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3

Answer:

One does not use integration or rationalizing to prove this. One proves this by changing only 1 side of the equation, until it is identical to the other side.

Explanation:

Prove:

#(1-sin(2x)) /(1+sin(2x))=tan^2(45^@-x)#

The identity for #tan(A-B)=(tan(A)-tan(B))/(1+tan(A)tan(B))#, therefore, we may substitute #((tan(45^@)-tan(x))/(1 + tan(45^@)tan(x)))^2# for #tan^2(45^@-x)#:

#(1-sin(2x)) /(1+sin(2x))=((tan(45^@)-tan(x))/(1 + tan(45^@)tan(x)))^2#

Use the fact that #tan(45^@) = 1#:

#(1-sin(2x)) /(1+sin(2x))=((1-tan(x))/(1 + tan(x)))^2#

Write #tan(x)# as #sin(x)/cos(x)#:

#(1-sin(2x)) /(1+sin(2x))=((1-sin(x)/cos(x))/(1 + sin(x)/cos(x)))^2#

Multiply the fraction inside the square by 1 in the form of #(cos(x)/cos(x))#:

#(1-sin(2x)) /(1+sin(2x))=((cos(x)/cos(x))(1-sin(x)/cos(x))/(1 + sin(x)/cos(x)))^2#

Perform the multiplication:

#(1-sin(2x)) /(1+sin(2x))=((cos(x)-cos(x)sin(x)/cos(x))/(cos(x) + cos(x)sin(x)/cos(x)))^2#

Please observe the cancellation of the embedded denominators:

#(1-sin(2x)) /(1+sin(2x))=((cos(x)-cancel(cos(x))sin(x)/cancel(cos(x)))/(cos(x) + cancel(cos(x))sin(x)/cancel(cos(x))))^2#

Write the equation without the cancelled factors:

#(1-sin(2x)) /(1+sin(2x))=((cos(x)-sin(x))/(cos(x) + sin(x)))^2#

Perform the multiplication implied by the square:

#(1-sin(2x)) /(1+sin(2x))=(cos^2(x)-2sin(x)cos(x)+sin^2(x))/(cos^2(x) +2sin(x)cos(x)+ sin^2(x))#

Use the identity #sin^2(x) + cos^2(x) = 1#:

#(1-sin(2x)) /(1+sin(2x))=(1-2sin(x)cos(x))/(1 +2sin(x)cos(x))#

Use the identity #sin(2x) = 2sin(x)cos(x)#:

#(1-sin(2x)) /(1+sin(2x))=(1-sin(2x))/(1 +sin(2x))# Q.E.D.

1

Answer:

Please see below.

Explanation:

To sketch the angle in standard position,

One side of the angle is the positive #x# axis (the right side of the horizontal axis). That is the initial side.

The terminal side has one end at the origin (the point #(0,0)#, also the intersection of the two axes) and goes through the point #(-3,1)#. So locate the point #(-3,1)#. Starting at the origin and count #3# to the left and up #1#. That will get you to the point #(-3,1)#. Put a dot there.
Now draw a line from the origin through the point #(-3,1)#. Your sketch should look a lot like this:

enter image source here

If we knew how the angle was made (which direction and how many times around the circle), we would show that also.

Give the angle a name. I will use #theta# (that is the Greek letter "theta")

Memorize this

If the point #(a,b)# lies on the terminal side of an angle in standard position, then let #r = sqrt(a^2+b^2)#

The angle has sine #b/r# and it has cosine #a/r#

For this question

We have #(a,b) = (-3,1)#, so

#r = sqrt((-3)^2+(1)^2) = sqrt(9+1) = sqrt10#.

So the sine of #theta# is

#sin(theta) = 1/sqrt10#

Many trigonometry teachers will insist that you write your answer with a rational number in the denominator. If this is something you have to do, multiply the answer by #sqrt10/sqrt10# to look like this:

#1/sqrt10 * sqrt10/sqrt10 = (1*sqrt10)/(sqrt10 * sqrt10) = sqrt10/10#

So we should answer

#sin(theta) = sqrt10/10#

The cosine of #theta# is #a/r# so we have

#cos(theta) = (-3)/sqrt10 = (-3)/sqrt10 * sqrt10/sqrt10 = -(3sqrt10)/10#

Our answer is
#cos(theta) = -(3sqrt10)/10#

1

Answer:

#"The Solution Set is "phi.#

Explanation:

