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 Aopposite side = x, hypotenuse = 1which makes adjacent = cot A = . |

e) arctan (x/2) is an angle Aopposite side = x, adjacent side = 2which makes hypotenuse = sec A = . |
f) arcsin ( ) is an angle A opposite side = x - 2, hypotenuse = 3so 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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