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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 θ) 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) = θ implies that sin θ = z and socorresponds to a right triangle with hypotenuse 1, with θ one of the acuteangles and z the length of the side opposite θ.
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 θ if
1 cos(θ) = 23
2 cos(θ) = 25
3 cos(θ) = 0.8.4 cos(θ) = 0.6.
Partial solutions.1 If cos(θ) = 2
3 then draw a triangle with legs of length 2,√5 and
hypotenuse of length 3. If the cosine of θ is 23 then the sine of θ is
√53 .
2 If cos(θ) = 25 then draw a triangle with legs of length 2,
√21 and
hypotenuse of length 5. The sine of θ is√215 .
3 If cos(θ) = 0.8 then draw a triangle with legs of length 3, 4 and
hypotenuse of length 5. The sine of θ is 35 .
4 If cos(θ) = 0.6 then draw a triangle with legs of length 3, 4 and
hypotenuse of length 5. The sine of θ 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 θ 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(θ) =√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(π8 )].2 arccos[sin(π8 )].3 cos[arcsin(13)].
Solutions.
1 Since arcsin is the inverse function of sine then arcsin[sin(π8 )] =π8 .
2 If θ is the angle π8 then the sine of θ is the cosine of the
complementary angle π2 −
π8 , which, after getting a common
denominator, simplifies to 3π8 . In other words, the sine of π
8 is the
cosine of 3π8 so arccos[sin(π8 )] =
3π8 . (Notice that I’ve solved this
problem this without ever having to figure out the value of sin(π8 ).)3 To simplify cos[arcsin(13)] we draw a triangle with hypotenuse of
length 3 and one side of length 1, placing the angle θ so thatsin(θ) = 1
3 . The other short side of the triangle must have length√8 = 2
√2 by the Pythagorean theorem so the cosine of θ is 2
√2
3 .
So cos[arcsin(13)] =2√2
3 .Smith (SHSU) Elementary Functions 2013 5 / 17
Some worked problems.
Simplify (without the use of a calculator) the following expressions:
4 arccos(sin(θ)), assuming that θ is in the interval [0, π2 ].
Solutions. To simplify arccos(sin(θ)), we draw a triangle (on the unitcircle, say) with an acute angle θ and short sides of lengths x, y andhypotenuse 1.
x
y1
θ
π2 − θ
The sine of θ is then y and the arccosine of y must be the complementaryangle π
2 − θ. So arccos(sin(θ)) = π2 − θ .
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 θis y and the cosine of π
2 − θ is y.
x
y1
θ
π2 − θ
So arcsin(y) = θ and arccos(y) = π2 − θ. Therefore
arccos(y) + arcsin(y) = θ + (π2 − θ) =π2 .
Indeed, the expression arccos(y) + arcsin(y) merely asks for the sumof two complementary angles! By definition, the sum of twocomplementary angles is π
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(cos−1(−34)) draw a triangle with legs 3,
√7 and
hypotenuse 4. The angle θ 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 θ should be√74 .
2 To compute tan(sin−1(−34)) draw a triangle with legs 3,
√7 and
hypotenuse 4. The tangent of the angle θ should be 3√7
. But the
angle θ is in the fourth quadrant so the final answer is − 3√7
.Smith (SHSU) Elementary Functions 2013 10 / 17
Drawing triangles to solve composite trig expressions
Simplify sin(2 arctan(−43)) (Use the trig identity sin 2θ = 2 sin θ cos θ.)
Solution. To compute sin(2 tan−1(−43)) = 2 sin θ cos θ where
tan(θ) = −43 draw a triangle with legs 3, 4 and hypotenuse 5. The cosine
of the angle θ is 35 and the sine of the angle θ is 4
5 . Since the originalproblem 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 −4
5 . Now we just plug these values into the “magical”identity given us:
sin(2θ) = 2 sin θ cos θ = 2(−4
5)(3
5) = −24
25 .
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
hypotenuse√1 + z2. The sine of the angle θ is z√
1+z2.
3 In the figure above, the cotangent of the angle θ is 1z .
4 The secant of the angle θ should be√1 + 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, let’s 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) =√y
and 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 θ with adjacent side of length x. The sideopposite of θ has length (by the Pythagorean theorem)
√1− x2. So the
cosine of θ is just√1− 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
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