mat 1320 a: calculus i 11th.pdf · mat 1320 a: calculus i trigonometry inverse trigonometric...

MAT 1320 A:Calculus I

Trigonometry

InverseTrigonometricFunctions

MAT 1320 A: Calculus I

Paul-Eugene ParentDepartment of Mathematics and Statistics

University of Ottawa

September 11th, 2013


Trigonometry


Outline

1 Trigonometry

2 Inverse Trigonometric Functions


Trigonometry


Consider a circle of radius r and some arc of length a alongits circumference.

DefinitionWe define the angle θ to be the ratio a

r .

Recall that angles are measured in radian. It’s a commonmistake to forget to put its calculator in radian beforeattempting any computations.

360◦ = 2π, 180◦ = π, 45◦ =π

4


Trigonometry


Trigonometric circle

Consider the circle of radius 1 centered at the origin.

The coordinates of the point A(θ) give rise to two functions,i.e.,

A(θ) = (cos(θ), sin(θ)).

Hence

• A(0) = (1, 0) which implies cos(0) = 1 and sin(0) = 0.

• A(π2 ) = (0, 1) which implies cos(π2 ) = 0 and sin(π2 ) = 1


Trigonometry






Hence

• A(0) = (1, 0) which implies

cos(0) = 1 and sin(0) = 0.



Trigonometry






Hence




Trigonometry






Hence


• A(π2 ) = (0, 1) which implies

cos(π2 ) = 0 and sin(π2 ) = 1


Trigonometry






Hence




Trigonometry


Properties

Clearly the function A(θ) is periodic of period 2π, i.e.,A(θ) = A(θ + 2π). Hence

cos(θ) = cos(θ + 2π) and sin(θ) = sin(θ + 2π).

Moreover, cos and sin being the coordinate functions, by thePythagorean Theorem, they must satisfy

cos2(θ) + sin2(θ) = 1.


Trigonometry

InverseTrigonometricFunctions As one can deduce by either their respective definition or

their associated graph,

• the function sin(θ) is an odd function, i.e.,

sin(−θ) = − sin(θ);

and

• the function cos(θ) is an even function, i.e.,

cos(−θ) = cos(θ).


Trigonometry

InverseTrigonometricFunctions As one can deduce by either their respective definition or

their associated graph,

• the function sin(θ) is an odd function, i.e.,

sin(−θ) = − sin(θ); and

• the function cos(θ) is an even function, i.e.,

cos(−θ) = cos(θ).


Trigonometry


y = sin(x)


Trigonometry


y = cos(x)


Trigonometry


Classical angles

θ 0 π6

π4

π3

π2 π 3π

2 2π

sin(θ) 0 12

√22

√32 1 0 −1 0

cos(θ) 1√32

√22

12 0 −1 0 1


Trigonometry


Other trigonometric functions

tan(θ) :=sin(θ)

cos(θ)and cot(θ) :=

cos(θ)

sin(θ)

and

sec(θ) :=1

cos(θ)and csc(θ) :=

1

sin(θ).


Trigonometry


Identities

From cos2(θ) + sin2(θ) = 1 one can deduce

• 1 + tan2(θ) = sec2(θ); and

• cot2(θ) + 1 = csc2(θ).

Other important identities

• sin(x ± y) = sin(x) cos(y)± sin(y) cos(x); and

• cos(x ± y) = cos(x) cos(y)∓ sin(x) sin(y).


Trigonometry


Identities


• 1 + tan2(θ) = sec2(θ); and

• cot2(θ) + 1 = csc2(θ).


• sin(x ± y) = sin(x) cos(y)± sin(y) cos(x);

and



Trigonometry


Identities


• 1 + tan2(θ) = sec2(θ); and

• cot2(θ) + 1 = csc2(θ).


