Examples from Section 6.8
Inverse Trigonometric Functions

p. 510

Common Values of Trigonometric Functions

Use right triangles with sides 1, 2, (Sqrt[3]) and angles 30° ((π/6)), 60° ((π/3)), 90° ((π/2)) or
sides 1, 1, (Sqrt[2]) and angles 45° ((π/4)), 45° ((π/4)), 90° ((π/2)), making adjustments based on the sign of the argument to place the angle in the appropriate quadrant.

1.

a)

ArcTan[1]

b)

ArcTan[-(Sqrt[3])]

c)

ArcTan[(1/(Sqrt[3]))]

3.

a)

ArcSin[-(1/2)]

b)

ArcSin[(1/(Sqrt[2]))]

c)

ArcSin[-((Sqrt[3])/2)]

5.

a)

ArcCos[(1/2)]

b)

ArcCos[-(1/2)]

c)

ArcCos[((Sqrt[3])/2)]

7.

a)

ArcSec[-(Sqrt[2])]

b)

ArcSec[(2/(Sqrt[3]))]

c)

ArcSec[-2]

Trigonometric Function Values

13.

Given that α  =  (sin^(-1)) (5/13), find cos (α), tan (α), sec (α), csc (α), and cot (α).
Solution. Draw a right triangle with hypothenuse 13 and the side opposite α eual to 5. The other side has length (Sqrt[(13^2) -  (5^2)])  =  12. You can read off the values of the other trig functions of α from this triangle.

α  =  ArcSin[5/13];

Cos[α]

Tan[α]

Sec[α]

Csc[α]

Finding Trigonometric Expressions

33.

Let α  =  (sin^(-1)) (x). Draw a right triangle with side x oppsite the angle α and hypothenuse 1, so that sin (α)  =  (x/1) =  x. The other side of the triangle has length (Sqrt[1  -  (x^2)]), so cos (α)  =  
        ((Sqrt[1  -  (x^2)])/1)  =  (Sqrt[1  -  (x^2)]).

Cos[ArcSin[x]]

35. Let α  =  
        (tan^(-1)) ((Sqrt[(x^2) -  2  x]))  =  
          (tan^(-1)) ((Sqrt[(((x - 1))^2) -  1])), x  ≥  2. Draw a right triangle with  hypothenuse x - 1 and the side adjacent to the angle α equal to 1 and  so that the side opposite α has length (Sqrt[(((x - 1))^2) -  1])and tan (α)  =  
        ((Sqrt[(((x - 1))^2) -  1])/1) =  (Sqrt[(((x - 1))^2) -  1]). Then   sin (α)  =  ((Sqrt[(((x - 1))^2) -  1])/(x - 1)).

Sin[ArcTan[(Sqrt[(x^2) -  2  x])]]

(The denominator here can be simplified to (Sqrt[1 - 2 x + (x^2)]) = (Sqrt[(((x - 1))^2)]) =   |  x - 1  |   =  x - 1, since x  ≥  2. )

Limits

41.

Plot[ArcSin[x],  {x, -1, 1}]

Limit[ArcSin[x],  x  ->  1,  Direction  ->  1]

(The option Direction -> 1 takes the limit from the left; Direction -> -1 takes the limit from the right.)

Grapher Explorations

67.

Newton's Serpentine.

Plot[((4  x)/((x^2) +  1)),  {x, -5, 5}]

Plot[2  Sin[2  ArcTan[x]],  {x, -5, 5}]

The two functions are the same:

TrigExpand[ 2  Sin[2  ArcTan[x]] ]

This can be verified using a trig identity. Draw a triangle with side opposite α equal to x and the adjacent side equal to 1 so the hypothnuse is (Sqrt[(x^2) +  1]) . Then tan (α)  =  x or α  =  (tan^(-1)) (x)and sin (2  α)  =  
        2 sin (α) cos (α)  =  
          2 (x/(Sqrt[(x^2) +  1])) (1/(Sqrt[(x^2) +  1]))  =  
            ((2  x)/((x^2) +  1)).