cos(π−θ)+sin(π2+θ)=0cos(π-θ)+sin(π2+θ)=0 Prove the identity.
- SCIENCE
Joints in train tracks are designed with extra space to allow for what?
All metals expand when heated. The expansion of each metal is a function of its thermal properties and vary from metal to metal. When laying down railway tracks, sufficient spacing is allowed…
1 educator answer
- TO KILL A MOCKINGBIRD
In the novel To Kill a Mockingbird, reflect on the discussion between Atticus and Scout regarding…
In Chapter 3, Scout has a rough first day of school and discusses the day’s misfortunes with her father. Scout tells her father that she doesn’t want to go to school anymore, and Atticus teaches…
1 educator answer
- THE TEMPEST
What are three themes in The Tempest?
The Tempest examines a number of subjects, including forgiveness, betrayal, and greed. Prospero uses magic to seek vengeance on those who have wronged him, but he eventually realizes that “the…
1 educator answer
- HISTORY
When did the Cold War begin? Who was in the Cold War? When did the Cold War end?
We generally request that you ask only one question in each post. However, since these three questions are so closely related and their answers are relatively short, I will answer all three. The…
1 educator answer
- JOHN KEATS
I must write an essay comparing the use of space in “Musee des Beaux Arts” and “Ode to a…
In “Musee des Beaux Arts,” by Auden, the speaker of the poem is looking at famous paintings by the “old Masters.” In these artworks, he notes that while miraculous things are occurring, other…
1 educator answer
- THE IMPORTANCE OF BEING EARNEST
What can you say about the adopted selves in Oscar Wilde’s The Importance of Being Earnest?
The adopted selves in The Importance of Being Ernest are traditionally thought of as Oscar Wilde’s commentary on the pretensions of Victorian society. Victorian society was highly moralistic and…
1 educator answer
- HISTORY
What was Cleopatra’s biggest goal in life?
Cleopatra’s main goals were to be the sole leader of Egypt and to expand the kingdom’s land area. Cleopatra took the throne of Egypt at age 18. She also shared the throne with her younger…
1 educator answer
- FAHRENHEIT 451
What does the allusion to Charles Darwin mean in the book Fahrenheit 451?
At the end of Fahrenheit 451 by Ray Bradbury, the main character, Montag, is being chased by the Hound as war is breaking out around him. His illicit storing of books has been revealed, and the…
1 educator answer
- TO KILL A MOCKINGBIRD
In To Kill a Mockingbird, Jem comes up with his own categories for people in the world–his own…
This is in Chapter 23. The conversation comes about because Aunt Alexandra has just told Scout that she cannot ever invite Walter Cunningham over because “he–is–trash” (256). Scout is about to…
1 educator answer
- HENRY V
To what degree is Henry V by Shakespeare historically accurate?
William Shakespeare’s play, Henry V, is loosely based on actual historical events, but also includes invented material and compresses the actual time sequence of the events. The single most…
1 educator answer
- SCIENCE
How might happiness be used as a dependent and independent variable in studies?
The independent variable in a study is the variable that’s manipulated or changed by the researcher. The dependent variable is the variable that changes in response to manipulation of the…
2 educator answers
- MATH
3^2x-1=5^x I am having trouble solving this since the bases are not the same and I am unclear how…
Hello! I think your equation is 32x−1=5x,32x-1=5x, not 32x−1=5x.32x-1=5x. For the first option yes, apply lnln to both sides (they are both positive). It is known (and not hard) that ln(ab)=b⋅ln(a).ln(ab)=b⋅ln(a)….
