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David Terr
Ph.D. Math, UC Berkeley

 

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4.4.Trigonometric Functions of Any Angle

In Section 4.2 we defined the six trigonometric functions for arbitrary angles by means of the unit circle. In Section 4.3 we used definitions in terms of right angles, which only work for angles up to π/2. In this section, we resume our original definition to investigate the behavior of the trigonometric functions for arbitrary angles.

Figure 4.2.2 tells the whole story, except for the secant and cosecant, which are better to think of in terms of Equations (4.2.5) and (4.2.6) respectively. By looking carefully at the figure, you might have noticed some symmetries. For instance, what happens if we interchange the roles of x and y? Then the angle θ gets replaced with its compliment, π/2 - θ and the roles of sine and cosine become switched. Thus we have the following identities:

  • (4.4.1) sin(π/2 - θ) = cos θ
  • (4.4.2) cos(π/2 - θ) = sin θ

While very useful, these identities do not tell us how to compute trigonometric functions of angles greater than 90 degrees. For that, we refer to Figure 4.2.2 again. What happens when we negate x? Now the angle θ gets replace with its suppliment, π - θ, and the point P = (x,y) gets reflected about the y-axis to the point P' = (-x,y) as shown.

quadrant 2 identities

Since we have x = cos θ and y = sin θ, we see that the coordinates of the point P' are x' = cos(π - θ) = -x = -cos θ and y' = -sin(π - θ) = y = sin θ. Thus we have proven the following identities:

  • (4.4.3) sin(π - θ) = sin θ
  • (4.4.4) cos(π - θ) = -cos θ

These formulas may be used to determine the sine and cosine of angles between π/2 and π, as the following examples show:

 

Example 1: Evaluate the following:

  • (a) sin 3π/4
  • (b) cos 3π/4
  • (c) tan 3π/4
  • (d) sin 5π/6
  • (e) cot 2π/3
  • (f) sec 5π/6

Solution: For all these, we make use of Equations (4.4.3) and (4.4.4).

  • (a) Using (4.4.3) we see that sin 3π/4 = sin(π - π/4) = sin π/4 = √2/2.
  • (b) Using (4.4.4) we see that cos 3π/4 = cos(π - π/4) = -cos π/4 = - √2/2.
  • (c) From the previous two examples, we see that tan 3π/4 = (sin 3π/4) / (cos 3π/4) = -1.
  • (d) Using (4.4.3) we see that sin 5π/6 = sin(π - π/6) = sin π/6 = 1/2.
  • (e) To evaluate cot 2π/3 we must first evaluate sin 2π/3 and cos 2π/3. From (4.4.3) we have sin 2π/3 = sin(π - π/3) = sin π/3 = √3/2. From (4.4.4) we have cos 2π/3 = cos(π - π/3) = -cos π/3 = -1 /2. Thus we have cot 2π/3 = (cos 2π/3) / (sin 2π/3) = -1/√3 = -√3/3.
  • (f) From (4.4.4) we have cos 5π/6 = cos(π - π/6) = -cos π/6 = -√3/2. Thus we have sec 5π/6 = 1 / (cos 5π/6) = -2/√3 = -2√3/3.

 

Thus far we have considered angles up to π. We need two more set of identities to compute trigonometric functions of all angles up to 2π. Formulas for sin(θ + π) and cos(θ + π) will allow us to compute the values of trigonometric functions of angles lying in the third quadrant, i.e. angles between π and 3π/2 . We derive these similarly to how we derived Equations (4.4.3) and (4.4.4). This time we consider how cosine and sine are transformed when we replace θ with θ + π, which amounts to negating both x and y. The following diagram shows the effect of this transformation.

quadrant 3 identities

As you can see, the point P = (x,y) = (cos θ, sin θ) now gets transformed to the point P' = (-x,-y) = (-cos θ, -sin θ), whence we have the following:

  • (4.4.5) sin(θ + π) = -sin θ
  • (4.4.6) cos(θ + π) = -cos θ

 

We still need to be able to compute trigonometric functions of angles in the fourth quadrant, i.e. angles between 3π/2 and 2π. By applying (4.4.3) and (4.4.5), we have sin(2π - θ) = sin((π - θ) + π) = -sin(π - θ) = -sin θ. Similarly, using (4.4.4) and (4.4.6), we have cos(2π - θ) = cos((π - θ) + π) = -cos(π - θ) = cos θ. Thus we have proven the following additional identities:

  • (4.4.7) sin(2π - θ) = -sin θ
  • (4.4.8) cos(2π - θ) = cos θ

 

