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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

Answer:

#sin(pi/4) = sqrt2/2#
#cos(pi/4) = sqrt2/2#
#tan(pi/4) = 1#

#csc(pi/4) = sqrt2#
#sec(pi/4) = sqrt2#
#cot(pi/4) = 1#

Explanation:

#pi/4# radians is the same as 90 degrees. So, first draw a right triangle with an angle of #pi/4#:
enter image source here

This creates a 45-45-90 triangle, also known as a right isosceles triangle. This is a very special triangle, and we know that both of its legs will be the same length, and the hypotenuse will be the length of one of the legs times #sqrt2#.

To make things easier for us, we will say the legs are both 1, and the hypotenuse, therefore, is #sqrt2#.

Now that we have a triangle with our 3 sides labeled (and our angle in red), we can find all 6 trig functions:

#sin(pi/4) = "opposite"/"hypotenuse" = 1/sqrt2 = sqrt2/2#

#cos(pi/4) = "adjacent"/"hypotenuse" = 1/sqrt2 = sqrt2/2#

#tan(pi/4) = "opposite"/"adjacent" = 1/1 = 1#

Now, the other three functions are just the inverses of the first three. Cosecant is the inverse of sine, secant is the inverse of cosine, and cotangent is the inverse of tangent. To see what I mean, look at this:

#csc(pi/4) = "hypotenuse"/"opposite" = sqrt2/1 = sqrt2#

#sec(pi/4) = "hypotenuse"/"adjacent" = sqrt2/1 = sqrt2#

#cot(pi/4) = "adjacent"/"opposite" = 1/1 = 1#

So there you have it! All 6 trig functions can be evaluated this way just by drawing a triangle and knowing its side lengths.

Final Answer

Answer:

A few thoughts...

Explanation:

#phi = 1/2+sqrt(5)/2 ~~ 1.6180339887# is known as the Golden Ratio.

It was known and studied by Euclid (approx 3rd or 4th century BCE), basically for many geometric properties...

It has many interesting properties, of which here are a few...

The Fibonacci sequence can be defined recursively as:

#F_0 = 0#

#F_1 = 1#

#F_(n+2) = F_n + F_(n+1)#

It starts:

#0, 1, 1, 2, 3, 5, 8, 13, 21, 34, 55, 89, 144, 233, 377, 610, 987,...#

The ratio between successive terms tends to #phi#. That is:

#lim_(n->oo) F_(n+1)/F_n = phi#

In fact the general term of the Fibonacci sequence is given by the formula:

#F_n = (phi^n - (-phi)^(-n))/sqrt(5)#

A rectangle with sides in ratio #phi:1# is called a Golden Rectangle. If a square of maximal size is removed from one end of a golden rectangle then the remaining rectangle is a golden rectangle.

This is related to both the limiting ratio of the Fibonacci sequence and the fact that:

#phi = [1;bar(1)] = 1+1/(1+1/(1+1/(1+1/(1+1/(1+...)))))#

which is the most slowly converging standard continued fraction.

If you place three golden rectangles symmetrically perpendicular to one another in three dimensional space, then the twelve corners form the vertices of a regular icosahedron. Hence we can calculate the surface area and volume of a regular icosahedron of given radius. See https://socratic.org/s/aFZyTQfn

An isosceles triangle with sides in ratio #phi:phi:1# has base angles #(2pi)/5# and apex angle #pi/5#. This allows us to calculate exact algebraic formulae for #sin(pi/10)#, #cos(pi/10)# and ultimately for any multiple of #pi/60# (#3^@#). See https://socratic.org/s/aFZztx8s

#Sec^2(π/16)+sec^2(3π/16)+sec^2(5π/16)+sec^2(7π/16)#

#=Sec^2(π/16)+sec^2((3π)/16)+sec^2(pi/2-(3π)/16)+sec^2(pi/2-π/16)#

#=Sec^2(π/16)+sec^2((3π)/16)+csc^2((3π)/16)+csc^2(π/16)#

#=Sec^2(π/16)+csc^2(π/16)+sec^2((3π)/16)+csc^2((3π)/16)#

#=1/cos^2(π/16)+1/sin^2(π/16)+1/cos^2((3π)/16)+1/sin^2((3π)/16)#

#=(sin^2(π/16)+cos^2(π/16))/(cos^2(π/16)sin^2(π/16))+(sin^2((3π)/16)+cos^2((3π)/16))/(cos^2((3π)/16)sin^2((3π)/16))#

#=1/(cos^2(π/16)sin^2(π/16))+1/(cos^2((3π)/16)sin^2((3π)/16))#

#=4/(2cos(π/16)sin(π/16))^2+4/(2cos((3π)/16)sin((3π)/16))^2#

#=4/sin^2(π/8)+4/sin^2((3π)/8)#

#=8/(2sin^2(π/8))+8/(2sin^2((3π)/8))#

#=8/(1-cos(pi/4))+8/(1-cos((3π)/4))#

#=8/(1-cos(pi/4))+8/(1-cos(pi-π/4))#

#=8/(1-cos(pi/4))+8/(1+cos(π/4))#

#=8/(1-1/sqrt2)+8/(1+1/sqrt2)#

#=8sqrt2(1/(sqrt2-1)+1/(sqrt2+1))#

#=8sqrt2((sqrt2+1+sqrt2-1)/((sqrt2-1)(sqrt2+1)))=32#

Answer:

graph{3sin(2(x-1)) [-10, 10, -5, 5]}

Explanation:

If we consider #Asin[B(x+C)]#, the first term A is increasing the amplitude of the sin graph. So if we make A = 3 we would get the following graph.

graph{3sinx [-10, 10, -5, 5]}

We will look at C next, this is the movement of the graph left or right, where a negative C value moves the graph to the right. So we move the whole graph 1 to the right in this case. #3sin(1(x-1))# give the following graph.

graph{3sin(x-1) [-10, 10, -5, 5]}

Finally B is stretching the graph parallel to the x axis by a factor of #1/B xx 2Pi#

So in your case B = 2, so #1/2 xx 2Pi = Pi# radians. This gives us the new period for your graph, this means a complete cycle occurs every #Pi# rads instead of every #2Pi# rads.

Then graphing this: #3sin(2(x-1))#

graph{3sin(2(x-1)) [-10, 10, -5, 5]}

Answer:

Graph #y=cos(x)# and shift everything to the left by #pi/6#

Desmos

Explanation:

We know that #sin# and #cos# has a period of #2pi#. That is to say that it repeats itself every #2pi# units.
I would assume you know how to graph a #f(x)=cos(x)# functions, if not, it should look like this:

Desmos

Now, you need to graph #f(x)=cos(x+pi/6)#.

Imagine you have a function #f(x)# and another function #g(x)=f(x+1)#.

What this means is that for any point #(x, y)# on the graph #g(x)#, it will take #x+1# units for #f(x)# to reach that same #y# value.
That is what this #g(x)=f(x+1)# is saying.

This means that all points on #g(x)# is occurring 1 unit earlier than #f(x)# so we shift #f(x)# to the left by 1 unit to obtain #g(x)#.

To generalize:
If #g(x)=f(x+n)# we shift #f(x) #n# units to the **left** to get #g(x)#. If #g(x)=f(x-n)# we shift #f(x) #n# units to the right to get #g(x)#.

Now, we can apply it to this question:

We have #f(x)=cos(x+pi/6)# which is basically saying we should shift #cos(x)# to the left by #pi/6# units.

Desmos

The blue curve is your #y=cos(x+pi/6)#
The red curve is your #y=cos(x)#

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