1 educator answer
- MATH
According to the power reducing formulas, you may re-wrute the expression such that: sin4(2x)⋅cos2x=1−cos2⋅(2x)2⋅1−cos2⋅(2x)2⋅1+cos2x2sin4(2x)⋅cos2x=1-cos2⋅(2x)2⋅1-cos2⋅(2x)2⋅1+cos2x2
1 educator answer
- MATH
According to the power reducing formulas, you may re-wrute the expression such that: sin2(2x)⋅cos2(2x)=1−cos2⋅(2x)2⋅1+cos2⋅(2x)2sin2(2x)⋅cos2(2x)=1-cos2⋅(2x)2⋅1+cos2⋅(2x)2
1 educator answer
- MATH
It is known that cos2(y)=(12)(1+cos(2y))cos2(y)=(12)(1+cos(2y)) and tan2(y)=1−cos(2y)1+cos(2y).tan2(y)=1-cos(2y)1+cos(2y). Repeating the first formula we obtain cos4(y)=(14)(1+2cos(2y)+cos2(2y))=cos4(y)=(14)(1+2cos(2y)+cos2(2y))=…
1 educator answer
- MATH
tan4(2x)tan4(2x) Use the power reducing formulas to rewrite the expression in terms of the first…
According to the power reducing formulas, you may re-wrute the expression such that: tan4(2x)=sin4(2x)cos4(2x)=(1−cos4x)2(1+cos4x)2tan4(2x)=sin4(2x)cos4(2x)=(1-cos4x)2(1+cos4x)2
1 educator answer
- MATH
sin4(2x)sin4(2x) Use the power reducing formulas to rewrite the expression in terms of the first…
According to the power reducing formulas, you may re-wrute the expression such that: sin4(2x)=sin2(2x)⋅sin2(2x)=1−cos2⋅(2x)2⋅1−cos2⋅(2x)2sin4(2x)=sin2(2x)⋅sin2(2x)=1-cos2⋅(2x)2⋅1-cos2⋅(2x)2sin4(2x)=(1−cos4x)24sin4(2x)=(1-cos4x)24
1 educator answer
- MATH
cos4(x)cos4(x) Use the power reducing formulas to rewrite the expression in terms of the first…
The suitable power reducing formula is cos2x=(12)(1+cos(2x)).cos2x=(12)(1+cos(2x)). Therefore cos4x=(14)(1+cos(2x))2=(14)(1+2cos(2x)+cos2(2x)).cos4x=(14)(1+cos(2x))2=(14)(1+2cos(2x)+cos2(2x)).Use the formula again for cos2(2x),cos2(2x),…
1 educator answer
- MATH
sec(u)=−2,π<u<3π2sec(u)=-2,π<u<3π2 Find the exact values of sin(2u), cos(2u), and tan(2u)…
Given sec(u)=−2,π<u<3π2sec(u)=-2,π<u<3π2 Angle u is in quadrant 3. If sec(u)=−2sec(u)=-2 then cos(u)=−12.cos(u)=-12. If a right triangle is drawn in quadrant 3, the side adjacent to angle u would be 1 and the…
1 educator answer
- MATH
tanu=35,0<u<π2tanu=35,0<u<π2 Find the exact values of sin(2u), cos(2u), and tan(2u) using…
Given tan(u)=35,0<u<π2tan(u)=35,0<u<π2 Angle u is in quadrant 1. A right triangle can be drawn is quadrant 1. Since tan(u)=35tan(u)=35 , the side opposite angle u is 3 and the side adjacent to angle u is 5….
1 educator answer
- MATH
cosu=−45,π2<u<πcosu=-45,π2<u<π Find the exact values of sin(2u), cos(2u), and tan(2u)…
By the double angle formulas cos(2u)=2cos2u−1,cos(2u)=2cos2u-1, sin(2u)=2sin(u)cos(u).sin(2u)=2sin(u)cos(u). By the first formula cos(2u)=2⋅(−45)2−1=725.cos(2u)=2⋅(-45)2-1=725. Because uu is in the second quadrant, its sinus is positive and is…
1 educator answer
- MATH
sin(u)=−35,3π2<u<2πsin(u)=-35,3π2<u<2π Find the exact values of sin(2u), cos(2u), and…
Given sin(u)=−35,3π2<u<2πsin(u)=-35,3π2<u<2π Angle u is in quadrant 4. A right triangle is drawn in quadrant 4. Sincesin(u)=−35sin(u)=-35 the side opposite from angle u is 3 and the hypotenuse is 5. Using the…
1 educator answer
- MATH
10sin2(x)−510sin2(x)-5 Use a double angle formula to rewrite the expression.
The corresponding double angle formula is cos(2x)=1−2sin2x.cos(2x)=1-2sin2x.Therefore 10sin2x−5=−5(1−2sin2x)=−5cos(2x).10sin2x-5=-5(1-2sin2x)=-5cos(2x).
1 educator answer
- MATH
4−8sin2(x)4-8sin2(x) Use a double angle formula to rewrite the expression.
The double angle formula states that cos(2x)=1−2sin2x,cos(2x)=1-2sin2x,therefore our expression is equal to 4−8sin2x=4(1−2sin2x)=4cos(2x).4-8sin2x=4(1-2sin2x)=4cos(2x).
1 educator answer
- MATH
cos2(x)−12cos2(x)-12 Use a double angle formula to rewrite the expression.
You need to use the double angle formula to re-write the expression, such that: cos2x−12=2cos2x−12=cos2x2cos2x-12=2cos2x-12=cos2x2 cos2x=cos(x+x)=cosx⋅cosx−sinx⋅sinxcos2x=cos(x+x)=cosx⋅cosx-sinx⋅sinx
1 educator answer
- MATH
6cos2(x)−36cos2(x)-3 Use a double angle formula to rewrite the expression.
