Calculus I Assignment # 4

This assignment covers
Inverse Trig Functions

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QUESTIONS

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1) SIMPLIFY: (make diagrams!)

 a) sin (arctan 3/4) b) tan (arcsin 12/13) c) d) cot (arcsin x) e) sec (arctan x/2) f) tan ( arcsin ) g) sin (2 arccos 1/2) h) sin (arcsin 1/2 - arccos 4/5) i) cos (arcsin 4/5 - arccos 1/2) j) tan ( arcsin )

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2) Find f '(x):

 a) f (x) = b) f (x) = c) d) f (x) = arctan e) f (x) = (tan x + arctan x)4

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3) The lower edge of a billboard that is 20 feet high stands on top of a 60 foot building.
Use inverse trig functions to find how far from the building a viewer should stand
(on flat ground) in order to maximize the angle between the lines of sight to the top
and bottom of the billboard. (Make a diagram).

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4) Find all points on the curve y = arcsin 3x where the
tangent is parallel to AB if A is (2, -3), B is (4, 7).

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SOLUTIONS

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1) SIMPLIFY: (make diagrams!)

 a) arctan (3/4) is an angle A opposite side = 3, adjacent side = 4 which makes hypotenuse = 5 sin A = 3/5. b) arcsin (12/13) is an angle A opposite side = 12, hypotenuse = 13 which makes adjacent = 5 tan A = 12/5. c) opposite side = -2, hypotenuse = which makes adjacent = ! 1 cos A = . d) arcsin x is an angle A opposite side = x, hypotenuse = 1 which makes adjacent = cot A = . e) arctan (x/2) is an angle A opposite side = x, adjacent side = 2 which makes hypotenuse = sec A = . f) arcsin ( ) is an angle A opposite side = x - 2, hypotenuse = 3 so adjacent = By now you can make the picture so:tan A =

 g) arccos (1/2) is an angle A adjacent = 1, hypotenuse = 2 this is the 30°, 60°, 90° triangle so opposite = (this means A = 60°) we want sin (2)(60°) = sin 120° = h) Let arcsin ½ = A and arccos 4/5 = B adjacent(A) = , opposite(B) = 3 sin (A - B) = sinA cosB - cos A sin B substituting the values gives us: i) Let arcsin 4/5 = A and arccos 1/2 = B adjacent(A) = 3, opposite(B) = . cos (A - B) = cosA cosB + sin A sin B substituting the values gives us: j) arcsin ( ) is angle A we have opposite and hypotenuse from sinA the adjacent = 3 so tan A = p/3.

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2) Find f '(x):

 a) f '(x) = b) f '(x) = c) d) f '(x) = . e) f '(x) = 4(tan x + arctan x) 3 (sec ² x + ).

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3) We want to maximize the angle (A - B), so we find an algebraic expression
to represent the angle difference, we differentiate and set it equal to zero.

Angle A = arctan (80/x) and angle B = arctan (60/x), therefore
(A - B) = arctan (80/x) - arctan (60/x)

this gives us x² = 4800, so x = 69.28 feet.

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4) Slope AB = 5 so we set y' = 5 and solve for x.

The points have x-coordinate = ! 4/15, and y-coordinate = arcsin (! 4/5).

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