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1

Answer:

Please see the explanation.

Explanation:

The slope, m, of the line is:

#m = (3- (-1))/(-2-4)#

#m = 4/-6#

#m = -2/3#

Use the point-slope form of the equation of a line:

#y = m(x-x_1)+y_1#

#y = -2/3(x-4)-1#

Here is a graph of that line:

Desmos.com

Multiply both sides by 3:

#3y = -2(x-4)-1#

Distribute the -2:

#3y = -2x+8-1#

Add 2x to both sides:

#2x+3y = 7#

Substitute #rcos(theta)# for x and #rsin(theta)# for y:

#2rcos(theta)+3rsin(theta) = 7#

Factor out r:

#r(2cos(theta)+3sin(theta)) = 7#

Divide both sides by #(2cos(theta)+3sin(theta))#

#r = 7/(2cos(theta)+3sin(theta))#

Here is a graph of that equation:

Desmos.com

3

Answer:

#cos(-870^o)=cos(-150^o)=-sqrt3/2#

Explanation:

Let's first notice that we're looking at #-870^o#. The negative sign means we'll rotate in a clockwise direction (and so go down through Q4 first).

There are #360^o# in a full circle, so when we go around once fully, we've gone 360 of the 870:

#-870^o +360^o=-610^o#

We need to go around once more time:

#-610^o +360^o=-150^o#

Now let's go around a third time, but not all the way. We'll go through Q4 (to do that takes #-90^o#) and end up in Q3 but not get to Q2 (that would take #-180^o#). Here's a visual of where we are:

www2.fiu.edu

Do you see the point for #-150^o#? It's in the same place as #+210^o#

Now that we know where we are, we can get down to finding the cosine!

The reference angle between #-150^o# and #180^o# is #30^o#. This means that the triangle we have is a 30-60-90 triangle and the lengths of the sides are #1, sqrt3, 2# for the short side (which is the opposite - the side that drops from the #x#-axis), the medium side (which is the adjacent - that's the #x#-axis), and the long side (which is the hypotenuse (the side that extends from the origin to the red dot on the circle)), respectively.

#cos="adj"/"hyp"=sqrt3/2#

One last thing to address and we're done - because the adjacent is along the negative #x#-axis, we make the cosine negative. And so:

#cos(-870^o)=cos(-150^o)=-sqrt3/2#

2

Answer:

See the Proof in the Explanation.

Explanation:

Given that, #sinx+siny=a rArr 2sin((x+y)/2)cos((x-y)/2)=a...(1).#

Similarly, #cosx+cosy=b rArr 2cos((x+y)/2)cos((x-y)/2)=b...(2).#

Hence, Squaring and Adding #(1) & (2),# we get,

#4cos^2((x-y)/2){sin^2((x+y)/2)+cos^2((x+y)/2)}=a^2+b^2,# i.e.,

#4cos^2((x-y)/2)=a^2+b^2, or, #

# 4{1-sin^2((x-y)/2)}=a^2+b^2,#

# rArr 4-4sin^2((x-y)/2)=a^2+b^2,#

# rArr 4sin^2((x-y)/2)=(4-a^2-b^2),# giving,

# sin((x-y)/2)=+-1/2(4-a^2-b^2)^(1/2).#

1

Answer:

#xapprox0.251#

Explanation:

#cosxcos3x=0.707#

#f(x)=cosxcos3x-0.707=0#

Use Newton method to find a zero of #f#

#x_(n+1)=x_n-(f(x_n))/(f'(x_n))#

#f'(x)=-2sin2x-2sin4x#

#0.707=cos(1/4pi)#, so a good #x_0# would be #(1/4pi)/3=0.262#

#x_1=0.262-(cos(0.262)cos(3(0.262))-0.707)/(2sin(2(0.262)-2sin(4(0.262))=0.253#

#x_2=0.251#

#x_3=0.251#

#x_4=0.251#

The iterations approach #0.251# to #0.251# is a solution to #cosxcos3x=0.707#

1

Answer:

#~~8.98^o#

Explanation:

So first up we need to understand the question completely. A helicopter starts at #A(6,9,3)#, and after 10 mins, it is at #B(3,10,2.5)#. We need to find the angle between the velocity vector (which would be #vec (AB)#) and the #XY# plane.

