Make the internet a better place to learn

2

Answer:

The explanation is written with the assumption that you meant #theta# for the argument of the cosine function.

Explanation:

Use the distributive property :

#2r - rcos(theta) = 2#

#2r = rcos(theta) + 2#

Substitute x for #rcos(theta)# and #sqrt(x^2 + y^2)# for r:

#2sqrt(x^2 + y^2) = x + 2#

Square both sides:

#4(x^2 + y^2) = x^2 + 4x + 4#

#4x^2 + 4y^2 = x^2 + 4x + 4#

#3x^2 - 4x + 4y^2 = 4#

Add #3h^2# to both sides:

#3x^2 - 4x + 3h^2 + 4y^2 = 3h^2+ 4#

Remove a factor of the 3 from the first 3 terms:

#3(x^2 - 4/3x + h^2) + 4y^2 = 3h^2+ 4#

Use the middle in the right side of the pattern #(x - h)^2 = x^2 - 2hx + h^2# and middle term of the equation to find the value of h:

#-2hx = -4/3x#

#h = 2/3#

Substitute the left side of the pattern into the equation:

#3(x - h)^2 + 4y^2 = 3h^2+ 4#

Substitute #2/3# for h and insert a -0 into the y term:

#3(x- 2/3)^2 + 4(y-0)^2 = 3(2/3)^2+ 4#

#3(x- 2/3)^2 + 4(y-0)^2 = 16/3#

Divide both sides by #16/3#

#3(x- 2/3)^2/(16/3) + 4(y-0)^2/(16/3) = 1#

#(x- 2/3)^2/(16/9) + (y-0)^2/(16/12) = 1#

Write the denominators as squares:

#(x- 2/3)^2/(4/3)^2 + (y-0)^2/(4/sqrt(12))^2 = 1#

The is standard Cartesian form of the equation of an ellipse with a center at #(2/3,0)#; its semi-major axis is #4/3# units long and is parallel to the x axis and its semi-minor axis is #4/sqrt(12)# units long and is parallel to the y axis.

2

Given
#y=f(x)=2sin(x-pi/4)#

To get x-intercepts we can put y=0 and solve for #x in [0,4pi]#

So #2sin(x-pi/4)=0#

#=>x-pi/4=npi" where " n in ZZ#

#=>x=npi+pi/4" where " n in ZZ#

Putting #n=0# we get

#x=pi/4#

Putting #n=1# we get

#x=(5pi)/4#

Putting #n=2# we get

#x=(9pi)/4#

Putting #n=3# we get

#x=(13pi)/4#

(a) So points of x-intercepts over #[0,4pi]# are

#(pi/4,0);((5pi)/4,0);((9pi)/4,0);((13pi)/4,0)#

b) For maximum value of y the value of #sin(x-pi/4)=1#

So #x-pi/4=pi/2#
#=>x= pi/4+pi/2=(3pi)/4#

For #x=(3pi)/4" " y=2#

The point(in (0,2pi) where the graph of this function reaches
maximum is #((3pi)/4,2)#

enter image source here

1

Answer:

Start from a "basic cycle" for the #tan# function to obtain the results below.

Explanation:

Starting with the "basic cycle" for #tan(theta)# i.e. for #theta in [-pi/2,+pi/2]#
enter image source here

Then consider what values of #x# would place #(x+pi)# in this same range, i.e. #(x+pi) in [-pi/2,+pi/2]#
(Yes; I know: because #tan# has a cycle length of #pi# we could recognize that the cycle will repeat at exactly the same place, but let's do this for the more general case.)
#(x+pi)in[-pi/2,pi/2]color(white)("X"rarrcolor(white)("X")x in [-(3pi)/2,-pi/2]#
Giving us the "basic cycle" for #tan(x+pi)#:
enter image source here

Adding #5# to this to get #y=5+tan(x+pi)# simply shifts the points upward #5# units:

In the image below, I have added the "non-basic cycles" as well as the "basic cycle" used for analysis:
enter image source here

2

Answer:

See explanation...

Explanation:

Consider a right angled triangle with an internal angle #theta#:

enter image source here

Then:

#sin theta = a/c#

#cos theta = b/c#

So:

#sin^2 theta + cos^ theta = a^2/c^2+b^2/c^2 = (a^2+b^2)/c^2#

By Pythagoras #a^2+b^2 = c^2#, so #(a^2+b^2)/c^2 = 1#

So given Pythagoras, that proves the identity for #theta in (0, pi/2)#

For angles outside that range we can use:

#sin (theta + pi) = -sin (theta)#

#cos (theta + pi) = -cos (theta)#

#sin (- theta) = - sin(theta)#

#cos (- theta) = cos(theta)#

So for example:

#sin^2 (theta + pi) + cos^2 (theta + pi) = (-sin theta)^2 + (-cos theta)^2 = sin^2 theta + cos^2 theta = 1#

#color(white)()#
Pythagoras theorem

Given a right angled triangle with sides #a#, #b# and #c# consider the following diagram:

enter image source here

The area of the large square is #(a+b)^2#

The area of the small, tilted square is #c^2#

The area of each triangle is #1/2ab#

So we have:

#(a+b)^2 = c^2 + 4 * 1/2ab#

That is:

#a^2+2ab+b^2 = c^2+2ab#

Subtract #2ab# from both sides to get:

#a^2+b^2 = c^2#

2

Answer:

Use the formula for a circle #(x^2+y^2=r^2)#, and substitute #x=rcostheta# and #y=rsintheta#.

