Section 3.3

#1. a) We want the reference angle for 210o. Since it is in the third quadrant, we find 210-180 = 30, and our reference angle is 30o.
b) We want the reference angle for 300o. Since it is in the fourth quadrant, we find 360-300 = 60, and our reference angle is 60o.
c) We want the reference angle for -120o. Since it is in the third quadrant, we compare it to -180o and observe that they are 60o apart, so our reference angle is 60o.

#3. a) We want the reference angle for 9π/4. Since 9/4 is reduced, our shortcut from class tells us the reference angle is just π/4.
b) We want the reference angle for 23π/6. Since 23/6 is reduced, our shortcut from class tells us the reference angle is just π/6.
c) We want the reference angle for -π/3. Our shortcut from class tells us the reference angle is just π/3.

#9. We want cos(135o). Since 135o is in the 2nd quadrant, the reference angle would be 180o - 135o = 45o. We know that cos(45o) = sqrt(2)/2. But we know that cosine is negative in the 2nd quadrant, so our answer is -sqrt(2)/2.

#11. We want tan(-60o). Since -60o is in the 4th quadrant, the reference angle would be 0o - -60o = 60o. We know that tan(60o) = sqrt(3). But we know that tangent is negative in the 4th quadrant, so our answer is -sqrt(3).

#16. We want sec(120 degrees). Since 120 deg is in quadrant 2, the reference angle would be 180 - 120 = 60 degrees. Drawing a 30-60-90 triangle, we can read the secant as hyp/adj, which is 2/1 = 2. Now we adjust for the quadrant, and since secant (and cosine) are negative in quadrant 2, our final answer is -2.

#19. We want sin(2π/3). The reference angle for 2π/3 is π/3. sin(π/3) = sqrt(3)/2. Since 2π/3 is in the 2nd quadrant, and sine is positive in the 2nd quadrant, sin(2π/3) = sqrt(3)/2.

#20. We want sin(5π/3). The reference angle for 5π/3 is π/3. sin(π/3) = sqrt(3)/2. Since 5π/3 is in the 4th quadrant, and sine is negative in the 4th quadrant, sin(5π/3) = -sqrt(3)/2.

#23. We want cos(-7π/3). The reference angle for -7π/3 is π/3. cos(π/3) = 1/2. Since -7π/3 is in the 4th quadrant, and cosine is positive in the 4th quadrant, cos(-7π/3) = 1/2.

#31. If we know that sine is negative, then we must be in quadrant 3 or 4. If we know cosine is negative, then we must be in quadrant 2 or 3. Since we need BOTH to be negative, we must be in quadrant 3.

#33. If we know that secant is positive, then we must be in quadrant 1 or 4. If we know tangent is negative, then we must be in quadrant 2 or 4. Since we need BOTH things to be true, we must be in quadrant 4.

#42. First, we observe that θ is in quadrant 3, which will help us adjust the plus & minus signs at the end. Now we construct a triangle that has cosine equal to 7/12, i.e. that has 7 on the adjacent and 12 on the hypotenuse. That would give sqrt(144 - 49) = sqrt(95) on the opposite. Now we can find the other trig functions, including adjusting for the quadrant:
sin(θ) = -opp/hyp = -sqrt(95)/12, cos(θ) = -adj/hyp = -7/12, tan(θ) = opp/adj = sqrt(95)/7,
csc(θ) = -hyp/opp = -12/sqrt(95), sec(θ) = -hyp/adj = -12/7, cot(θ) = adj/opp = 7/sqrt(95) .

#43. First, since tangent is negative and cosine is positive, we must be in quadrant 4. Now we construct a triangle that has tangent equal to 3/4, i.e. that has 3 on the opposite and 4 on the adjacent. That would give 5 on the hypotenuse. Now we can find the other trig functions, including adjusting for the quadrant:
sin(θ) = -opp/hyp = -3/5, cos(θ) = +adj/hyp = 4/5, tan(θ) = -opp/adj = -3/4,
csc(θ) = -hyp/opp = -5/3, sec(θ) = +hyp/adj = 5/4, cot(θ) = -adj/opp = -4/3 .

#44. First, since θ is in quadrant 1, all of our trig values will be positive. Now we construct a triangle that has cosecant equal to 2 (or 2/1), i.e. that has 2 on the hypotenuse and 1 on the opposite. That would give sqrt(3) on the adjacent. Now we can find the other trig functions:
sin(θ) = opp/hyp = 1/2, cos(θ) = adj/hyp = sqrt(3)/2, tan(θ) = opp/adj = 1/sqrt(3),
csc(θ) = hyp/opp = 2/1, sec(θ) = hyp/adj = 2/sqrt(3), cot(θ) = adj/opp = sqrt(3)/1 .