#costheta/(1+sintheta)+sintheta/(1+costheta)=2sectheta.#

#:. {costheta(1+costheta)+sintheta(1+sintheta)}/{(1+sintheta)(1+costheta)}=2/costheta.#

#:.(costheta+cos^2theta+sintheta+sin^2theta)/{(1+sintheta)(1+costheta)}=2/costheta.#

#:. (1+costheta+sintheta)costheta=2(1+sintheta)(1+costheta).#

#:. costheta+cos^2theta+sinthetacostheta=2+2sintheta+2costheta+2sinthetacostheta.#

#:. cos^2theta=2+2sintheta+costheta+sinthetacostheta.#

#:.1-sin^2theta=2(1+sintheta)+costheta(1+sintheta).#

#:.(1+sintheta)(1-sintheta)=(1+sintheta)(2+costheta).#

#:.(1+sintheta){(1-sintheta)-(2+costheta)}=0.#

#:.(1+sintheta)(-1-sintheta-costheta)=0.#

#:. (1+sintheta)=0, or, costheta+sintheta=-1.#

#"But, "because (1+sintheta)=0,# makes the given eqn. meaningless,

#:. costheta+sintheta=-1,.#

Remembering that, #0^@ < theta < 90^@,# we have,

# 0 < sintheta < 1, and, 0 < costheta <1,#

and, adding these give,

# 0 < sintheta+costheta < 2.#

Obviously, no such #theta in (0^@,90^@)# exists.

Hence, #"The Solution Set is "phi.#

Enjoy Maths.!

1

Answer:

#75.8# miles

Explanation:

Draw a sketch first and fill in what you know.

Start with a rectangle and draw in one diagonal.
The jetliner is at a vertical height of #35,000# feet.
This is the breadth of the rectangle.

The jetliner is at the top of the rectangle where the diagonal starts.

The diagonal is the direct line from the jetliner to the coast ahead and below, and is the hypotenuse of a right-angled triangle.

The angle of depression is the angle between the horizontal length and the diagonal. It can be shown in two places, because the angle of elevation is equal to the angle of depressions, because they are alternate angles on parallel lines.

enter image source here

These angles are always measured FROM THE HORIZONTAL!

We need to find the length of the rectangle.

#("height (opp)")/("length" (adj)) = tan 5°#

#l/35000 = 1/(tan5°)" "larr# invert to have #l# on top

#l = 35000/(tan5°)=35000/0.08748866#

#l = 400,052 feet#

Now to convert to miles:

#1 yard = 3 feet" "larr div 3# to get yards

#1 " mile " = 1760 yards" "larr div 1760# to get miles.

#400,052 div 3 div 1760 = 75.77 " miles"#

1

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]}

2

Answer:

#sec(u+v)=24/(sqrt62-sqrt60)# and #tan(u-v)=(sqrt434-sqrt10)/(sqrt62+sqrt70)#

Explanation:

As #sinu=-sqrt5/6# and #u# is in #Q4#, #cosu# is positive and it is given by #cosu=sqrt(1-(-sqrt5/6)^2)=sqrt(1-5/36)=sqrt31/6#

Further, as #cosv=sqrt2/4# and #v# is in #Q4#, #sinu# is neegative and it is given by #sinv=-sqrt(1-(sqrt2/4)^2)=-sqrt(1-2/16)=-sqrt14/4#

Hence #cos(u+v)=cosucosv-sinusinv#

= #sqrt31/6xxsqrt2/4-(-sqrt5/6)xx(-sqrt14/4)=(sqrt62-sqrt60)/24#

and #sec(u+v)=24/(sqrt62-sqrt60)#

#cos(u-v)=cosucosv+sinusinv#

= #sqrt31/6xxsqrt2/4+(-sqrt5/6)xx(-sqrt14/4)=(sqrt62+sqrt70)/24#

and #sin(u-v)=sinucosv-cosusinv#

= #(-sqrt5/6)xxsqrt2/4-sqrt31/6xx(-sqrt14/4)=(-sqrt10+sqrt434)/24#

= #(sqrt434-sqrt10)/24#

Hence, #tan(u-v)=sin(u-v)/cos(u-v)=(sqrt434-sqrt10)/(sqrt62+sqrt70)#

3

Answer:

# B-C=30^@.#

In fact, #B=75^@, and, C=45^@.#

Explanation:

We use the following property known as Napier's Rule (NR) :

