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# TRIGONOMETRIC FUNCTIONS OF ANY ANGLE

The corresponding acute angle

The main theorem

cos (−θ) and sin (−θ)

Polar coördinates

Proof of the main theorem

HOW SHALL WE EVALUATE tan 118°, 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
. 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.

a)  110°   70°           b)  225°   45°

c)  −30°   30°         d)  380°   20°

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 15).  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.

sin θ = sin (180° θ).

cos θ = −cos (180° θ).

Problem 3.   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 = −½ .  x = 3π/4 or 5π/4.

Problem 4.   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  (r/y)  is negative.  Therefore, csc (−30)° = −csc 30° = −2.

See Topic 4.

We come now to the following theorem:

 cos (−θ)  =  cos θ.   sin (−θ)  =  −sin θ. No matter in which quadrant θ falls, both θ and −θ have the same corresponding acute angle.  What is more, both 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) and cos θ will have the same sign:

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.

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

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

f)   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.  For they will be vertical angles, which are equal.  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 (xy).  Since

 xr = cos θ

(Topic 15), then

 x   =  r cos θ.

And since

 yr = sin θ,
 y   =  r sin θ.

Example 5.   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? Answer. x = r cos θ  =  8 cos 30°  =  8 · 2 =  4 .

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? 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 .

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

Proof of the main 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. 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 = br

 sin φ = br

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

Similarly,

 cos θ = − ar = −cos φ tan θ = − ba = −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. Next, if θ is a third quadrant angle, so that r terminates at (−a, −b), then

 sin θ = − br = −sin φ cos θ = − ar = −cos φ
 tan θ = −b−a = ba = tan φ

And so on, so that again, each function of θ is plus or minus that same function of φ. Finally, if θ is a fourth quadrant angle, so that r terminates at (a, −b), then

 sin θ = − br = −sin φ cos θ = ar = cos φ tan θ = − ba = −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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