Assignment 1
Math 69.007A, Winter, 1997
Instructor: Prof. A. Mingarelli
Due: Jan. 24, 1997
NOTATION: Pi = 3.14159... is our usual Greek number
1. [10] Evaluate the following quantities exactly (i.e., not as numbers found
by a calculator):
a)sin (11 Pi/12)
Solution:1. a) [sqrt(3)-1]/[2sqrt(2)] ... Note that sin(11Pi/12) = sin(Pi/12) by the sin-difference formula and because 11Pi/12 = Pi - Pi/12. Now, write Pi/12 = Pi/3 - Pi/4, use the sin-difference formula and simplify.
b) cos (13 Pi/12)
Solution:1. b) -sqrt(2+sqrt(3))/2 ... Use the cos-sum formula: Note that 13Pi/12 = Pi + Pi/12...so that cos(13Pi/12) = -cos(Pi/12). But we found sin(Pi/12) in a) above. So, cos(Pi/12) = sqrt(2+sqrt(3))/2, and thus our answer is the negative of the latter.
c) tan (-5 Pi/12)
Solution: 1. c) -sqrt{[2+sqrt(3)/(2-sqrt(3)]} ... Note that -5Pi/12 = Pi/12 - 6Pi/12 = Pi/12 - Pi/2. So we can convert tan into sin and cos and use the fact that we know these functions for the angle Pi/12, from a) and b) above. You'll find that tan(-5Pi/12) = -cot(Pi/12) which is the desired answer.
d) sin(Pi/4 - Pi/3)
Solution: 1. d) {sqrt(2) - sqrt(6)}/4 ... Use the sin-difference formula and simplify.
2. [10] If 2 sin(x - y) = sin(x + y) prove that tan(x) = 3 tan(y).
Solution:The simplest way of doing this problem is by expanding both sides with the sin-difference and sin-sum formulae separately, recombining terms and note that you'll obtain: sin(x) cos(y) = 3 cos(x) sin(y). Dividing both sides by cos(x) cos(y) we get the result.
3. [10] Solve the following equation for x: 2 cos2(x) + sin(x) = 2 where x lies between "0" and "2 Pi".
Solution: Pi/6 and 5Pi/6...By inspection we see that x = 0 and x = 2Pi are solutions but they are not the ones we want, right? (because we want 0 < x < 2Pi for our solutions). Use of the trig. identity cos2(x) = 1 - sin2(x), we get, after some simplification and rearrangement, -2 sin2(x) + sin(x) = 0. Now, factor out the sin-term here...You'll find that sin(x) (1 - 2 sin(x) ) = 0, i.e., sin(x) = 1/2 (or sin(x) = 0, which we know gives nothing new). Finally, the only solutions of sin(x) = 1/2 are Pi/6 and 5Pi/6 for x in the given interval.
4. [10] Find the approximate "size" of our sun in the sky given that the mean distance from the earth to the (surface of the) sun is about 150,000,000 km. and the solar radius is 700,000 km. (The size is calculated by finding the angle subtended by the two tangents to the sun from a point at the center of our eye).
Solution: 0.009 radians or 0.53 degrees ...This question is from my ABC's of Calculus, (Module on Inverse Functions, 1994). Consider the right-triangle where x is the angle subtended by a tangent to the surface of the sun and a line to the center of the sun from our eye. Then, by trig., sin(x) = 7 x 105 / {(1500 x 105) + (7 x 105)} = 7/1507. We need to double this angle to get the size, though. So sin(x) = 7/1507 means x = 0.009 rads (use your calculator to find this "inverse").
Total: 40 points