#" NR : In "DeltaABC, tan((B-C)/2)=(b-c)/(b+c)cot(A/2).#

It can be proved very easily using the Sine Rule for #Delta.#

In our Problem, we have, #b/c=(sqrt3+1)/2.#

By the Componedo-Dividendo, we get,

#(b-c)/(b+c)=(sqrt3+1-2)/(sqrt3+1+2)=(sqrt3-1)/(sqrt3+3).#

#:.," by, NR, "tan((B-C)/2)=(sqrt3-1)/(sqrt3+3)*cot(60^@/2),#

#=(sqrt3-1)/(sqrt3+3)*sqrt3=(sqrt3-1)/(sqrt3(1+sqrt3))*sqrt3,#

#=(sqrt3-1)/(sqrt3+1),#

#:. tan((B-C)/2)=tan15^@.#

#:. B-C=30^@.#

Foot Note : In fact, #B, and C# can also be found individually.

#A=60^@ rArr B+C=180^@-60^@=120^@...........(1).#

# B-C=30^@...............................................................(2).#

Solving, #(1) and (2), B=75^@, and, C=45^@.#

Enjoy Maths.!

2

Answer:

#sin 45^@/(cos 30^@+sin 60^@) = sqrt(6)/6#

Explanation:

This is essentially a guess, but note that all of these angles have exact algebraic expressions for the their trigonometric values. So the question probably wants you to express this quotient as an exact algebraic expression.

Since all of the angles are positive but less than #90^@#, they can be treated directly as angles in right angled triangles.

To find the value of #sin 45^@#, consider a right angled triangle with sides #1, 1, sqrt(2)#, being half of a unit square ...

enter image source here

Then:

#sin 45^@ = "opposite"/"hypotenuse" = 1/sqrt(2) = sqrt(2)/2#

For the other two trigonometric values, consider a right angled triangle with sides #1, sqrt(3), 2#, which is one half of an equilateral triangle ...

enter image source here

We have:

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

#sin 60^@ = "opposite"/"hypotenuse" = sqrt(3)/2#

#color(white)()#
Having obtained our trigonometric values, here's the algebra:

#sin 45^@/(cos 30^@+sin 60^@) = (sqrt(2)/2)/(sqrt(3)/2+sqrt(3)/2)#

#color(white)(sin 45^@/(cos 30^@+sin 60^@)) = sqrt(2)/(2sqrt(3))#

#color(white)(sin 45^@/(cos 30^@+sin 60^@)) = (sqrt(2)sqrt(3))/6#

#color(white)(sin 45^@/(cos 30^@+sin 60^@)) = sqrt(6)/6#

1

Answer:

In order to answer this I have assumed a vertical shift of #+7#

#color(red)(3cos(2theta)+7)#

Explanation:

The standard cos function #color(green)(cos(gamma))# has a period of #2pi#

If we want a period of #pi# we need to replace #gamma# with something that will cover the domain "twice as fast" e.g. #2theta#.
That is #color(magenta)(cos(2theta))# will have a period of #pi#.

To get an amplitude of #3# we need to multiply all values in the Range generated by #color(magenta)(cos(2theta))# by #color(brown)3# giving
#color(white)("XXX")color(brown)(3cos(2theta))#

There is to be no horizontal shift, so the argument for #cos# will not be modified by any further addition/subtraction.

In order to achieve the vertical shift (that I assumed would be #color(red)(+7)# [substitute your own value]) we will need to add #color(red)7# to all values in our modified range:
#color(white)("XXX")color(red)(3 cos(2theta) +7)#

enter image source here

2

#cos (x/2) -sin(x/2) =1-tan (x/2)#

#=>cos (x/2) -sin(x/2) =1-sin (x/2)/cos(x/2)#

#=>cos (x/2) -sin(x/2) =(cos(x/2)-sin (x/2))/cos(x/2)#

#=>(cos (x/2) -sin(x/2)) -(cos(x/2)-sin (x/2))/cos(x/2)=0#

#=>(cos (x/2) -sin(x/2))(1 -1/cos(x/2))=0#

When

#=>(cos (x/2) -sin(x/2)) =0#

#=>tan(x/2)=0#

#=>x/2=npi" where "n inZZ#

#=>x=2npi" where "n inZZ#

when

#(1 -1/cos(x/2))=0#

#=>cos(x/2)=1#

#=>x/2=2npi" where "n inZZ#

#=>x=4npi" where "n inZZ#

Hence combining we have

#x=2npi" where "n inZZ#