Make the internet a better place to learn

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#

2

Answer:

#f_0 = 1/(2pi)" Hz"#

Explanation:

Given: #f(t)= sin(4t) - cos(7t)# where t is seconds.

Use this reference for Fundamental Frequency

Let #f_0# be the fundamental frequency of the combined sinusoids, in Hz (or #"s"^-1#).

#omega_1 = 4" rad/s"#
#omega_2 = 7" rad/s"#

Using the fact that #omega = 2pif#

#f_1 = 4/(2pi) = 2/pi " Hz"# and #f_2 = 7/(2pi)" Hz"#

The fundamental frequency is the greatest common divisor of the two frequencies:

#f_0 = gcd(2/pi " Hz", 7/(2pi)" Hz")#

#f_0 = 1/(2pi)" Hz"#

Here is a graph:

graph{y = sin(4x) - cos(7x) [-10, 10, -5, 5]}

Please observe that it repeats every #2pi#

2

Answer:

#x^2+y^2=((x^2-y^2)/(x^2+y^2))^2#

Explanation:

Here is the graph of the original equation #r = cos(2theta)#:

Desmos.com

Use the identity #cos(2theta) = cos^2(theta)-sin^2(theta)#

#r = cos^2(theta)-sin^2(theta)#:

Substitute the #sqrt(x^2+y^2)# for r, #x^2/(x^2+y^2)# for #cos^2(theta)#, and #y^2/(x^2+y^2)# for #sin^2(theta)#:

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

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

Here is the graph of that equation:

graph{sqrt(x^2+y^2)=x^2/(x^2+y^2)-y^2/(x^2+y^2) [-10, 10, -5, 5]}

It is only half of the loops

Square both sides:

#x^2+y^2=((x^2-y^2)/(x^2+y^2))^2#

Here is the graph of that equation:

graph{x^2+y^2=((x^2-y^2)/(x^2+y^2))^2 [-10, 10, -5, 5]}

This looks like the polar equation.

4

Answer:

#color(red)[34^o]#

Explanation:

enter image source here

We can construct this diagram from the given information.

Let length of #AB# be #l_1# and length of #BC# be #l_2#.

It is given that #l_1/l_2 = 2/3# #-># 1.

Also, observe that #BD = FE# and #BF=DE#. #-># 2.

We have to find #angle CAE# which will give us elevation of C from A.

Let #angle CAE# be #theta#.

From #triangle ABF# we observe that:-

#[BF]/ [AB] = sin 25^o => BF = AB*sin 25^o = 0.423*l_1#

#therefore# using 2., #DE=0.423*l_1# #-># 3.

#[AF]/[AB] = cos 25^o => AF = AB*cos25^o = 0.906*l_1# #-># 4.

From #triangle BCD# we find:-

#[CD]/[BC] = sin40^o => CD = BC*sin40^o = 0.643*l_2# #-># 5.

#[BD]/[BC] = cos40^0 => BD = BC*cos40^o = 0.766*l_2#

#therefore# form 2., #FE=0.766*l_2# #-># 6.

Now,
#tan theta = [CE]/[AE] = [CD+DE]/[AF+FE]#

Substituting values of #DE, AF, CD, FE# from 3., 4., 5., 6.

#tan theta = [0.643*l_2 + 0.423*l_1]/[0.766*l_2 + 0.906*l_1]#

Dividing the numerator and deniminator by #l_2#

#=> tan theta = [0.643*cancel(l_2/l_2) + 0.423*l_1/l_2]/[0.766*cancel(l_2/l_2) + 0.906*l_1/l_2]#

Substituting value of #l_1/l_2# from 1.

#=> tan theta = [0.643 + 0.423*2/3]/[0.766+0.906*2/3] = 0.675#

#therefore theta = tan^-1 0.675 approx 34^o#

#therefore color(red)[angleCAE = 34^o] # which is the angle of elevation of #C# from #A#.

2

Answer:

Please see the explanation.

Explanation:

Given: #8=(3x-y)^2+2y+6x#

Here is a graph of the Cartesian equation:
Desmos.com

Expand the square:

#9x^2-6xy+y^2+6x+2y-8 = 0#

Substitute #rcos(theta)# for every x:

#9(rcos(theta))^2-6(rcos(theta))y+y^2+6(rcos(theta))+2y-8 = 0#

Substitute #rsin(theta)# for every y:

#9(rcos(theta))^2-6(rcos(theta))(rsin(theta))+(rsin(theta))^2+6(rcos(theta))+2(rsin(theta))-8 = 0#

There is a common factor of #r^2# in the first 3 terms:

#(9cos^2(theta)-6cos(theta)sin(theta)+sin^2(theta))r^2+6(rcos(theta))+2(rsin(theta))-8 = 0#

There is a common factor of r in the next 2 terms:

