cal-1assign 4

Calculus I Assignment # 4

This assignment covers
Inverse Trig Functions

QUESTIONS

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 )

2) Find f '(x):

a) f (x) = b) f (x) =

c) d) f (x) = arctan

e) f (x) = (tan x + arctan x)⁴

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).

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).

SOLUTIONS

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.

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 + ).

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.

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).

Back to Calculus I MathRoom Index

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 )

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

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.

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