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Elementary FunctionsPart 4, Trigonometry

Lecture 4.7a, Solving Problems with Inverse Trig Functions

Dr. Ken W. Smith

Sam Houston State University

2013

Smith (SHSU) Elementary Functions 2013 1 / 17

Inverse trig functions create right triangles

An inverse trig function has an angle (y or ) as its output. That anglesatisfies a certain trig expression and so we can draw a right triangle thatrepresents that expression.

One can always draw a right triangle with an inverse trig function andthink of the output as a certain angle in that triangle.

For example, the equation arcsin(z) = implies that sin = z and socorresponds to a right triangle with hypotenuse 1, with one of the acuteangles and z the length of the side opposite .

Smith (SHSU) Elementary Functions 2013 2 / 17

Some Worked Problems on Inverse Trig Functions

We will practice this idea with some worked problems....

1 Draw a right triangle with the appropriate lengths and use thattriangle to find the sine of the angle if

1 cos() = 232 cos() = 253 cos() = 0.8.4 cos() = 0.6.

Partial solutions.1 If cos() = 23 then draw a triangle with legs of length 2,

5 and

hypotenuse of length 3. If the cosine of is 23 then the sine of is53 .

2 If cos() = 25 then draw a triangle with legs of length 2,21 and

hypotenuse of length 5. The sine of is215 .

3 If cos() = 0.8 then draw a triangle with legs of length 3, 4 and

hypotenuse of length 5. The sine of is 35 .

4 If cos() = 0.6 then draw a triangle with legs of length 3, 4 and

hypotenuse of length 5. The sine of is 45 .Smith (SHSU) Elementary Functions 2013 3 / 17

Some Worked Problems on Inverse Trig Functions

When we work with inverse trig functions it is especially important to drawa triangle since the output of the inverse trig function is an angle of aright triangle.

Indeed, one could think of inverse trig functions as creating righttriangles. The angle in the drawing below is arcsin(z). Notice that thePythagorean theorem then gives us the third side of the triangle (writtenin blue); its length is

1 z2. This allows us to simplify expressions like

cos(arcsin z), recognizing that

cos(arcsin z) = cos() =1 z2.

In a similar manner, we can simplify tan(arcsin z) to

tan(arcsin(z)) =z

1 z2.

Smith (SHSU) Elementary Functions 2013 4 / 17

Some Worked Problems on Inverse Trig Functions

Simplify (without use of a calculator) the following expressions

1 arcsin[sin(8 )].2 arccos[sin(8 )].3 cos[arcsin(13)].

Solutions.

1 Since arcsin is the inverse function of sine then arcsin[sin(8 )] =8 .

2 If is the angle 8 then the sine of is the cosine of thecomplementary angle 2

8 , which, after getting a common

denominator, simplifies to 38 . In other words, the sine of8 is the

cosine of 38 so arccos[sin(8 )] =

38 . (Notice that Ive solved this

problem this without ever having to figure out the value of sin(8 ).)3 To simplify cos[arcsin(13)] we draw a triangle with hypotenuse of

length 3 and one side of length 1, placing the angle so thatsin() = 13 . The other short side of the triangle must have length8 = 2

2 by the Pythagorean theorem so the cosine of is 2

2

3 .

So cos[arcsin(13)] =22

3 .Smith (SHSU) Elementary Functions 2013 5 / 17

Some worked problems.

Simplify (without the use of a calculator) the following expressions:

4 arccos(sin()), assuming that is in the interval [0, 2 ].

Solutions. To simplify arccos(sin()), we draw a triangle (on the unitcircle, say) with an acute angle and short sides of lengths x, y andhypotenuse 1.

x

y1

2

The sine of is then y and the arccosine of y must be the complementaryangle 2 . So arccos(sin()) =

2 .

Smith (SHSU) Elementary Functions 2013 6 / 17

Some worked problems.

5 Simplify arccos(y) + arcsin(y).

Solution. Notice in the triangle in the figure below, that the sine of is y and the cosine of 2 is y.

x

y1

2

So arcsin(y) = and arccos(y) = 2 . Therefore

arccos(y) + arcsin(y) = + (2 ) =2 .

Indeed, the expression arccos(y) + arcsin(y) merely asks for the sumof two complementary angles! By definition, the sum of twocomplementary angles is 2 !

Smith (SHSU) Elementary Functions 2013 7 / 17

Some worked problems.

In the next presentation, we will solve more problems with inverse trigfunctions.

