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3

Given that #f(t)# the value of a stock is modeled as

#f(t)=mt+b+Asin((pit)/6)# where t is in months since Jan1

Let us consider

#"At Jan1 "t= 0 and f(0)=$20.00#

#"At Apr1 "t= 3and f(3)=$37.50#

#"At Jul1 "t= 6 and f(6)=$32.50#

#"At Oct1 "t= 9 and f(6)=$35.00#

#"At Jan1 "t= 12 and f(6)=$50.00#

so
#"At Jan1 "t= 0 and f(0)=$20.00#

#=>mxx0+b+Asin(0)=20=>b=20.......(1)#

#"At Apr1 "t= 3and f(3)=$37.50#

#=>mxx3+20+Asin((pi*3)/6)=37.50#

#=>3m+20+A=37.50#

#=>3m+A=37.50-20.00#

#=>3m+A=17.50............................(2)#

#"At Jul1 "t= 6 and f(6)=$32.50#

#=>6xxm+20+Asin((pi*6)/6)=35.00#

#=>6xxm+20+0=35#

#=>m=15/6=2.5..........................(3)#

From (2) and (3) we get

#=>3xx2.5+A=17.50#

#=>A=10.00#

Hence
(a)
#color(red)(m=2.50" "b=20 and A=10)#
(b)

Now the given equation becomes

#f(t)=2.5t+20+10sin((pit)/6)#

Differentiating this equation w,r to t we get the rate of change of the price of stock as

#f'(t)=d/(dt)(2.5t)+d/(dt)(20)+d/(dt)(10sin((pit)/6))#

#=>f'(t)=2.5+(10pi)/6cos((pit)/6)#
In this equation the value of #cos((pit)/6)# varies with time.
#cos((pit)/6)# has got maximum positive value at t=0 and t =12 as #cos 0 and cos (2pi) = 1#
So rate of positive change of price of stock will occur in this period i.e. Jan every year this means the stock appreciate s most in this period.

(c)

Taking #f'(t)=0# we have

#2.5+(10pi)/6cos((pit)/6)=0#

#=>cos((pit)/6)=-1.5/pi#

#=>t=6/picos^-1(-1.5/pi)~~3.95~~4->color(red)"month of May"#

when t =4

#f''(4)=-(10pi^2)/36sin((pi*4)/6)<0->"indicates maximum"#

Again #t = 6/pi(2pi-cos^-1(-1.5/pi))=12-6/picos^-1(-1.5/pi)#

#=12-3.95=8.05->" indicates month of September" #

when t =8

#f''(8)=-(10pi^2)/36sin((pi*8)/6)>0->color(red)"indicates Minimum"#

So during the period May -September growth of price gets diminished i.e the price of the stock is actually losing in this period.
enter image source here

2

drawn

The equation #d(t) = 3.5 cos ((pit)/6)+4.5# models the water at the beach. The variation of water level with time as per the given equation has been shown in the above figure.graphically.
Graph has been plotted taking 12 O clock night as zero hour i,e, t = 0

The condition to dive off safely is that the water must be at least #3m# deep.
So inserting #d(t)=3# in the given equation we get

#3.5 cos ((pit)/6)+4.5=3#

#=>3.5 cos ((pit)/6)=3-4.5=-1.5#

#=>(pit)/6=2npi+-cos^-1(-1.5/3.5)#

#=>t=12n+-6/picos^-1(-1.5/3.5)" where " n in NN#

#=>t=12n+-3.84#

when n =0

#=>t=12n+3.84=3.84 hr#

when n =1

#=>t=12*1-3.84=8.16 hr->color(red)(8hr" " 9min " "36 sec)#

and

#=>t=12*1+3.84=8.16 hr->color(red)(15hr" " 50min " "24sec)#

Hence from above calculation as well as from graphical presentation of variation of depth of water with time it is obvious that during daytime the water level remains at least 3m deep for the period

#color(red)(8hr" " 9min " "36 sec) to color(red)(15hr" " 50min " "24sec) #
and this period is safe to dive off .

2

Answer:

Alternate solution for part (c)

Explanation:

For Solution posted by @dk_ch
The modelling equation becomes

#f(t)=2.5t+20+10sin((pit)/6)#

Using inbuilt graphics tool the plotted equation looks like

enter image source here
The maximum and minimum of the curve are located for #t=3.951# and #8.049# respectively. These values correspond to months of May and September.
As such the stock actually lost value from May to August.

--.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-..-

For Solution posted by @Cesareo R

#f(t)=2.25t+18.49+7.29sin((pit)/6)#, values rounded to two decimal places

#m = a_1 = 2.25439#, #b = a_2 = 18.49#, #A = a_3 = 7.29285#
enter image source here
Using inbuilt graphics tool the plotted equation looks like

The maximum and minimum of the curve are located for #t=4.204# and #7.796# respectively. These values correspond to months of May and September.
As such the stock actually lost value from May to August.