So first we need to find the velocity vector which would be #vec (AB)#:

To find the vector between two points #M(x_1,y_1,z_1)# and #N(x_2,y_2,z_2)#, use:

#vec(MN)=((x_2-x_1),(y_2-y_1),(z_2-z_1))#

So from our points, we have:

#A(6,9,3)# and #B(3,10,2.5)#

#=>vec(AB)=((3-6),(10-9),(2.5-3))#

#=>vec(AB)=((-3),(1),(-0.5))#

This vector is the velocity vector of the helicopter in terms of km per 10 minutes (as it travels from #A# to #B# in 10 minutes)

This vector would look like:

Geogebra.com

(the red dot is A and the Blue dot is B)

So we need to find the angle that #vec(AB)# makes with the XY plane.

To do this find the angle between #vec(AB)# and the vector on the XY plane which is the same as #vec(AB)#, but the #z# value is zero. Let's call this vector #a#.

#vec(AB)=((-3),(1),(-0.5))#

#=>veca=((-3),(1),(0))#

To find the angle between the two vectors #vecb# and #vecb#, use:

#costheta=(|vecb*vecc|)/(|vecb||vecc|)#

In this equation, the numerator is the modulus of the dot product and the denominator is the length of each vector.

So when we are finding the angle between #vec(AB)# and #veca#, we would use:

#costheta=(|vec(AB)*veca|)/(|vec(AB)||veca|)#

#=>costheta=(|((-3),(1),(-0.5))*((-3),(1),(0))|)/(|((-3),(1),(-0.5))||((-3),(1),(0))|)#

#=>costheta=(-3*-3+1*1+-0.5*0)/(sqrt((-3)^2+1^2+(-0.5)^2)sqrt((-3)^2+1^2+(0)^2))#

#=>costheta=(9+1+0)/(sqrt(9+1+0.25)sqrt(9+1+0))#

#=>costheta=10/(sqrt10.25sqrt10)#

#=>costheta=10/(sqrt(10.25*10)#

#=>costheta=10/sqrt102.5#

#=>theta=cos^-1(10/sqrt102.5)#

#=>theta~~8.98# (to two s.f.)

2

#cos^3((2pi)/7)+cos^3((4pi)/7)+cos^3((8pi)/7)#

Using formula

#cos^3A=1/4(cos3A+3cosA)# the given expreession becomes

#=1/4(cos((6pi)/7)+3cos((2pi)/7)+cos((12pi)/7)+3cos((4pi)/7)+cos((24pi)/7)+3cos((8pi)/7))#

#=1/4(cos(pi-pi/7)+3cos((2pi)/7)+cos(2pi-(2pi)/7)+3cos(pi-(3pi)/7)+cos(3pi+(3pi)/7)+3cos(pi+pi/7))#

#=1/4(-cos(pi/7)+3cos((2pi)/7)+cos((2pi)/7)-3cos((3pi)/7)-cos((3pi)/7)-3cos(pi/7))#

#=1/4(4cos((2pi)/7)-4cos((3pi)/7)-4cos(pi/7))#

#=cos((2pi)/7)-cos((3pi)/7)-cos(pi/7)#

#=1/(2sin(pi/7))(2sin(pi/7)cos((2pi)/7)-2sin(pi/7)cos((3pi)/7)-2sin(pi/7)cos(pi/7))#

