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Featured 2 months ago

First use the cosine angle-addition formula:

#cos(a+b)=cos(a)cos(b)-sin(a)sin(b)#

Then the original expression equals:

#=cos(tan^-1(2))cos(tan^-1(3))-sin(tan^-1(2))sin(tan^-1(3))#

Note that if

So, when

#cos(tan^-1(2))=cos(theta)="adjacent"/"hypotenuse"=1/sqrt5#

#sin(tan^-1(2))=sin(theta)="opposite"/"hypotenuse"=2/sqrt5#

Now let angle

#{("opposite"=3),("adjacent"=1),("hypotenuse"=sqrt10):}#

Then:

#cos(tan^-1(3))=cos(phi)="adjacent"/"hypotenuse"=1/sqrt10#

#sin(tan^-1(3))=sin(phi)="opposite"/"hypotenuse"=3/sqrt10#

Plugging these into the expression from earlier we get:

#=1/sqrt5(1/sqrt10)-2/sqrt5(3/sqrt10)#

Note that

#=(1-6)/(5sqrt2)=-1/sqrt2#

Featured 2 months 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

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 1 month 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.

Featured yesterday

I like getting rid of the phase shift (the

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

We have:

#y = 3(cosxcos(pi) - sinxsinpi) - 3#

#y = 3(cosx(-1) - 0) - 3#

#y = -3cosx - 3#

Now you need a little bit of knowledge on the basic cosine function,

graph{y = cosx [-10, 10, -5, 5]}

Whenever there is a coefficient

In the graph of

The

Finally, the

graph{y = -3cosx - 3 [-10, 10, -5, 5]}

Hopefully this helps!

Featured yesterday

(a)

[Taking

Where

(b)

(i)

The value of

So

(ii)Here

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