2

You have to start by finding a point that lies on this line. I think the simplest would be #(-6, -1)# (since we are in quadrant III, both the x and y axis have to have negative values).

We now draw an imaginary triangle, as shown in the following diagram.

enter image source here

We can now clearly see that our side opposite #theta# measures #1# unit and our side adjacent #theta# measures #6# units.

We must finally find the hypotenuse prior to determining the ratios. By pythagorean theorem:

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

#(-1)^2 + (-6)^2= c^2#

#1 + 36 = c^2#

#c^2 = 37#

#c = +-sqrt(37)#

However, the hypotenuse can never have a negative length, so we can only accept the positive solution. We can now find our ratios.

#sintheta = "opposite"/"hypotenuse" = -1/sqrt(37) = -sqrt(37)/37#

#csctheta = 1/sintheta = 1/("opposite"/"hypotenuse") = "hypotenuse"/"opposite" = sqrt(37)/(-1) = -sqrt(37)#

#costheta = "adjacent"/"hypotenuse" = -6/sqrt(37) = (-6sqrt(37))/37#

#sectheta = 1/costheta = 1/("adjacent"/"hypotenuse") = "hypotenuse"/"adjacent" = sqrt(37)/(-6) = -sqrt(37)/6#

#tantheta = "opposite"/"adjacent" = -1/(-6) = 1/6#

#cottheta = 1/tantheta = 1/("opposite"/"adjacent") = "adjacent"/"opposite" = -6/(-1) = 6#

Hopefully this helps!

1

Answer:

See below:

Explanation:

With a triangle with #sectheta=2#, let's first remember that:

#sectheta="hypotenuse"/"adjacent"=2/1#

We could run through the pythagorean theorem for the third side, but we can also remember that these two measurements are part of a 30-60-90 triangle, and so the third side is #sqrt3#. To prove it, let's go ahead and do Pythagorean Theorem:

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

#1^2+(sqrt3)^2=2^2#

#1+3=4#

This gives us the 6 trig ratios:

#sintheta=sqrt3/2#

#costheta=1/2#

#tantheta=(sqrt3/2)/(1/2)=(sqrt3/2)(2/1)=sqrt3#

#csctheta=2/sqrt3=(2sqrt3)/3#

#sectheta=2/1=2#

#cottheta=1/sqrt3=sqrt3/3#

The angle of the triangle we're looking at is #60=pi/3#, with the opposite being the middle length of #sqrt3#, the adjacent length of 1, and hypotenuse of 2.

freemathhelp.com

2

Answer:

The angle #-3.4°# lies in quadrant 4 (#Q_"IV"#).

Explanation:

When we talk about the quadrant an angle is in, we're really talking about the quadrant the terminal arm of the angle is in, when the angle is drawn with its initial arm along the #+x# axis (a.k.a. standard position).

To draw an angle, imagine a clock centered at the origin, with the minute hand pointing to the right (at the "3"). Then, rotate the hand counterclockwise (towards the "12") by the amount of the angle. Since the angle #-3.4°# is negative, we end up rotating clockwise (towards the "6") by 3.4 degrees.

Now, #-3.4°# is quite small compared to a full #360°# circle, so we will only rotate a little bit (not even to the "4" on our clock). So we have our terminal arm in the bottom-right quadrant.

Which quadrant is this? The quadrants are named in increasing order, starting with quadrant 1 (#Q_"I"#) in the top right, and moving counterclockwise to #Q_"II"# in the top left, etc. In this way, higher numbered quadrants contain bigger angles (until #360°#, of course).

enter image source here

So the angle #-3.4°# lies in #Q_"IV"#.

3

Answer:

#cos^-1 (0.6)= 53.1°#

Explanation:

Note that #cos^-1 # does not mean #1/cos# as we are used to in algebra.

#cos^-1# is the notation used for arc-cos.

#Cos 30° = 0.866 " "hArr" "cos^-1(0.866) = 30°#

In this case #cos^-1 (0.60)# is asking the question..

"Which angle has a #Cos# value of 0.60?"

The only way to determine this is with a calculator or tables.
Using a graph is possible, but not accurate enough.

Depending on which calculator you have, the following are possible key presses:

D.A.L. calculators:

shift #cos^-1 (0.6) =" larr# ( ) might not be needed on some models

The answer will be #53.1°#

Or #0.6" shift" cos^-1" "larr =# sign not needed, the answer #color(white)(...............................................)#appears immediately

The answer will be #53.1°#

Note that this is only the acute angle, the 1st quadrant angle.
In the 4th quadrant, #Cos 306.9° = 0.6#

3

Answer:

See below:

Explanation:

To help follow what's going on I'll put in red text things that have changed from the line before.