Explanation:

The formula for a circle centred at the origin is

#x^2+y^2=r^2#

That is, the distance from the origin to any point #(x,y)# on the circle is the radius #r# of the circle.

Picture a circle of radius #r# centred at the origin, and pick a point #(x,y)# on the circle:

graph{(x^2+y^2-1)((x-sqrt(3)/2)^2+(y-0.5)^2-0.003)=0 [-2.5, 2.5, -1.25, 1.25]}

If we draw a line from that point to the origin, its length is #r#. We can also draw a triangle for that point as follows:

graph{(x^2+y^2-1)(y-sqrt(3)x/3)((y-0.25)^4/0.18+(x-sqrt(3)/2)^4/0.000001-0.02)(y^4/0.00001+(x-sqrt(3)/4)^4/2.7-0.01)=0 [-2.5, 2.5, -1.25, 1.25]}

Let the angle at the origin be theta (#theta#).

Now for the trigonometry.

For an angle #theta# in a right triangle, the trig function #sin theta# is the ratio #"opposite side"/"hypotenuse"#. In our case, the length of the side opposite of #theta# is the #y#-coordinate of our point #(x,y)#, and the hypotenuse is our radius #r#. So:

#sin theta = "opp"/"hyp" = y/r" "<=>" "y=rsintheta#

Similarly, #cos theta# is the ratio of the #x#-coordinate in #(x,y)# to the radius #r#:

#cos theta = "adj"/"hyp"=x/r" "<=>" "x = rcostheta#

So we have #x=rcostheta# and #y=rsintheta#. Substituting these into the circle formula gives

#"      "x^2"     "+"      "y^2"     "=r^2#
#(rcostheta)^2+(rsintheta) ^2 = r^2#
#r^2cos^2theta + r^2 sin^2 theta = r^2#

The #r^2#'s all cancel, leaving us with

#cos^2 theta + sin^2 theta = 1#

This is often rewritten with the #sin^2# term in front, like this:

#sin^2 theta + cos^2 theta = 1#

And that's it. That's really all there is to it. Just as the distance between the origin and any point #(x,y)# on a circle must be the circle's radius, the sum of the squared values for #sin theta# and #cos theta# must be 1 for any angle #theta#.

2

We are to turn the given equation into a form of #cosx/sinx="number"#

Let me try to have a solution.

Given equation is

#6sin^2x+7cosx=8#

Dividing both sides by sin^2x we get

#6+(7cosx)/sin^2x=8/sin^2x#

#=>6+7cotxcscx=8csc^2x#

#=>6+7cotxcscx=8(1+cot^2x)#

#=>6+7cotxcscx=8+8cot^2x#

#=>8cot^2x+8-6=7cotxcscx#

#=>8cot^2x+2=7cotxcscx#

#=>(8cot^2x+2)^2=7^2cot^2xcsc^2x#

#=>(8cot^2x+2)^2=7^2cot^2x(1+cot^2x)#

#=>64cot^4x+32cot^2x+4-49cot^4x-49cot^2x=0#

#=>15cot^4x-17cot^2x+4=0#

#=>15cot^4x-12cot^2x-5cot^2x+4=0#

#=>3cot^2x(5cot^2x-4)-1(5cot^2x-4)=0#

#=>(3cot^2x-1)(5cot^2x-4)=0#

When #3cot^2x-1=0#

#=>cotx=pm1/sqrt3#

#=>cosx/sinx=pm1/sqrt3#

Again when

#5cot^2x-4=0#

#=>cotx=pm2/sqrt5#

#=>cosx/sinx=pm2/sqrt5#

1

Given #tanu=5/12#
Both #u and v # are in III qudrant
So
#tanu->+ve#

#tanv->+ve#

#sinu->-ve#

#sinv->-ve#

#cosu->-ve#

#cosv->-ve#

#sinu=1/cscu=-1/sqrt(1+cot^2u)=-1/sqrt(1+12^2/5^2)=-5/13#

#cosu=sinu/tanu=-(5/13)/(5/12)=-12/13#

Given #sinv=-3/5#

#cosv#
#=-sqrt(1-sin^2v)=-sqrt(1-9/25)=-4/5#

#tanv=sinv/cosv=3/4#

#sin(u+v)#

#=sinucosv+cosusinv#

#=(-5/13)(-4/5)+(-12/13)(-3/5)#

#=20/65+36/65=56/65#

#tan(u+v)=(tanu+tanv)/(1-tanutanv)#

#=(5/12+3/4)/(1-5/12*3/4)=(14/12)/(11/16)=7/6*16/11=56/33#

#cos(u+v)#

#=cosucosv-sinusinv#

#=(-12/13)(-4/5)-(-5/13)(-3/5)#

#=48/65-15/65=33/65#

1

Answer:

#{x| x!= (3pi)/4 + pin, x in RR, "where " n in ZZ}#

Explanation:

For this problem, we have to ask ourselves a couple of questions.

