Featured Answers

1
Active contributors today

Answer:

#cos(tan^-1(2)+tan^-1(3))=-1/sqrt2#

Explanation:

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 #theta=tan^-1(2)#, then #tan(theta)#. That is, a right triangle with angle #theta# has #tan(theta)=2#, which is the triangle with the leg opposite #theta# being #2# and the leg adjacent to #theta# being #1#. The Pythagorean theorem tells us that the hypotenuse of this triangle is #sqrt5#.

So, when #theta=tan^-1(2)#, we see that:

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

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

Now let angle #phi# be defined by #phi=tan^-1(3)#, such that #tan(phi)=3#. This is the right triangle with:

#{("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 #sqrt5(sqrt10)=sqrt50=5sqrt2#:

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

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

#cosx+cos3x=0#?

HSBC244
HSBC244
Featured 1 month 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!

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.

I like getting rid of the phase shift (the #x + pi# part) using the sum and difference formulas. The one that is applicable here is

#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, #y = cosx#. Here's the graph:

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

Whenever there is a coefficient #a# next to the cosine, you have an altered amplitude, which is the distance between the centre (the line #y = 0#) and the top or bottom of the curve.

In the graph of #y = cosx#, the amplitude is simply #1#. In the graph of #y = -3cosx - 3#, the amplitude will be #3#.

The #-# is in front of the #3# to signify a reflection over the x-axis.

Finally, the #-3# to the far right of the equation signifies a vertical transformation of #3# units down. We are left with the following graph:

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

Hopefully this helps!

enter image source here

(a)

#sqrt15sin(2x)+sqrt5cos(2x)#

#=sqrt20(sqrt15/sqrt20sin(2x)+sqrt5/sqrt20cos(2x))#

[Taking #sqrt15/sqrt20=cosalpha and sqrt5/sqrt20=sinalpha#]

#=sqrt20(cosalphasin(2x)+sinalphacos(2x))#

#=2sqrt5sin(2x+alpha)#

Where #alpha=tan^-1(sqrt5/sqrt15)=tan^-1(1/sqrt3)=pi/6#

(b)

(i)
#f(x)=2/(5+sqrt15sin(2x)+sqrtcos(2x))#

#=>f(x)=2/(5+2sqrt5sin(2x+alpha))#

The value of #f(x)# will be maximum when #sin(2x+alpha)# is minimum i.e. #sin(2x+alpha)=-1#

So

#f(x)_"max"=2/(5-2sqrt5)#

#=>f(x)_"max"=(2(5+2sqrt5))/((5-2sqrt5)(5+2sqrt5))#

#=>f(x)_"max"=(10+4sqrt5)/(25-20)=2+4/5sqrt5#

(ii)Here #f(x)# is maximum when

#sin(2x+alpha)=-1#

#=>sin(2x+pi/6)=sin((3pi)/2)#

#=>2x+pi/6=(3pi)/2#

#=>2x=(3pi)/2-pi/6=(4pi)/3#

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

View more
Questions
Ask a question Filters
Loading...
  • N N. answered · Yesterday
This filter has no results, see all questions.
×
Question type

Use these controls to find questions to answer

Unanswered
Need double-checking
Practice problems
Conceptual questions