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Section exercises

Verbal

Explain the basis for the cofunction identities and when they apply.

The cofunction identities apply to complementary angles. Viewing the two acute angles of a right triangle, if one of those angles measures x , the second angle measures π 2 x . Then sin x = cos ( π 2 x ) . The same holds for the other cofunction identities. The key is that the angles are complementary.

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Is there only one way to evaluate cos ( 5 π 4 ) ? Explain how to set up the solution in two different ways, and then compute to make sure they give the same answer.

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Explain to someone who has forgotten the even-odd properties of sinusoidal functions how the addition and subtraction formulas can determine this characteristic for f ( x ) = sin ( x ) and g ( x ) = cos ( x ) . (Hint: 0 x = x )

sin ( x ) = sin x , so sin x is odd. cos ( x ) = cos ( 0 x ) = cos x , so cos x is even.

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Algebraic

For the following exercises, find the exact value.

sin ( 11 π 12 )

6 2 4

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tan ( 19 π 12 )

2 3

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For the following exercises, rewrite in terms of sin x and cos x .

sin ( x 3 π 4 )

2 2 sin x 2 2 cos x

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cos ( x + 2 π 3 )

1 2 cos x 3 2 sin x

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For the following exercises, simplify the given expression.

sec ( π 2 θ )

csc θ

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tan ( π 2 x )

cot x

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sin ( 2 x ) cos ( 5 x ) sin ( 5 x ) cos ( 2 x )

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tan ( 3 2 x ) tan ( 7 5 x ) 1 + tan ( 3 2 x ) tan ( 7 5 x )

tan ( x 10 )

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For the following exercises, find the requested information.

Given that sin a = 2 3 and cos b = 1 4 , with a and b both in the interval [ π 2 , π ) , find sin ( a + b ) and cos ( a b ) .

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Given that sin a = 4 5 , and cos b = 1 3 , with a and b both in the interval [ 0 , π 2 ) , find sin ( a b ) and cos ( a + b ) .

sin ( a b ) = ( 4 5 ) ( 1 3 ) ( 3 5 ) ( 2 2 3 ) = 4 6 2 15 cos ( a + b ) = ( 3 5 ) ( 1 3 ) ( 4 5 ) ( 2 2 3 ) = 3 8 2 15

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For the following exercises, find the exact value of each expression.

sin ( cos 1 ( 0 ) cos 1 ( 1 2 ) )

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cos ( cos 1 ( 2 2 ) + sin 1 ( 3 2 ) )

2 6 4

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tan ( sin 1 ( 1 2 ) cos 1 ( 1 2 ) )

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Graphical

For the following exercises, simplify the expression, and then graph both expressions as functions to verify the graphs are identical. Confirm your answer using a graphing calculator.

cos ( π 2 x )

sin x

Graph of y=sin(x) from -2pi to 2pi.
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tan ( π 3 + x )

cot ( π 6 x )

Graph of y=cot(pi/6 - x) from -2pi to pi - in comparison to the usual y=cot(x) graph, this one is reflected across the x-axis and shifted by pi/6.
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tan ( π 4 x )

cot ( π 4 + x )

Graph of y=cot(pi/4 + x) - in comparison to the usual y=cot(x) graph, this one is shifted by pi/4.
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sin ( π 4 + x )

sin x 2 + cos x 2

Graph of y = sin(x) / rad2 + cos(x) / rad2 - it looks like the sin curve shifted by pi/4.
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For the following exercises, use a graph to determine whether the functions are the same or different. If they are the same, show why. If they are different, replace the second function with one that is identical to the first. (Hint: think 2 x = x + x . )

f ( x ) = sin ( 4 x ) sin ( 3 x ) cos x , g ( x ) = sin x cos ( 3 x )

They are the same.

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f ( x ) = cos ( 4 x ) + sin x sin ( 3 x ) , g ( x ) = cos x cos ( 3 x )

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f ( x ) = sin ( 3 x ) cos ( 6 x ) , g ( x ) = sin ( 3 x ) cos ( 6 x )

They are the different, try g ( x ) = sin ( 9 x ) cos ( 3 x ) sin ( 6 x ) .

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f ( x ) = sin ( 4 x ) , g ( x ) = sin ( 5 x ) cos x cos ( 5 x ) sin x

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f ( x ) = sin ( 2 x ) , g ( x ) = 2 sin x cos x

They are the same.

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f ( θ ) = cos ( 2 θ ) , g ( θ ) = cos 2 θ sin 2 θ

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f ( θ ) = tan ( 2 θ ) , g ( θ ) = tan θ 1 + tan 2 θ

They are the different, try g ( θ ) = 2 tan θ 1 tan 2 θ .

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f ( x ) = sin ( 3 x ) sin x , g ( x ) = sin 2 ( 2 x ) cos 2 x cos 2 ( 2 x ) sin 2 x

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f ( x ) = tan ( x ) , g ( x ) = tan x tan ( 2 x ) 1 tan x tan ( 2 x )

They are different, try g ( x ) = tan x tan ( 2 x ) 1 + tan x tan ( 2 x ) .

