16 ## TRIGONOMETRIC FUNCTIONS OF ANY ANGLEHOW 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 In each quadrant, φ is the corresponding acute angle of θ. φ is the shortest angular distance to the (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. 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 (We will prove that below.) Example 1. Evaluate tan 120°.
tan 120° = ±tan 60° = ± Now in the second quadrant, tan θ ( tan 120° = − Example 2. Evaluate cos 195°.
cos 195° = ±cos 15° = ± .966, from the Table. In the third quadrant, cos θ ( cos 195° = − .966 Example 3. Evaluate sec (−45°).
sec 45° = ±sec 45° = ± In the fourth quadrant, sec θ ( 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
The supplement, 180°
sin θ = sin (180°
cos θ = −cos (180°
Problem 3. What radian angle less than 2π is a) sin b) tan c) cos 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 ( We come now to the following theorem:
No matter in which quadrant θ falls, both θ and −θ have the same corresponding acute angle. What is more, both fall in the cos (−θ) = cos θ. On the other hand, θ and −θ fall in sin (−θ) = −sin θ. That is what we wanted to prove. In the language of functions, cos 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.
same corresponding acute angle. And θ + π will fall in the cos (θ + π) = −cos θ.
Problem 6. cos (θ + 5π) = −cos θ. Explain why.
θ plus any
Polar coördinates We can specify the position of a point P by giving its distance But the polar coördinates are easily related to the rectangular coördinates (
(Topic 15), then
And since
Example 5. A radius of 8 cm sweeps out an angle of 30° in standard position. What are the rectangular coordinates (
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, 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,
Proof of the main theorem A function of any angle will equal plus or minus that same function First, if θ is a second quadrant angle, then
while in the first quadrant,
Therefore the sine of θ is equal to the sine of the corresponding acute angle. Similarly,
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
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
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. Please make a donation to keep TheMathPage online. Copyright © 2014 Lawrence Spector Questions or comments? E-mail: themathpage@nyc.rr.com |