#(9cos^2(theta)-6cos(theta)sin(theta)+sin^2(theta))r^2+(6cos(theta)+2sin(theta))r-8 = 0#

The above is a quadratic equation in standard form where:

r in the independent variable

#a = 9cos^2(theta)-6cos(theta)sin(theta)+sin^2(theta)#

#b = 6cos(theta)+2sin(theta)#

#c = -8#

We use the identity #cos^2(theta) + sin^2(theta) = 1 to simplify a:

#a = 8cos^2(theta)-6cos(theta)sin(theta)+1#

We can use the quadratic formula to write r and a function of #theta# but, because radii should only be positive we will discard the negative root:

#r = (-b+sqrt(b^2-4(a)(c)))/(2a)#

Substitute for every a:

#r = (-b+sqrt(b^2-4(8cos^2(theta)-6cos(theta)sin(theta)+1)(c)))/(2(8cos^2(theta)-6cos(theta)sin(theta)+1))#

Substitute for c:

#r = (-b+sqrt(b^2+32(8cos^2(theta)-6cos(theta)sin(theta)+1)))/(2(8cos^2(theta)-6cos(theta)sin(theta)+1))#

Substitute for every b:

#r = (-(6cos(theta)+2sin(theta))+sqrt((6cos(theta)+2sin(theta))^2+32(8cos^2(theta)-6cos(theta)sin(theta)+1)))/(2(8cos^2(theta)-6cos(theta)sin(theta)+1))#

Here is a graph of the polar equation:

Desmos.com

The graphs are identical, therefore, the conversion is complete.

Note: I am sure that the expression under the radical can be simplified but I will leave that to you.

2

drawn The given function representing the height #h# in cm of the person above the ground with time #t# is given by

#h=15cos((2pi)/3t)+65#

(a) h is a cosine function of t, So it will have maximum value when #cos((2pi)/3t)=1=cos0=>t=0# s
and the maximum height of the swing becomes

#h_(max)=15cos((2pi)/3xx0)+65=80cm#

(b) it takes #t=0#sec to reach maximum height after the person starts.

c) The minimum height of the swing will be achieved when #cos((2pi)/3t)=-1#

Minimum height

#h_(min)=15xx(-1)+65=50cm#

(d). Again #cos((2pi)/3t)=-1=cos(pi)=>t=3/2=1.5#s

Here #t=1.5#s represents the minimum time required to achieve the minimum height after start.

e) For #h=60# the equation becomes

#60=15cos((2pi)/3t)+65#

#=>60-65=15cos((2pi)/3t)#

#=>cos((2pi)/3t)=-1/3#

#=>t=[1/120(2nxx180pmcos^-1(-1/3))] sec" where " n in ZZ#

#=>t=(3npm0.912)#sec

when #n=0, t=0.912s# this is the minimum time to reach at height #h=60 cm # after start. just after that height #h=60 cm # will be achieved again for #n=1 #

#=>t=(3xx1-0.912)=2.088#sec

So in #(2.088-0.912)=1.176s# sec (the minimum time difference) within one cycle the swing will be less than 60 cm above the ground.

f) The height of the swing at 10 s can be had by inserting #t=10#s in the given equation

#h_(10)=15cos((2pi)/3xx10)+65=57.5#cm

2

Answer:

#x~~0.3398#, #x~~0.8481#, #x=2.294#, and #x=2.802# radians.

Explanation:

Strategy: Rewrite this equation as a quadratic equation using #u=sin(x)#. Solve the quadratic equation for #u# by factoring. Then replace #u# with #sin(x)# again and solve using #arcsin#.

Step 1. Rewrite this equation as a quadratic using #color(red)u=color(red)sin(x)#

You are given
#12(color(red)(sin(x)))^2-13(color(red)(sin(x)))+3=0#

Replace #color(red)(sin(x))# with #color(red)(u)#
#12color(red)(u)^2-13color(red)(u)+3=0#

Step 2. Factor the quadratic equation.
#(3u-1)(4u-3)=0#

Solving gives us
#3u-1=0# and #4u-3=0#
#u=1/3# and #u=3/4#

Step 3. Replace #u# with #sin(x)# again and solve with #arcsin#
#sin(x)=1/3# and #sin(x)=3/4#

#sin^(-1)(sin(x))=sin^(-1)(1/3)# and #sin^(-1)(sin(x))=sin^(-1)(3/4)#

#x~~0.3398# radians and #x~~0.8481# radians

These answer work because we were asked to find the solutions in #[0,2pi]~~[0,6.2832]#

However, the graph of #y=12(sin(x))^2-13(sin(x))+3# is

Desmos.com and MS Paint

Which shows four solutions in the interval #[0,2pi]#, not two.

We must find the other solutions by recognizing that sine functions are periodic with respect to #pi#

So, #x=pi-0.8481=2.294# and #x=pi-0.3398=2.802#

The other two solutions are
#x=2.294# and #x=2.802#