•When does #sinx + cosx# equal #0#.
•What is the domain of #3^(x + 1)#

I'll start by answering the first one. Solve the trigonometric equation.

#sinx + cosx = 0#

Square both sides

#(sinx + cosx)^2 = 0^2#

#sin^2x + 2sinxcosx + cos^2x = 0#

Apply #sin^2x + cos^2x = 1#:

#1 + 2sinxcosx = 0#

Use #2sinxcosx = sin2x#:

#1 + sin2x = 0#

#sin2x = -1#

#2x = arcsin(-1)#

#2x = (3pi)/2 + 2pin# because the sine function has a period of #2pi#

#x = (3pi)/4 + pin#

This means that whenever #x = (3pi)/4 + pin#, #n# an integer, the graph of #g(x) = 3^(x + 1)/(sinx + cosx)# will have vertical asymptotes.

Let's answer the second question.

#3^(x + 1)# is your run of the mill exponential function ; it will have a domain of all the real numbers, but will have a restricted range (we aren't dealing with range in this problem, though, so I won't go into detail there).

This means that the domain of #g(x)# is #{x| x!= (3pi)/4 + pin, x in RR, "where " n in ZZ}#.

Hopefully this helps!

3

#cosx+cos3x=0#?

HSBC244
HSBC244
Featured 3 weeks ago

Answer:

#x = pi/2, (3pi)/2, pi/4, (3pi)/4, (5pi)/4 and (7pi)/4#

Explanation:

Note that #cos3x# can be rewritten as #cos(2x + x)#.

#cosx + cos(2x + x) = 0#

Now use #cos(A + B) = cosAcosB - sinAsinB#.

#cosx + cos2xcosx - sin2xsinx = 0#

Apply #cos2x = 2cos^2x -1# and #sin2x = 2sinxcosx#.

#cosx + (2cos^2x - 1)cosx - 2sinxcosx(sinx) = 0#

#cosx + 2cos^3x - cosx - 2sin^2xcosx = 0#

Use #sin^2x + cos^2x = 1#:

#cosx + 2cos^3x - cosx - 2(1 - cos^2x)cosx = 0#

#cosx + 2cos^3x - cosx - 2cosx + 2cos^3x = 0#

#4cos^3x - 2cosx = 0#

Factor:

#2cosx(2cos^2x - 1) = 0#

We have

#cosx = 0#

#x = pi/2, (3pi)/2#

AND

#cosx = +-1/sqrt(2)#

#x = pi/4, (3pi)/4, (5pi)/4 and (7pi)/4#

Hopefully this helps!

2

Answer:

Here is the graph of the function (it's not drawn to scale):
enter image source here
I will explain how I graph it below:

Explanation:

Your given equation is #y = 1 +5cos(x-pi)# .

I would write the given equation in the standard form according to #y = AcosB(x-C)+D#

You get: #y = 5cos(x-pi) + 1#

This way, it is easier to define your amplitude, phase shift, vertical shift, and period multiplier.

In order to graph, we need to know the following:

  1. Amplitude , which is #|A|#
  2. Period Multiplier , which is B
  3. Phase Shift , which is C
  4. Vertical Shift, which is D
  5. Period, which is #Period = (2pi)/B#

Let's get all our required values from our equation: #y = 5cos(x-pi)+1#

  1. Amplitude = #|5|# = 5
  2. Period Multiplier = B = 1
  3. Phase Shift = #-pi# = #pi# units to the right
  4. Vertical Shift = #+1# = 1 units up
  5. Period = #Period = (2pi)/1# = #2pi#

It is important to note that B is not apparent in this equation because it is equal to 1. The actual equation should really be #y = 5cos1(x-pi)+1# .

Our period multiplier is important for determining our period, and it also affects our phase shift. However, we do not have to worry about it for this equation since it's just equal to 1.

Since we have both an amplitude and a vertical shift, we get new minimum and maximum y values for the graph:

Maximum Y Value = Vertical Shift + Amplitude = #1 + 5 = 6#

Minimum Y Value = Vertical Shift - Amplitude = #1 - 5 = -4#

Most importantly, this graph is a positive cosine graph because our equation is 5cos.. and not -5cos.. . A positive cosine graph always starts at the maximum value.

Also, our vertical shift is +1, so our new x-axis start has shifted up by 1. We are starting at y = 1 and not y = 0.

As with our phase shift, it is #pi# units to the right. How I like to think of the phase shift is that it's just a new start to the graph. You can treat the phase shift as a place to actually begin your graph.

So.. with all this information, we can now actually graph the equation of #y = 5cos(x-pi) + 1# .

Here are some visual cues.

enter image source here

enter image source here

My art is pretty bad, sographing the graph on a website should be much more different.

I hope this helps.