What about angles that are not between 0 and 2π? We have not discussed these yet. What do we mean by such an angle? The way we defined the angular measure in Section 4.1, such angles would not be well-defined. Nevertheless, one can imagine an arbitrary angluar measure by considering rotary motion, such as the turning of a wheel or the hands of a clock. Thus, for instance, in two hours, the minute hand of a clock turns by an angle of 720 degrees, or 4π radians. (Since angles are defined mathematically as being positive when going counterclockwise, strictly speaking, the minute hand of a clock actually turns by an angle of -720 degrees, or -4π radians.) Similarly, the earth rotates about its axis by an angle of 2π every day. (This time this is a positive angle, since it rotates counterclockwise.) Thus, in one week, the earth rotates by an angle of 14π.

Clearly, every time something rotates by an angle of 2π it returns to its initial configuration, since it has completed a full circle. Thus, the trigonometric functions all remain unchanged when multiples of 2π is added to their arguments. This is expressed mathematically by the following formulas. (Here n represents an arbitrary integer.)

  • (4.4.9) sin(θ + 2πn) = sin θ
  • (4.4.10) cos(θ + 2πn) = cos θ

Another way of saying this is that the functions sine and cosine (as well as the other four trigonometric functions) are periodic with period 2π. In fact, it can easily be shown that the tangent and cotangent functions are periodic with period π, i.e. we have

  • (4.4.11) tan(θ + πn) = tan θ
  • (4.4.12) cot(θ + πn) = cot θ

We leave the proofs of (4.4.11) and (4.4.12) as an exercise.

 

Now we have enough munition to be able to compute trigonometric functions of arbitrary angles from our knowledge of theses trigonometric functions of the associated angles between 0 and π/2.

Example 2: Use the appropriate identities to evaluate the following trigonometric expressions.

  • (a) sin 5π/4
  • (b) cos 7π/4
  • (c) tan 7π/6
  • (d) cot 11π/6
  • (e) sec π
  • (f) csc 3π/2
  • (g) sin 5π/2
  • (h) cos(-π/4)
  • (i) tan(-11π/3)
  • (j) sec(627π/4)

Solution:

  • (a) Since 5π/4 lies in the third quadrant, we use Equation (4.4.5) with θ=π/4. We obtain sin 5π/4 = sin(π/4 + π) = -sin π/4 = -√2/2.
  • (b) Since 7π/4 lies in the fourth quadrant, we use Equation (4.4.8) with θ=π/4. We obtain cos 7π/4 = cos(2π - π/4) = cos π/4 = √2/2.
  • (c) Note that 7π/6 = π/6 + π, we see from Equation (4.4.11) with θ=π/6 and n=1 that tan 7π/6 = tan(π/6 + π) = tan π/6 = √3/3.
  • (d) Since 11π/6 lies in the fourth quadrant, we use Equations (4.4.8) and (4.4.9) with θ=π/6. We first find sin 11π/6 = sin(2π - π/6) = -sin π/6 = -1/2 and cos 11π/6 = cos(2π - π/6) = cos π/6 = √3/2. Thus we have tan 11π/6 = (sin 11π/6) / (cos 11π/6) = -1/√3 = -√3/3.
  • (e) First we compute cos π, which by (4.4.6) is equal to cos(0+π) = -cos 0 = -1. Thus we have sec π = 1 / cos π = 1 / -1 = -1.
  • (f) First we use (4.4.3) to compute sin 3π/2 = sin(π - π/2) = sin π/2 = 1. Thus we have csc 3π/2 = 1 / (sin 3π/2) = 1/1 = 1.
  • (g) Applying (4.4.9) with θ=π/2 and n=1, we have sin 5π/2 = sin(π/2 + 2π) = sin π/2 = 1.
  • (h) Applying (4.4.10) with θ=7π/4 and n=-1, we have cos(-π/4) = cos(7π/4 + 2π) = cos 7π/4, which we saw from (b) is equal to √2/2.
  • (i) Applying (4.4.11) with θ=π/3 and n=-4, we have tan(-11π/3) = tan(π/3 - 4π) = tan π/3 = √3.
  • (j) Applying (4.4.10) with θ=3π/4 and n=78, we first find cos(627π/4) = cos(3π/4 + 156π) = cos 3π/4, which by (4.4.4) is equal to cos(π-π/4) = -cos π/4 = -1/√2, whence we have sec(627π/4) = 1 / cos(627π/4) = -√2.

 

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