#cot^2theta/(sintheta+costheta)=(cos^2thetasintheta-cos^3theta)/(2sin^4theta-sin^2theta)#

I'll first use #cot^2theta=cos^2theta/sin^2theta# on the left side and factor the numerator and denominator on the right side. We can also see a #sintheta-costheta# term setting up in the right numerator and so I'll multiply through on the left using that term:

#((color(red)(cos^2theta/sin^2theta))/(sintheta+costheta))color(red)(((sintheta-costheta)/(sintheta-costheta)))=color(red)((cos^2theta(sintheta-costheta))/(sin^2theta(2sin^2theta-1))#

#cos^2theta/((sin^2theta)(sintheta+costheta))((sintheta-costheta)/(sintheta-costheta))=(cos^2theta(sintheta-costheta))/(sin^2theta(2sin^2theta-1)#

#(cos^2thetacolor(red)((sintheta-costheta)))/((color(red)(sin^2theta))(sintheta+costheta)color(red)((sintheta-costheta)))=(cos^2theta(sintheta-costheta))/(sin^2theta(2sin^2theta-1)#

#(cos^2theta(sintheta-costheta))/((sin^2theta)color(red)((sin^2theta-cos^2theta)))=(cos^2theta(sintheta-costheta))/(sin^2theta(2sin^2theta-1)#

We can now use #sin^2theta+cos^2theta=1 => cos^2theta=1-sin^2theta#

#(cos^2theta(sintheta-costheta))/((sin^2theta)(sin^2theta-color(red)((1-sin^2theta))))=(cos^2theta(sintheta-costheta))/(sin^2theta(2sin^2theta-1)#

#(cos^2theta(sintheta-costheta))/((sin^2theta)(sin^2thetacolor(red)(-1+sin^2theta)))=(cos^2theta(sintheta-costheta))/(sin^2theta(2sin^2theta-1)#

#(cos^2theta(sintheta-costheta))/((sin^2theta)color(red)((2sin^2theta-1)))=(cos^2theta(sintheta-costheta))/(sin^2theta(2sin^2theta-1)#

2

Answer:

#(i-3)/(5i+2)~~0.587*(cos(1.630)+i * sin(-1.630))#

Explanation:

#color(red)("~~~ Complex Conversion: Rectangular " harr " Tigonometric~~~")#
#color(white)("XX ")color(blue)(a+bi harr r * [cos(theta)+i * sin(theta)])#
#color(white)("XXX")color(blue)("where " r=sqrt(a^2+b^2))#
#color(white)("XXX")color(blue)("and "theta={("arctan"(b/a),"if "(a,bi) in "Q I or Q IV"),("arctan"(b/a)+pi,"if " (a,bi) in "Q II or Q III):})#

#color(red)("~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~")#

#color(red)("~~~ Trigonometric Division ~~~")#
#color(white)("XX ")color(blue)((cos(theta)+i * sin(theta))/(cos(phi)+i * sin(phi)))#

#color(white)("XXX")color(blue)(= [color(green)((cos(theta) * cos(phi) + sin(theta) * sin(phi))] +i * [color(magenta)(sin(theta) * cos(phi)-cos(theta) * sin(phi))]#

#color(red)("~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~")#

If #A=i-3=-3+1i#
then
#color(white)("XXX")r_A=sqrt((-3)^2+1^2)=sqrt(10)#
and
#color(white)("XXX")theta_A=arctan(-1/3)+pi~~2.819842099#

If #B=5i+2=2+5i#
then
#color(white)("XXX")r_B=sqrt(2^2+5^2)=sqrt(29)#
and
#color(white)("XXX")theta_B=arctan(5/2)~~1.19028995#

#A/B=(i-3)/(5+2i)#

#color(white)("XXX")=sqrt(10)/sqrt(29) * ([cos(theta_A)+isin(theta_A))/(cos(theta_B)+isin(theta_B)))#

#color(white)("XXX")=sqrt(10)/sqrt(29) *([color(green)(cos(theta_A)*cos(theta_B)+sin(theta_A) * sin(theta_B))+i[color(magenta)(sin(theta_A) * cos(theta_B) - cos(theta_A) * sin(theta_B))])#

Plugging in the values for #theta_A# and #theta_B# and using a calculator/spreadsheet:
#color(white)("XXX")~~-0.034482759+0.586206897i#

If we call this value #C#
then
#color(white)("XXX")r_C=sqrt((-0.03448...)^2+(0.5862...)^2)~~0.58722022#
and (since #C# is in Quadrant 2)
#color(white)("XXX")theta_C=arctan((0.5862...)/(-0.03448...))+pi~~1.62955215#

#A/B=C=r_C(cos(theta_C)+i * sin(theta_C))#