By the double angle formula for cosine, cos(2x)=2cos2x−1.cos(2x)=2cos2x-1.Therefore our expression is equal to 6cos2(x)−3=3(2cos2(x)−1)=3cos(2x).6cos2(x)-3=3(2cos2(x)-1)=3cos(2x).
1 educator answer
- MATH
sin(x)cos(x)sin(x)cos(x) Use a double angle formula to rewrite the expression.
You need to use the double angle formula to re-write the expression, such that: sin2x=2sinx⋅cosx⇒sinx⋅cosx=sin2x2sin2x=2sinx⋅cosx⇒sinx⋅cosx=sin2x2 Hence, using the double angle formula to re-write the…
1 educator answer
- MATH
6sin(x)cos(x)6sin(x)cos(x) Use a double angle formula to rewrite the expression.
The corresponding double angle formula is sin(2x)=2sin(x)cos(x).sin(2x)=2sin(x)cos(x). Therefore our expression is equal to 6sin(x)cos(x)=3⋅(2sin(x)cos(x))=3sin(2x).6sin(x)cos(x)=3⋅(2sin(x)cos(x))=3sin(2x).
1 educator answer
- MATH
tan(2x)−2cos(x)=0,0≤x≤2πtan(2x)-2cos(x)=0,0≤x≤2π tan(2x)−2cos(x)=0tan(2x)-2cos(x)=0 sin(2x)cos(2x)−2cos(x)=0sin(2x)cos(2x)-2cos(x)=0 sin(2x)−2cos(2x)cos(x)=0sin(2x)-2cos(2x)cos(x)=0 2sin(x)cos(x)−2cos(2x)cos(x)=02sin(x)cos(x)-2cos(2x)cos(x)=0 2cos(x)(sin(x)−cos(2x))=02cos(x)(sin(x)-cos(2x))=0 using the…
1 educator answer
- MATH
tan(2x)−cot(x)=0tan(2x)-cot(x)=0 Find the exact solutions of the equation in the interval [0, 2pi).
tan(2x)−cot(x)=0tan(2x)-cot(x)=0 express in terms of sin and cos, sin(2x)cos(2x)−cos(x)sin(x)=0sin(2x)cos(2x)-cos(x)sin(x)=0 sin(x)sin(2x)−cos(x)cos(2x)cos(2x)sin(x)=0sin(x)sin(2x)-cos(x)cos(2x)cos(2x)sin(x)=0 sin(x)sin(2x)−cos(x)cos(2x)=0sin(x)sin(2x)-cos(x)cos(2x)=0…
1 educator answer
- MATH
You need to evaluate the solution to the equation (cos2x+sin2x)2=1(cos2x+sin2x)2=1 , such that: cos22x+2cos2x⋅sin2x+sin22x=1cos22x+2cos2x⋅sin2x+sin22x=1 You need to use the formula cos22x+sin22x=1cos22x+sin22x=1 , such…
1 educator answer
- MATH
By a double angle formula sin(4x)=2sin(2x)cos(2x).sin(4x)=2sin(2x)cos(2x). Therefore our equation may be rewritten as 2sin(2x)cos(2x)+2sin(2x)=0,2sin(2x)cos(2x)+2sin(2x)=0,or sin(2x)(cos(2x)+1)=0.sin(2x)(cos(2x)+1)=0. So sin(2x)=0sin(2x)=0 or cos(2x)=−1.cos(2x)=-1.The…
1 educator answer
- MATH
cos(2x)+sin(x)=0cos(2x)+sin(x)=0 Find the exact solutions of the equation in the interval [0, 2pi).
You need to evaluate the solution to the equation cos2x+sinx=0cos2x+sinx=0 , hence, you need to use the formula of double angle for cos2xcos2x , such that: 1−2sin2x+sinx=01-2sin2x+sinx=0 You need to…
1 educator answer
- MATH
cos(2x)−cos(x)=0cos(2x)-cos(x)=0 Find the exact solutions of the equation in the interval [0, 2pi).
