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Calculus I                                                 Quiz  #4                                       September 17, 2003

Name____________________          R.  Hammack                                            Score ______
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(1)
(a)  cos(π) =-1

(b)  cos(π/3) =1/2

(c)  sin(π/3) =3^(1/2)/2

(d)  tan(π/3) =sin(π/3)/cos(π/3) = 3^(1/2)/2/1/2 = 3^(1/2)

(e)  sin((7π)/4) =-1/2^(1/2)



(2) Find all values of θ for which tan(θ) = 3^(1/2).
From part d of the above problem, we see that one such value of θ is π/3, which is a radian measure in the first quadrant. Adding π to θ = π/3 puts the angle into the third quadrant, but still tan(π/3 + π) = 3^(1/2).
In general, you could add any multiple of π. Thus the answer is θ = π/3 + n π, where n is an integer.

(3) Suppose 0≤x≤π/2 and sec(x) = 5/2. Find sin(x).
From 1/cos(x) = sec(x) = 5/2, we get cos(x) = 2/5.
Now apply the identity cos^2(x) + sin^2(x) = 1to get
(2/5)^2 + sin^2(x) = 1
sin^2(x) = 1 - 4/25 = 21/25
Then sin(x) = 21/25^(1/2) = 21^(1/2)/5

(4) Consider the following triangle. Find cos(θ).
[Graphics:HTMLFiles/quiz4sol_30.gif]
By the Pythagorean Theorem, we have HYP = (1^2 + 3^2)^(1/2) = 10^(1/2).
Thus cos(θ) = ADJ/HYP = 1/10^(1/2).