quiz3-sol - AMS 151 (Fall, 2009) Joe Mitchell Applied...

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AMS 151 (Fall, 2009) Joe Mitchell Applied Calculus I Quiz # 3 – Solution Notes 1. Evaluate the following expressions: (a). sin(cos 1 ( 5 11 )) = Draw a right triangle and let θ be one of the two angles that is not π/ 2. Then, θ = cos 1 ( 5 11 ) if the side adjacent to θ is of length 5 and the hypotenuse is of length 11. Then, the side opposite θ is of length 11 2 - 5 2 = 96. Thus, sin(cos 1 ( 5 11 )) = sin θ = 96 11 . (b). tan(cos 1 ( 3 2 )) = Draw a right triangle and let θ be one of the two angles that is not π/ 2. Then, θ = cos 1 ( 3 2 ) if the side adjacent to θ is of length 3 and the hypotenuse is of length 2. Then, the side opposite θ is of length 2 2 - 3 = 1. (In fact, we know that θ = π/ 6, 30 degrees.) Thus, tan(cos 1 ( 3 2 )) = tan θ = 1 3 . (c). cos( 40 π 3 ) = Note that 40 π 3 = 13 π + π 3 (i.e., it is 6.5 revolutions, plus an additional angle π/ 3 (60 degrees). Thus, cos( 40 π 3 ) = cos(4 π/ 3) =
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This note was uploaded on 02/07/2010 for the course AMS 151 taught by Professor Zhang during the Fall '08 term at SUNY Stony Brook.

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quiz3-sol - AMS 151 (Fall, 2009) Joe Mitchell Applied...

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