#=1/(2sin(pi/7))(sin((3pi)/7)-sin(pi/7)-sin((4pi)/7)+sin((2pi)/7)-sin((2pi)/7))#

#=1/(2sin(pi/7))(sin((3pi)/7)-sin(pi/7)-sin(pi-(3pi)/7))#

#=1/(2sin(pi/7))(sin((3pi)/7)-sin(pi/7)-sin((3pi)/7))#

#=1/(2sin(pi/7))xx(-sin(pi/7))#

#=-1/2#

3

Q-1

Let #x=4sin9cos9=2sin18=2xx(sqrt5-1)/4=1/2(sqrt5-1)#

Again

#t_1=4sin63cos63=2sin126=2sin(180-54)=2sin54=1/2(sqrt5+1)#

So #x xxt_1=1/2(sqrt5-1)xx1/2(sqrt5+1)=1/4xx4=1#

Otherwise without using exact values

#x*t_1=4sin9cos9*4sin63cos63#

#=>x*t_1=2sin18*2sin126#

#=>x*t_1=4sin18*sin(180-54)#

#=>x*t_1=2/cos18*2sin18*cos18*sin54#

#=>x*t_1=1/cos18*2sin36*sin54#

#=>x*t_1=(cos(54-36)-cos(54+36))/cos18#

#=>x*t_1=(cos18-cos90)/cos18#

#=>x*t_1=(cos18-0)/cos18=1#

Hence #x=1/t_1#

Now

#log_(t_1)4sin9cos9#

#=log_(t_1)x#

#=log_(t_1)(1/t_1)#

#=log_(t_1)(t_1)^(-1)#

#=-log_(t_1)(t_1)=-1#

Q-2

#l=((cot^2xcos^2x)/(cot^2x-cos^2x))^2#

#=>l=((cot^2xcos^2x)/(1/tan^2x-1/sec^2x))^2#

#=>l=((cot^2x*tan^2x*cos^2x*sec^2x)/(sec^2x-tan^2x))^2#

#=>l=((1*1)/1)^2=1#

Now

#m=a^(log_(sqrta)abs(2cos(y/2))#

#=>m=a^(log_(sqrta)abs(2cos((4pi)/2))#

#=>m=a^(log_(sqrta)abs(2cos(2pi))#

#=>m=a^(log_(sqrta)2#

#=>m=((sqrta)^2)^(log_(sqrta)2#

#=>m=(sqrta)^(2log_(sqrta)2#

#=>m=(sqrta)^(log_(sqrta)2^2)=4#

So

#l^2+m^2=1^2+4^2=17#

5

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

2

As per problem the movement of the London Eye is periodic one and its height at #t# th instant #h(t)# is given by the following equation

#h(t)=a+bcos(ct)......[1]#, where #h(t)# in meter and t in sec.

Here #c# should be the angular velocity associated with the periodic motion. Again it is also given that time taken by the London Eye to complete one cycle is 3o min. This means time period #T=30min#

Hence #"Angular velocity"=c=omega=(2pi)/T=(2pi)/30=pi/15# rad/min.

So the equation[1] takes the following form

#h(t)=a+bcos((2pi)/30t)......[2]#

Now at #t=0# the London Eye is at the bottom i.e.#h(0)=0#

Applying this condition on [2] we get

#h(0)=a+bcos((2pi)/30xx0)#

#0=a+b=> a+b=0.....[3]#

Since at #t=0# it is at the bottom and it completes the cycle returning at the bottom at #t=30# min,
then we can say that at #t=15# min it will reach at maximum height 135 m i.e. #h(15)=135#m
So we can write

#h(15)=135=a+bcos((2pi)/30xx15)#

#=>a+bcos(pi)=135#

#=>a-b=135.....[4]#

Now adding [3] and [4] we have #2a=135=>a=67.5# m

Subtracting [4] from [3] we get #2b=-135=>b=-67.5#m

So finally the given equation takes the following form

#color(red)(h(t)=67.5-67.5cos((2pi)/30t))#

The variation of height with time can be represented by following graph.

drawn

1

Answer:

#tan 15^@ = 2 - sqrt(3)#

Explanation:

Use:

#sin 30^@ = 1/2#

#cos 30^@ = sqrt(3)/2#

#sin 45^@ = cos 45^@ = sqrt(2)/2#

#sin(alpha-beta) = sin alpha cos beta - sin beta cos alpha#

#cos(alpha-beta) = cos alpha cos beta + sin alpha sin beta#

#tan theta = sin theta / cos theta#

Then:

#sin 15^@ = sin (45^@ - 30^@)#

#color(white)(sin 15^@) = sin 45^@ cos 30^@ - sin 30^@ cos 45^@#

#color(white)(sin 15^@) = sqrt(2)/2 sqrt(3)/2 - 1/2 sqrt(2)/2#

#color(white)(sin 15^@) = sqrt(2)/4(sqrt(3)-1)#

#color(white)()#

#cos 15^@ = cos (45^@ - 30^@)#

#color(white)(cos 15^@) = cos 45^@ cos 30^@ + sin 45^@ sin 30^@#

#color(white)(cos 15^@) = sqrt(2)/2 sqrt(3)/2 + sqrt(2)/2 1/2#

#color(white)(cos 15^@) = sqrt(2)/4(sqrt(3)+1)#

#color(white)()#

#tan 15^@ = sin 15^@ / cos 15^@#

#color(white)(tan 15^@) = (color(red)(cancel(color(black)(sqrt(2)/4)))(sqrt(3)-1)) / (color(red)(cancel(color(black)(sqrt(2)/4)))(sqrt(3)+1))#

#color(white)(tan 15^@) = (sqrt(3)-1) / (sqrt(3)+1)#

#color(white)(tan 15^@) = ((sqrt(3)-1)^2) / ((sqrt(3)+1)(sqrt(3)-1))#

#color(white)(tan 15^@) = (3-2sqrt(3)+1) / (3-1)#

#color(white)(tan 15^@) = 2 - sqrt(3)#

#color(white)()#
Footnote

We can see the original values of #sin 30^@#, etc used above by considering the right angled triangles:

enter image source here

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