Make the internet a better place to learn

2

Answer:

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

Explanation:

The maximum and minimum of any #cos# graph on its own will be #+1# or #-1#. It doesn't matter so much what's inside the #cos# function, only what's outside.

Multiply these by #-2#, the outside of the equation, to get maximum/minimum points of #-2# and #+2#.

The maximum is therefore #+2#.

To find at which point this occurs, work backwards.

#+2 = -2cos(x-pi/2)#

Divide both sides by #-2#

#-1 = cos(x-pi/2)#

Do a backwards #cos# (#cos^-1#) to get rid of the forwards #cos#

#cos^-1(-1)=x-pi/2#
#pi = x-pi/2#
#x = (3pi)/2#

Now we have our #x# and #y# values:

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

1

Answer:

#x_1=npi, n=2k+-1, k in ZZ#

#x_2=n/3pi,mod(n,2)=0#

Explanation:

#cos2x+3cosx=-2#

Use the double angle formula for cosine to expand #cos2x# and rewrite the equation in standard form

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

#2cos^2x+3cosx+1=0#

Let #cosx=y#

#y=(-3+-sqrt(9-4(2)(1)))/(2(2))=(-3+-1)/4#

#y=-1# or #y=-1/2#

#cosx=-1#

#x=npi, n=2k+-1, k in ZZ#

#cosx=-1/2#

#x_2=n/3pi,mod(n,2)=0#

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.

2

# sum_(r=1)^(r=4) {[sin((2r-1)pi/8)]^4 +[cos((2r-1)pi/8]^4}#

Taking #(2r-1)pi/8=theta#

# {[sin((2r-1)pi/8)]^4 +[cos((2r-1)pi/8]^4}#

#= {[sintheta]^4 +[costheta]^4}.#

#={[sin^2theta +cos^2theta]^2-2sin^2thetacos^2theta}#

#={1^2-1/2(2sinthetacostheta)^2}#

#={1^2-1/2sin^2 2theta}#

# ={1-1/2sin^2 (2r-1)pi/4}#

# ={1-1/2sin^2 ((rpi)/2-pi/4))}#

# ={1-1/2(pm1/sqrt2)^2}color(red)"*"#

# ={1-1/4 }=3/4#

#[color(red)"*"=>sin((rpi)/2-pi/4)=pmsin(pi/4) or pmcos(pi/4)" for " r in ZZ^"+"]#
So

# sum_(r=1)^(r=4) {[sin((2r-1)pi/8)]^4 +[cos((2r-1)pi/8]^4}#

# =sum_(r=1)^(r=4) {3/4}=4xx3/4=3#

3

Answer:

#(2pi)/3, (4pi)/3, (8pi)/3#

Explanation:

#2cosx+1=0 #

#cosx= -1/2 #

Now we know from this handy "CAST" nmemonic that cos is negative in Q2+Q3:

Rembrandt

We can even draw the triangles:

Picasso

So in Q1: #cos x = 1/2 implies x = pi/3#

The corresponding angle in Q2 is #pi - pi/3 = (2pi)/3#

The corresponding angle in Q3 is #pi + pi/3 = (4pi)/3#

Now because it is #x in[0,3pi)#, we are not done yet.

After spinning round a full #2 pi#, we have another #pi# to go which takes us back into Q2.

The second corresponding angle in Q2 is #2 pi + (pi - pi/3) = (8pi)/3#

1

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!

1

Answer:

Period of #y = 1+4 \sin(2x)# is #n \pi# and amplitude is #4#

Explanation:

GENERAL WAVE EQUATION

#f(x) = a \sin(bx+c) + d#,

where #\pm c#, #\pm d# are wave right (left) or up (down) shifts along #x "-axis"# and #y"-axis"# respectively.

SOLUTION (for period)

Ignoring the shifts along axes i.e. assuming #c = d = 0# we can have our period and amplitude without any sort of error.

#f(x) = y=1+4 \sin(2x)#

#\Rightarrow y -1 = 4 \sin(2x)#

#\Rightarrow g(x) = 4 \sin(2x)#

We know from fundamental trigonometric ideas that the period of sine function is #2 n \pi# i.e.

#\sin(x) = \sin(2n \pi +x)#,

Let,

#\sin(u) = \sin(2n \pi + u)#

Then for #u=2x#, we have;

#\sin(2x) = \sin(2n \pi +2x)#

Multiplying both sides by #4#,

#4 \sin(2x) = 4\sin 2(n \pi + x)#

#\equiv g(x) = g(n\pi+x)#

Hence period is #n \pi#

SOLUTION (amplitude)

In general the amplitude of sine function is given as;

If #y=\sin(x)# then amplitude is #1# unit long. as followed by graph

graph{y=\sin(x)[-3,3,-1.5,1.5]}

Our equation is a sort of #y=asin(bx)#

Which becomes;

#y= a[\sin(bx)]#

And that expands the graph of ordinary sine function i.e. #sin(x)# up to #|a|# times along #y"-axis"#

graph{y=4\sin(2x)}

Therefore our amplitude is #4#

The graph of our real function #y=1+ 4 \sin(2x) # is shifted upwards #1# unit

graph{y=1+4 \sin(2x)}

But amplitude remains same.

2

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#

1

Answer:

Here is the graph of #y=sin^(-1)x#.

enter image source here

Explanation:

Step 1: Sketch the graph of #y=sin x# on #[-pi/2,pi/2]#.

enter image source here

Step 2: Sketch the line #y=x#.

enter image source here

Step 3: Reflect the graph of #y=sin x# about the line #y=x#.

enter image source here

1

Answer:

#csc(2arctan(3/4))=25/24#

Explanation:

Because #sin# and #csc# are reciprocals:

#csc(2arctan(3/4))=1/sin(2arctan(3/4))#

Using the double angle identity #sin(2theta)=2sin(theta)cos(theta)#:

#=1/(2color(blue)(sin(arctan(3/4)))color(red)(cos(arctan(3/4)))#

We can find the values of #sin(arctan(3/4))# and #cos(arctan(3/4))# using a similar method.

Note that when #theta=arctan(3/4)#, then #tan(theta)=3/4#. That is, where #theta# is an angle in a right triangle, the side opposite #theta# is #3# and the leg adjacent to #theta# is #4#. The Pythagorean theorem tells us that the hypotenuse is #5#.

enter image source here

We then see that:

#color(blue)(sin(arctan(3/4)))=sin(theta)="opposite"/"hypotenuse"=3/5#

#color(red)(cos(arctan(3/4)))=cos(theta)="adjacent"/"hypotenuse"=4/5#

So the original expression is:

#=1/(2color(blue)((3/5))color(red)((4/5)))#

#=25/24#