cos(2x)−cos(x)=0,0≤x≤2πcos(2x)-cos(x)=0,0≤x≤2π using the identity cos(2x)=2cos2(x)−1cos(2x)=2cos2(x)-1 cos(2x)−cos(x)=0cos(2x)-cos(x)=0 2cos2(x)−1−cos(x)=02cos2(x)-1-cos(x)=0 Let cos(x)=y, 2y2−y−1=02y2-y-1=0 solving using the quadratic formula,…
1 educator answer
- MATH
sin(2x)sin(x)=cos(x),0≤x≤2πsin(2x)sin(x)=cos(x),0≤x≤2π sin(2x)sin(x)−cos(x)=0sin(2x)sin(x)-cos(x)=0 =2sin(x)cos(x)sin(x)−cos(x)=0=2sin(x)cos(x)sin(x)-cos(x)=0 =cos(x)(2sin2(x)−1)=0=cos(x)(2sin2(x)-1)=0 =cos(x)(2–√sin(x)−1)(2–√sin(x)+1)=0=cos(x)(2sin(x)-1)(2sin(x)+1)=0 solving each part…
1 educator answer
- MATH
sin(2x)−sin(x)=0sin(2x)-sin(x)=0 Find the exact solutions of the equation in the interval [0, 2pi).
sin(2x)−sin(x)=0,0≤x≤2πsin(2x)-sin(x)=0,0≤x≤2π using the identity sin(2x)=2sin(x)cos(x)sin(2x)=2sin(x)cos(x) sin(2x)−sin(x)=0sin(2x)-sin(x)=0 =2sin(x)cos(x)−sin(x)=0=2sin(x)cos(x)-sin(x)=0 =(sin(x))(2cos(x)−1)=0=(sin(x))(2cos(x)-1)=0solving each part separately, sin(x)=0sin(x)=0 or…
1 educator answer
- MATH
We know that sin(x+π2)=cos(x).sin(x+π2)=cos(x). Therefore our equation may be written as cos(x)−cos2(x)=0,cos(x)-cos2(x)=0, or cos(x)⋅(1−cos(x))=0.cos(x)⋅(1-cos(x))=0.This means that cos(x)=0cos(x)=0 or cos(x)=1.cos(x)=1. The roots in the interval
1 educator answer
- MATH
tan(x+π)+2sin(x+π)=0,0≤x≤2πtan(x+π)+2sin(x+π)=0,0≤x≤2π using tan(x+π)=tan(x),sin(x+π)=−sin(x)tan(x+π)=tan(x),sin(x+π)=-sin(x) tan(x)−2sin(x)=0tan(x)-2sin(x)=0 sin(x)cos(x)−2sin(x)=0sin(x)cos(x)-2sin(x)=0 sin(x)−2sin(x)cos(x)=0sin(x)-2sin(x)cos(x)=0 sin(x)(1−2cos(x))=0sin(x)(1-2cos(x))=0 solving each…
1 educator answer
- MATH
sin(x+π6)−sin(x−7π6)=3–√2sin(x+π6)-sin(x-7π6)=32 Find all solutions of the equation in the interval…
Use the formula of difference of sinuses sin(a)−sin(b)=2sin(a−b2)cos(a+b2)sin(a)-sin(b)=2sin(a-b2)cos(a+b2) and obtain 2sin(2π3)cos(x−π2)=3–√2,2sin(2π3)cos(x-π2)=32, or 2⋅3–√2⋅cos(x−π2)=3–√2,2⋅32⋅cos(x-π2)=32, or cos(x−π2)=12.cos(x-π2)=12. The…
1 educator answer
- MATH
cos(x+π4)−cos(x−π4)=1cos(x+π4)-cos(x-π4)=1 Find all solutions of the equation in the interval [0,…
cos(x+π4)−cos(x−π4)=1,0≤x≤2πcos(x+π4)-cos(x-π4)=1,0≤x≤2π We will use the following identity, cos(A+B)=cosAcosB−sinAsinBcos(A+B)=cosAcosB-sinAsinB cos(x+π4)−cos(x−π4)=1cos(x+π4)-cos(x-π4)=1
1 educator answer
- MATH
It is known that cos(x+π)=−cos(x).cos(x+π)=-cos(x). So the equation is equivalent to −cos(x)−cos(x)−1=0,-cos(x)-cos(x)-1=0, or cos(x)=−12.cos(x)=-12. The solutions in the interval [0,2π)[0,2π) are x=2π3x=2π3 and x=4π3.x=4π3.
1 educator answer
- MATH
It is known that sin(x+π)=−sin(x).sin(x+π)=-sin(x). So the equation is equivalent to −sin(x)−sin(x)+1=0,-sin(x)-sin(x)+1=0, or sin(x)=12.sin(x)=12. The solutions in the interval [0,2π)[0,2π) are x=π6x=π6 and x=5π6.5π6.
