Featured 1 month ago

The explanation is written with the assumption that you meant

Use the distributive property :

Substitute x for

Square both sides:

Add

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

Use the middle in the right side of the pattern

Substitute the left side of the pattern into the equation:

Substitute

Divide both sides by

Write the denominators as squares:

The is standard Cartesian form of the equation of an ellipse with a center at

Featured 1 month ago

Given

To get x-intercepts we can put y=0 and solve for

So

Putting

Putting

Putting

Putting

(a) So points of x-intercepts over

b) For maximum value of y the value of

So

For

The point(in (0,2pi) where the graph of this function reaches

maximum is

Featured 1 month ago

Start from a "basic cycle" for the

Starting with the "basic cycle" for

Then consider what values of

(Yes; I know: because

Giving us the "basic cycle" for

Adding

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

Featured 1 month ago

See explanation...

Consider a right angled triangle with an internal angle

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

So given Pythagoras, that proves the identity for

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#

**Pythagoras theorem**

Given a right angled triangle with sides

The area of the large square is

The area of the small, tilted square is

The area of each triangle is

So we have:

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

That is:

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

Subtract

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

Featured 1 month ago

Use the formula for a circle

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

Picture a circle of radius

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

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 (

Now for the trigonometry.

For an angle

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

Similarly,

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

So we have

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

#(rcostheta)^2+(rsintheta) ^2 = r^2#

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

The

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

This is often rewritten with the

#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

Featured 3 weeks ago

We are to turn the given equation into a form of

Let me try to have a solution.

Given equation is

Dividing both sides by sin^2x we get

When

Again when

Featured 3 weeks ago

Given

Both

So

Given

Featured 3 weeks ago

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

#1 + 2sinxcosx = 0#

Use

#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

Let's answer the second question.

This means that the domain of

Hopefully this helps!

Featured 3 weeks ago

Note that

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

Now use

#cosx + cos2xcosx - sin2xsinx = 0#

Apply

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

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

Use

#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!

Featured 3 weeks ago

Here is the graph of the function (it's not drawn to scale):

I will explain how I graph it below:

Your given equation is

I would write the given equation in the standard form according to

You get:

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:

**Amplitude**, which is#|A|# **Period Multiplier**, which is**B****Phase Shift**, which is**C****Vertical Shift,**which is**D**- Period, which is
#Period = (2pi)/B#

Let's get all our required values from our equation:

**Amplitude**=#|5|# =**5****Period Multiplier**= B =**1****Phase Shift**=#-pi# =#pi# units to the right**Vertical Shift**=#+1# =**1 units up****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

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 =

**Minimum Y Value** = Vertical Shift - Amplitude =

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

So.. with all this information, we can now actually graph the equation of

Here are some visual cues.

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

I hope this helps.