2

Problem 6.   Use this figure to prove:  tan (−θ) = −tan θ.

cos minus x, sin minus x

tan θ = b/a.  tan (−θ) = −b/a.  Therefore, tan (−θ) = −tan θ.

tan θ is an odd function.

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).

Polar coordinates

But the polar coördinates are easily related to the rectangular coördinates (xy).  Since

x
r
  =   cos θ

(Topic 15), then

x   =  r cos θ.

And since

y
r
  =   sin θ,
y   =  r sin θ.

Example 4.   A radius of 8 cm sweeps out an angle of 30° in standard position. What are the rectangular coordinates (x, y) of the endpoint of the radius?

rectangular coordinates
  Answer.     x = r cos θ  =  8 cos 30°  =  8 ·  square root of 3
 2
  =  4square root of 3.

y = r sin θ  =  8 sin 30°   =  8 · ½  =  4.

Problem 7.   Radius AB of a unit circle sweeps out an angle 135°.  What are the coordinates of B?

Rectangular coordinates?

The corresponding acute angle of 135° is 45°.  In the second quadrant, the cosine is negative and the sine, positive.  Therefore, x = r cos θ = 1· cos 135° = −cos 45° = −½square root of 2.
y = r sin θ = 1· sin 135° = sin 45° = ½square root of 2.

Problem 8.   Radius AB of length 2 sweeps out an angle of −60°.  What are the coordinates of B?

rectangular coordinates

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(−½square root of 3) = −square root of 3.

Proof of the 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.

A second quadrant angle

First, if θ is a second quadrant angle, then r will terminate at a point
(−a, b).  The corresponding acute angle is φ, which is also shown in its first quadrant position.  In the second quadrant,

sin θ   =  b
r

while in the first quadrant,

sin φ   =  b
r

Therefore the sine of θ is equal to the sine of the corresponding acute angle.

Similarly,

cos θ   = − a
r
 =  −cos φ 
tan θ   = − b
a
 =  −tan φ 

And so on, for the remaining functions, so that in every case, a function of θ is plus or minus that same function of φ.

A third quadrant angle

Next, if θ is a third quadrant angle, so that r terminates at (−a, −b), then

sin θ   = − b
r
 =  −sin φ 
 
cos θ   = − a
r
 =  −cos φ
tan θ  =  b
a
 =  b
a
  = tan φ

And so on, so that again, each function of θ is plus or minus that same function of φ.

A fourth quadrant angle

Finally, if θ is a fourth quadrant angle, so that r terminates at (a, −b), then

sin θ   = − b
r
 =  −sin φ 
 
cos θ   =  a
r
 =  cos φ
 
tan θ  = − b
a
 =  −tan φ

Therefore again, each function of θ is plus or minus that same function of the corresponding acute angle φ;  which is what we wanted to prove.

Next Topic:  Line Values


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