• sin(x ± y) = sin(x) cos(y)± sin(y) cos(x); and



Trigonometry


Exercises at home

Show that

• sin(2x) = 2 sin(x) cos(x);

and

•

cos(2x) = cos2(x)− sin2(x)

= 2 cos2(x)− 1

= 1− 2 sin2(x)


Trigonometry


Exercises at home

Show that

• sin(2x) = 2 sin(x) cos(x); and

•

cos(2x) = cos2(x)− sin2(x)

= 2 cos2(x)− 1

= 1− 2 sin2(x)


Trigonometry


Exercise

Find all x ∈ [0, 2π] such that sin(x) = sin(2x).

Solution: Recall that sin(2x) = 2 sin(x) cos(x), hence weneed to solve

sin(x) = 2 sin(x) cos(x) equivalently sin(x)(1−2 cos(x)) = 0.

We have two cases

• sin(x) = 0 which implies that x = 0, π, or 2π; or

• 1− 2 cos(x) = 0 which implies cos(x) = 12 or

equivalently x = π3 or 5π

3 .

Conclusion: The solutions are x = 0, π3 , π,5π3 , or 2π.


Trigonometry


Exercise


Solution: Recall that sin(2x) = 2 sin(x) cos(x),

hence weneed to solve


We have two cases




3 .



Trigonometry


Exercise



sin(x) = 2 sin(x) cos(x)

equivalently sin(x)(1−2 cos(x)) = 0.

We have two cases




3 .



Trigonometry


Exercise




We have two cases




3 .



Trigonometry


Exercise




We have two cases

• sin(x) = 0 which implies that x = 0, π, or 2π;

or



3 .



Trigonometry


Exercise




We have two cases




3 .



Trigonometry


The problem

We want to “invert” the trigonometric functions. Recall thatthe domain and the range of sin(x) and cos(x) are the realnumbers. Two problems arise:

• both functions are periodic. Hence they are notone-to-one; and

• both functions are not onto as their images are [−1, 1].

We then need to consider restricted versions of thosefunctions if we want to proceed.


Trigonometry


The problem


• both functions are periodic. Hence they are notone-to-one;

and




Trigonometry


The problem


• both functions are periodic. Hence they are notone-to-one; and




Trigonometry


y = Sin(x)

In the case of sin(x) we consider the classical domain[−π

2 ,π2 ] and the restricted function

Sin : [−π2,π

2] −→ [−1, 1]

x 7→ sin(x)


Trigonometry


Green graph is y = Sin(x)


Trigonometry

InverseTrigonometricFunctions Hence the function y = Sin(x) is now onto and one-to-one

and thus admits an inverse function

arcsin : [−1, 1] −→ [−π2,π

2].

In particular

• arcsin(Sin(x)) = x for all x ∈ [−π2 ,

π2 ];

• Sin(arcsin(x)) = x for all x ∈ [−1, 1]; and

• y = arcsin(x) if and only if Sin(y) = x .


Trigonometry



arcsin : [−1, 1] −→ [−π2,π

2].

In particular


π2 ];

• Sin(arcsin(x)) = x for all x ∈ [−1, 1];

and



Trigonometry



arcsin : [−1, 1] −→ [−π2,π

2].

In particular


π2 ];

• Sin(arcsin(x)) = x for all x ∈ [−1, 1]; and



Trigonometry


y = arcsin(x)


Trigonometry


y = Cos(x) and y = Tan(x)

We can play this game with the cos(x) and tan(x) functions.To make them both onto and one-to-one we restrict theirrespective domain and range as

• Cos : [0, π] −→ [−1, 1];

and

• Tan : ]− π2 ,

π2 [−→ R.


Trigonometry


y = Cos(x) and y = Tan(x)

We can play this game with the cos(x) and tan(x) functions.To make them both onto and one-to-one we restrict theirrespective domain and range as

• Cos : [0, π] −→ [−1, 1]; and

• Tan : ]− π2 ,

π2 [−→ R.