#45. First, since sine is negative and secant is positive, we must be in quadrant 4. Now we construct a triangle that has secant equal to 5 (or 5/1), i.e. that has 5 on the hypotenuse and 1 on the adjacent. That would give sqrt(24) on the opposite. Now we can find the other trig functions, including adjusting for the quadrant:
sin(θ) = -opp/hyp = -sqrt(24)/5, cos(θ) = +adj/hyp = 1/5, tan(θ) = -opp/adj = -sqrt(24)/1,
csc(θ) = -hyp/opp = -5/sqrt(24), sec(θ) = +hyp/adj = 5/1, cot(θ) = -adj/opp = -1/sqrt(24) .

#46. First, since sine is negative and cotangent is positive, we must be in quadrant 3. Now we construct a triangle that has cotangent equal to 1/4, i.e. that has 1 on the adjacent and 4 on the opposite. That would give sqrt(17) on the hypotenuse. Now we can find the other trig functions, including adjusting for the quadrant:
sin(θ) = -opp/hyp = -4/sqrt(17), cos(θ) = -adj/hyp = -1/sqrt(17), tan(θ) = +opp/adj = 4/1,
csc(θ) = -hyp/opp = -sqrt(17)/4, sec(θ) = -hyp/adj = -sqrt(17)/1, cot(θ) = +adj/opp = 1/4 .

#50. From the formula in the book, the area is 0.5 ab sin(θ). Our given info was 7, 9 and 72o, so the area is (0.5)(7)(9)sin(72o), which equals (31.5)(0.9511) = 30.0.

#51. From the formula in the book, the area is 0.5 ab sin(θ). Our given info was 10, 22 and 10o, so the area is (0.5)(10)(22)sin(10o), which equals (110)(0.1736) = 19.1 .

#53. From the formula in the book, area A = 0.5 ab sin(θ). Our given info was area = 16, and the two sides are 5 & 7. So we have 16 = 0.5(5)(7)sin(θ), which rearranges to sin(θ) = 16/17.5 = 0.9143. Using the calculator we get that θ = 66.1o.

#54. We need to find the area of the sector, then subtract the area of the triangle.
Since the area of the circle is π(r)2, and since 120o is 1/3 of the circle, the area of the sector must be π(2)2/3 = 4π/3, which is approximately 4.187.
The area of the triangle is 0.5 ab sin(T), so 0.5(2)(2)sin(120o) = 2 sqrt(3)/2 = sqrt(3), which is approximately 1.732.
That makes the desired area equal to 4.187 - 1.732 = 2.455 .

#55a. The picture below is the same as in the textbook, except I've labeled a few more things. First, I've labeled the two segments that actually make up the pipe x & y. Eventually I will need to find x+y. I also labeled the upper angle (between the y & the 6) as θ. That is because that upper θ angle and the lower θ angle are complements of the same angle, so they must be equal. Finally, I also drew an extra horizontal line across the hall (near the bend), but its length should also be 9.
Looking at the lower left triangle, we can see that csc(θ) = x/9, so x = 9 csc(θ). Looking at the upper right triangle, we can see that sec(θ) = y/6, so y = 6 sec(θ).
And now we can see that x+y = 9 csc(θ) + 6 sec(θ) , as desired.

trig pic

#56a. The picture below is the same as in the textbook, except I've drawn and labeled a few more things. First, I've closed off the top of the rain gutter, which then shows a trapezoid. (Our goal is to find the area of that trapezoid.) I've labeled the height of the trapezoid with h, and I labeled the ends of the upper base of the trapezoid with a couple x's. Lastly I labeled where theta fell inside the trapezoid. (Notice the two right triangles.) What we need to do is identify x & h in terms of theta.
On either triangle we can see that sin(θ) = h/10, so h=10 sin(θ). We can also see that cos(θ) = x/10, so x=10 cos(θ). Now we can see that the height of the trapezoid is h = 10 sin(θ), and the upper base of the trapezoid is 10+20cos(θ).
The formula for area of a trapezoid is: (the average of the two bases)x(the height). For the average of the two bases we have (10+2x+10)/2 = x+10 = 10 cos(θ) + 10. Since the height is 10 sin(θ), the area of the trapezoid must be:
((10+10 cos(θ))(10 sin(θ), which multiplies out to 100 sin(θ) + 100 sin(θ) cos(θ), which is the desired answer.

trig pic



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