(End)

Smith (SHSU) Elementary Functions 2013 8 / 17

Elementary FunctionsPart 4, Trigonometry

Lecture 4.7b, Inverse Trig Expressions Create Triangles

Dr. Ken W. Smith

Sam Houston State University

2013

Smith (SHSU) Elementary Functions 2013 9 / 17

Drawing triangles to solve composite trig expressions

Some problems involving inverse trig functions include the composition ofthe inverse trig function with a trig function. If the inverse trig functionoccurs first in the composition, we can simplify the expression bydrawing a triangle.

Worked problems. Do the following problems without a calculator.Find the exact value of

1 sin(arccos(34))2 tan(arcsin(34))

Solutions.

1 To compute sin(cos1(34)) draw a triangle with legs 3,7 and

hypotenuse 4. The angle needs to be in the second quadrant so thecosine will be negative. In this case, the sine will be positive. So the

sine of the angle should be74 .

2 To compute tan(sin1(34)) draw a triangle with legs 3,7 and

hypotenuse 4. The tangent of the angle should be 37

. But the

angle is in the fourth quadrant so the final answer is 37

.Smith (SHSU) Elementary Functions 2013 10 / 17

Drawing triangles to solve composite trig expressions

Simplify sin(2 arctan(43)) (Use the trig identity sin 2 = 2 sin cos .)

Solution. To compute sin(2 tan1(43)) = 2 sin cos wheretan() = 43 draw a triangle with legs 3, 4 and hypotenuse 5. The cosineof the angle is 35 and the sine of the angle is

45 . Since the original

problem has a negative sign in it, and we are working with the arctangentfunction, then we must be working with an angle in the fourth quadrant,so the sine is really 45 . Now we just plug these values into the magicalidentity given us:

sin(2) = 2 sin cos = 2(45)(3

5) = 2425 .

Smith (SHSU) Elementary Functions 2013 11 / 17

Drawing triangles to solve composite trig expressions

Simplify the following expressions involving arctangent:tan(arctan(z)), sin(arctan(z)), cot(arctan(z)), sec(arctan(z)).

Solutions.

1 To compute tan(arctan(z)) just recognize that tan z and arctan z are

inverse functions and so tan(arctan(z)) = z .

2 To compute sin(arctan(z)) draw a right triangle with sides 1, z and

hypotenuse1 + z2. The sine of the angle is z

1+z2.

3 In the figure above, the cotangent of the angle is 1z .

4 The secant of the angle should be1 + z2 .

Smith (SHSU) Elementary Functions 2013 12 / 17

More on inverting composite trig functions

Just like other functions, we can algebraically manipulate expressions tocreate an inverse function.

Some worked problems. Find the inverse function of y = sin(x) + 2

Solutions. To find the inverse function of y = sin(x) + 2, lets exchange

inputs and outputs:x = sin(

y) + 2

and then solve for y by subtracting 2 from both sides

x 2 = sin(y),applying the arcsin to both sides,

arcsin(x 2) = yand then squaring both sides

(arcsin(x 2))2 = y

so that the answer is is y = (arcsin(x 2))2.Smith (SHSU) Elementary Functions 2013 13 / 17

More on inverting composite trig functions

Find the inverse function of y = sin(x+ 2)

Solutions. We setx = sin(

y + 2),

take the arcsine of both sides:

arcsin(x) =y + 2),

square both sides(arcsin(x))2 = y + 2,

and then subtract 2 from both sides.

The inverse function of y = sin(x+ 2) is y = (arcsinx)2 2.

Smith (SHSU) Elementary Functions 2013 14 / 17

More on inverting composite trig functions

Find the inverse function of y = esin(x+2)

Solutions. We setx = esin(

y+2)

take the natural log of both sides:

ln(x) = sin(y + 2),

then take the arcsine of both sides

arcsin(ln(x)) =y + 2,

and then subtract 2 from both sides

arcsin(ln(x)) 2 = y,

and finally square both sides.

The inverse function of y = esin(x+2)is y = (arcsin(lnx) 2)2.

Smith (SHSU) Elementary Functions 2013 15 / 17

More on inverting composite trig functions

Find the inverse function of y = sin(arccosx)

Solutions.First we simplify sin(arccosx). Draw a right triangle with a hypotenuse oflength 1 and an acute angle with adjacent side of length x. The sideopposite of has length (by the Pythagorean theorem)

1 x2. So the

cosine of is just1 x2.

We have simplified y = sin(arccosx) to y =1 x2.

It happens that the inverse function of y =1 x2 obeys the equation

x =1 y2 so x2 = 1 y2 so y2 = 1 x2 so y =

1 x2. (That is

y =1 x2 is its own inverse function!)

Smith (SHSU) Elementary Functions 2013 16 / 17

Using inverse trig functions

REMEMBER: When faced with an inverse trig function, think about thetriangle the function creates!

In the next presentation, we will look at trig identities and equations.

(End)

Smith (SHSU) Elementary Functions 2013 17 / 17