Trigonometry


Blue graph is y = Cos(x)


Trigonometry


Blue graph is y = Tan(x)


Trigonometry

InverseTrigonometricFunctions Hence both y = Cos(x) and y = Tan(x) admit inverses, i.e.,

• arccos : [−1, 1] −→ [0, π];

and

• arctan : R −→ ]− π2 ,

π2 [

with the usual identities

• arccos(Cos(x)) = x for all x ∈ [0, π];

• Cos(arccos(x)) = x for all x ∈ [−1, 1];

• arctan(Tan(x)) = x for all x ∈]− π2 ,

π2 [; and

• Tan(arctan(x)) = x for all x ∈ R.


Trigonometry


• arccos : [−1, 1] −→ [0, π]; and

• arctan : R −→ ]− π2 ,

π2 [





π2 [; and



Trigonometry


• arccos : [−1, 1] −→ [0, π]; and

• arctan : R −→ ]− π2 ,

π2 [





π2 [;

and



Trigonometry


• arccos : [−1, 1] −→ [0, π]; and

• arctan : R −→ ]− π2 ,

π2 [





π2 [; and



Trigonometry


y = arccos(x)


Trigonometry


y = arctan(x)


Trigonometry


Exercises

Simplify the expression cos(arctan(x)).

Solution:

On one hand we know that

y = arctan(x)⇐⇒ Tan(y) = x for y ∈]− π

2,π

2[.

On the other hand we have the trigonometric identity

sec2 y = 1 + tan2 y

= 1 + x2 why?

Finally

cos(arctan(x)) = cos(y) =1

sec(y)=

1√1 + x2

... why?


Trigonometry


Exercises

Simplify the expression cos(arctan(x)).

Solution: On one hand we know that

y = arctan(x)⇐⇒ Tan(y) = x for y ∈]− π

2,π

2[.

On the other hand we have the trigonometric identity

sec2 y = 1 + tan2 y

= 1 + x2 why?

Finally

cos(arctan(x)) = cos(y) =1

sec(y)=

1√1 + x2

... why?


Trigonometry


Compute exactly tan(arcsin(1/3))

Solution:

On one hand we know that

θ = arcsin(1/3)⇐⇒ Sin(θ) = 1/3 when θ ∈ [−π2,π

2].

On the other hand we have the trigonometric identitycos2 θ = 1− sin2 θ.Since θ ∈ [−π

2 ,π2 ], we thus have cos θ =

√1− sin2 θ.

Finally tan(arcsin(1/3)) = tan θ = sin θcos θ = sin θ√

1−sin2 θ=

1/3√1−(1/3)2

=√24 .


Trigonometry





2].



√1− sin2 θ.


1−sin2 θ=

1/3√1−(1/3)2

=√24 .


Trigonometry





2].

On the other hand we have the trigonometric identitycos2 θ = 1− sin2 θ.

Since θ ∈ [−π2 ,

π2 ], we thus have cos θ =

√1− sin2 θ.


1−sin2 θ=

1/3√1−(1/3)2

=√24 .


Trigonometry





2].



√1− sin2 θ.


1−sin2 θ=

1/3√1−(1/3)2

=√24 .


Trigonometry





2].



√1− sin2 θ.

Finally tan(arcsin(1/3)) = tan θ

= sin θcos θ = sin θ√

1−sin2 θ=

1/3√1−(1/3)2

=√24 .


Trigonometry





2].



√1− sin2 θ.

Finally tan(arcsin(1/3)) = tan θ = sin θcos θ

= sin θ√1−sin2 θ

=

1/3√1−(1/3)2

=√24 .


Trigonometry





2].



√1− sin2 θ.


1−sin2 θ

=

1/3√1−(1/3)2

=√24 .


Trigonometry





2].



√1− sin2 θ.


1−sin2 θ=

1/3√1−(1/3)2

=√24 .

mat 1320 a: calculus i 11th.pdf · mat 1320 a: calculus i trigonometry inverse trigonometric...

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