1 educator answer
- MATH
sin(x+y)+sin(x−y)=2sin(x)cos(y)sin(x+y)+sin(x-y)=2sin(x)cos(y) Prove the identity.
sin(x+y)+sin(x−y)=2sin(x)cos(y)sin(x+y)+sin(x-y)=2sin(x)cos(y) We will make use of the following formulas to prove the identity, sin(A+B)=sinAcosB+cosAsinBsin(A+B)=sinAcosB+cosAsinB sin(A−B)=sinAcosB−cosAsinBsin(A-B)=sinAcosB-cosAsinB LHS=sin(x+y)+sin(x−y)sin(x+y)+sin(x-y)…
1 educator answer
- MATH
cos(x+y)cos(x−y)=cos2(x)−sin2(y)cos(x+y)cos(x-y)=cos2(x)-sin2(y) Prove the identity.
cos(x+y)cos(x−y)=cos2(x)−sin2(y)cos(x+y)cos(x-y)=cos2(x)-sin2(y) We will use the following product formulas to prove the identity, 2cosAcosB=cos(A+B)+cos(A−B)2cosAcosB=cos(A+B)+cos(A-B) LHS=cos(x+y)cos(x−y)cos(x+y)cos(x-y) =cos(x+y+x−y)+cos(x+y−(x−y))2=cos(x+y+x-y)+cos(x+y-(x-y))2…
1 educator answer
- MATH
tan(π4−θ)=1−tan(θ)1+tan(θ)tan(π4-θ)=1-tan(θ)1+tan(θ) Prove the identity.
tan(π4−θ)=1−tan(θ)1+tan(θ)tan(π4-θ)=1-tan(θ)1+tan(θ) we will use the following formula to prove the identity, tan(A−B)=tanA−tanB1+tanAtanBtan(A-B)=tanA-tanB1+tanAtanB LHS=tan(π4−θ)tan(π4-θ)…
1 educator answer
- MATH
cos(π−θ)+sin(π2+θ)=0cos(π-θ)+sin(π2+θ)=0 Prove the identity.
You need to use the formula cos(a−b)=cosa⋅cosb+sina⋅sinbcos(a-b)=cosa⋅cosb+sina⋅sinb and sin(a+b)=sina⋅cosb+sinb⋅cosa,sin(a+b)=sina⋅cosb+sinb⋅cosa, such that: cos(π−θ)=cosπ⋅cosθ+sinπ⋅sinθcos(π-θ)=cosπ⋅cosθ+sinπ⋅sinθ Since
1 educator answer
- MATH
cos(5π4−x)=−2–√2(cos(x)+sin(x))cos(5π4-x)=-22(cos(x)+sin(x)) Prove the identity.
Prove the identity cos(5π4+x)=−2–√2(cos(x)+sin(x))cos(5π4+x)=-22(cos(x)+sin(x)) Use the formula cos(u)cos(v)+sin(u)sin(v)cos(u)cos(v)+sin(u)sin(v) on the left side of the equation. cos(5π4+x)cos(5π4+x)…
1 educator answer
- MATH
sin(π6+x)=(12)(cos(x)+3–√sin(x))sin(π6+x)=(12)(cos(x)+3sin(x)) Prove the identity.
Prove the identity sin(π6+x)=12(cos(x)+3–√sin(x))sin(π6+x)=12(cos(x)+3sin(x)) Use the formula sin(u+v)=sin(u)cos(v)+cos(u)sin(v)sin(u+v)=sin(u)cos(v)+cos(u)sin(v) on the left side of the equation. sin(π6+x)sin(π6+x) =sin(π6)cos(x)+cos(π6)sin(x)=sin(π6)cos(x)+cos(π6)sin(x)…
1 educator answer
- MATH
sin(π2+x)=cos(x)sin(π2+x)=cos(x) Prove the identity.
Use the formula of sine of the sum of angles: sin(a+b)=sin(a)cos(b)+cos(a)sin(b)sin(a+b)=sin(a)cos(b)+cos(a)sin(b) for a=xa=x and b=π2.b=π2.Recall that sin(π2)=1sin(π2)=1 and cos(π2)=0:cos(π2)=0:…
1 educator answer
- MATH
sin(π2−x)=cos(x)sin(π2-x)=cos(x) Prove the identity.
Use the formula for the sine of the sum of angles: sin(a−b)=sin(a)cos(b)−cos(a)sin(b)sin(a-b)=sin(a)cos(b)-cos(a)sin(b) for a=π2a=π2 and b=x.b=x. It gives sin(π2−x)=sin(π2)cos(x)−cos(π2)sin(x).sin(π2-x)=sin(π2)cos(x)-cos(π2)sin(x). Because sin(π2)=1sin(π2)=1 and…
1 educator answer