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Technology

For the following exercises, find the exact value algebraically, and then confirm the answer with a calculator to the fourth decimal point.

sin ( 195° )

3 1 2 2 , or  0.2588

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cos ( 345° )

1 + 3 2 2 , or 0.9659

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Extensions

For the following exercises, prove the identities provided.

tan ( x + π 4 ) = tan x + 1 1 tan x

tan ( x + π 4 ) = tan x + tan ( π 4 ) 1 tan x tan ( π 4 ) = tan x + 1 1 tan x ( 1 ) = tan x + 1 1 tan x

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tan ( a + b ) tan ( a b ) = sin a cos a + sin b cos b sin a cos a sin b cos b

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cos ( a + b ) cos a cos b = 1 tan a tan b

cos ( a + b ) cos a cos b = cos a cos b cos a cos b sin a sin b cos a cos b = 1 tan a tan b

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cos ( x + y ) cos ( x y ) = cos 2 x sin 2 y

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cos ( x + h ) cos x h = cos x cos h 1 h sin x sin h h

cos ( x + h ) cos x h = cos x cosh sin x sinh cos x h = cos x ( cosh 1 ) sin x sinh h = cos x cos h 1 h sin x sin h h

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For the following exercises, prove or disprove the statements.

tan ( u + v ) = tan u + tan v 1 tan u tan v

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tan ( u v ) = tan u tan v 1 + tan u tan v

True

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tan ( x + y ) 1 + tan x tan x = tan x + tan y 1 tan 2 x tan 2 y

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If α , β , and γ are angles in the same triangle, then prove or disprove sin ( α + β ) = sin γ .

True. Note that sin ( α + β ) = sin ( π γ ) and expand the right hand side.

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If α , β , and y are angles in the same triangle, then prove or disprove tan α + tan β + tan γ = tan α tan β tan γ

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Questions & Answers

f(x)=x/x+2 given g(x)=1+2x/1-x show that gf(x)=1+2x/3
Ken Reply
proof
AUSTINE
sebd me some questions about anything ill solve for yall
Manifoldee Reply
how to solve x²=2x+8 factorization?
Kristof Reply
x=2x+8 x-2x=2x+8-2x x-2x=8 -x=8 -x/-1=8/-1 x=-8 prove: if x=-8 -8=2(-8)+8 -8=-16+8 -8=-8 (PROVEN)
Manifoldee
x=2x+8
Manifoldee
×=2x-8 minus both sides by 2x
Manifoldee
so, x-2x=2x+8-2x
Manifoldee
then cancel out 2x and -2x, cuz 2x-2x is obviously zero
Manifoldee
so it would be like this: x-2x=8
Manifoldee
then we all know that beside the variable is a number (1): (1)x-2x=8
Manifoldee
so we will going to minus that 1-2=-1
Manifoldee
so it would be -x=8
Manifoldee
so next step is to cancel out negative number beside x so we get positive x
Manifoldee
so by doing it you need to divide both side by -1 so it would be like this: (-1x/-1)=(8/-1)
Manifoldee
so -1/-1=1
Manifoldee
so x=-8
Manifoldee
SO THE ANSWER IS X=-8
Manifoldee
so we should prove it
Manifoldee
x=2x+8 x-2x=8 -x=8 x=-8 by mantu from India
mantu
lol i just saw its x²
Manifoldee
x²=2x-8 x²-2x=8 -x²=8 x²=-8 square root(x²)=square root(-8) x=sq. root(-8)
Manifoldee
I mean x²=2x+8 by factorization method
Kristof
I think x=-2 or x=4
Kristof
x= 2x+8 ×=8-2x - 2x + x = 8 - x = 8 both sides divided - 1 -×/-1 = 8/-1 × = - 8 //// from somalia
Mohamed
1KI POWER 1/3 PLEASE SOLUTIONS
Prashant Reply
hii
Amit
how are you
Dorbor
well
Biswajit
can u tell me concepts
Gaurav
Find the possible value of 8.5 using moivre's theorem
Reuben Reply
which of these functions is not uniformly cintinuous on (0, 1)? sinx
Pooja Reply
which of these functions is not uniformly continuous on 0,1
Basant Reply
solve this equation by completing the square 3x-4x-7=0
Jamiz Reply
X=7
Muustapha
=7
mantu
x=7
mantu
3x-4x-7=0 -x=7 x=-7
Kr
x=-7
mantu
9x-16x-49=0 -7x=49 -x=7 x=7
mantu
what's the formula
Modress
-x=7
Modress
new member
siame
what is trigonometry
Jean Reply
deals with circles, angles, and triangles. Usually in the form of Soh cah toa or sine, cosine, and tangent
Thomas
solve for me this equational y=2-x
Rubben Reply
what are you solving for
Alex
solve x
Rubben
you would move everything to the other side leaving x by itself. subtract 2 and divide -1.
Nikki
then I got x=-2
Rubben
it will b -y+2=x
Alex
goodness. I'm sorry. I will let Alex take the wheel.
Nikki
ouky thanks braa
Rubben
I think he drive me safe
Rubben
how to get 8 trigonometric function of tanA=0.5, given SinA=5/13? Can you help me?m
Pab Reply
More example of algebra and trigo
Stephen Reply
What is Indices
Yashim Reply
If one side only of a triangle is given is it possible to solve for the unkown two sides?
Felix Reply
cool
Rubben
kya
Khushnama
please I need help in maths
Dayo Reply
Okey tell me, what's your problem is?
Navin

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Source:  OpenStax, Algebra and trigonometry. OpenStax CNX. Nov 14, 2016 Download for free at https://legacy.cnx.org/content/col11758/1.6
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