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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#

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))#

1

Answer:

#cos(x+y)=(6sqrt2-5)/(3sqrt34)#

Explanation:

Use the cosine angle addition formula:

#cos(x+y)=cosxcosy-sinxsiny#

We need to determine #sinx#, #cosx#, and #cosy# from whatever we have.

Determining #mathbf(sinx,cosx: )#

#tan^2x+1=sec^2x" "" "" "#use #tanx=5/3#

#25/9+1=sec^2x#

#secx=sqrt(34/9)=sqrt34/3#

Then:

#color(blue)(cosx=1/secx=3/sqrt34#

Now using:

#sin^2x+cos^2x=1" "" "" "#where #cosx=3/sqrt34#

#sin^2x+9/34=1#

#color(blue)(sinx=sqrt(25/34)=5/sqrt34#

Determining #mathbf(cosy: )#

#sin^2y+cos^2y=1" "" "" "#use #siny=1/3#

#1/9+cos^2y=1#

#color(blue)(cosy=sqrt(8/9)=(2sqrt2)/3#

Returning to #mathbf(cos(x+y): )#

#cos(x+y)=cosxcosy-sinxsiny#

#color(white)(cos(x+y))=3/sqrt34((2sqrt2)/3)-5/sqrt34(1/3)#

#color(white)(cos(x+y))=(6sqrt2-5)/(3sqrt34)#

Note these are all assuming that #x# and #y# are in the first quadrant (everything is positive).

2

Answer:

The explanation is written with the assumption that you meant #theta# for the argument of the cosine function.

Explanation:

Use the distributive property :

#2r - rcos(theta) = 2#

#2r = rcos(theta) + 2#

Substitute x for #rcos(theta)# and #sqrt(x^2 + y^2)# for r:

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

Square both sides:

#4(x^2 + y^2) = x^2 + 4x + 4#

#4x^2 + 4y^2 = x^2 + 4x + 4#

#3x^2 - 4x + 4y^2 = 4#

Add #3h^2# to both sides:

#3x^2 - 4x + 3h^2 + 4y^2 = 3h^2+ 4#

Remove a factor of the 3 from the first 3 terms:

#3(x^2 - 4/3x + h^2) + 4y^2 = 3h^2+ 4#

Use the middle in the right side of the pattern #(x - h)^2 = x^2 - 2hx + h^2# and middle term of the equation to find the value of h:

#-2hx = -4/3x#

#h = 2/3#

Substitute the left side of the pattern into the equation:

#3(x - h)^2 + 4y^2 = 3h^2+ 4#

Substitute #2/3# for h and insert a -0 into the y term:

#3(x- 2/3)^2 + 4(y-0)^2 = 3(2/3)^2+ 4#

#3(x- 2/3)^2 + 4(y-0)^2 = 16/3#

Divide both sides by #16/3#

#3(x- 2/3)^2/(16/3) + 4(y-0)^2/(16/3) = 1#

#(x- 2/3)^2/(16/9) + (y-0)^2/(16/12) = 1#

Write the denominators as squares:

#(x- 2/3)^2/(4/3)^2 + (y-0)^2/(4/sqrt(12))^2 = 1#

The is standard Cartesian form of the equation of an ellipse with a center at #(2/3,0)#; its semi-major axis is #4/3# units long and is parallel to the x axis and its semi-minor axis is #4/sqrt(12)# units long and is parallel to the y axis.

2

Answer:

#cos(tan^-1(2)+tan^-1(3))=-1/sqrt2#

Explanation:

First use the cosine angle-addition formula:

#cos(a+b)=cos(a)cos(b)-sin(a)sin(b)#

Then the original expression equals:

#=cos(tan^-1(2))cos(tan^-1(3))-sin(tan^-1(2))sin(tan^-1(3))#

Note that if #theta=tan^-1(2)#, then #tan(theta)#. That is, a right triangle with angle #theta# has #tan(theta)=2#, which is the triangle with the leg opposite #theta# being #2# and the leg adjacent to #theta# being #1#. The Pythagorean theorem tells us that the hypotenuse of this triangle is #sqrt5#.

So, when #theta=tan^-1(2)#, we see that:

#cos(tan^-1(2))=cos(theta)="adjacent"/"hypotenuse"=1/sqrt5#

#sin(tan^-1(2))=sin(theta)="opposite"/"hypotenuse"=2/sqrt5#

Now let angle #phi# be defined by #phi=tan^-1(3)#, such that #tan(phi)=3#. This is the right triangle with:

#{("opposite"=3),("adjacent"=1),("hypotenuse"=sqrt10):}#

Then:

#cos(tan^-1(3))=cos(phi)="adjacent"/"hypotenuse"=1/sqrt10#

#sin(tan^-1(3))=sin(phi)="opposite"/"hypotenuse"=3/sqrt10#

Plugging these into the expression from earlier we get:

#=1/sqrt5(1/sqrt10)-2/sqrt5(3/sqrt10)#

Note that #sqrt5(sqrt10)=sqrt50=5sqrt2#:

#=(1-6)/(5sqrt2)=-1/sqrt2#

1

Given
#y=f(x)=2sin(x-pi/4)#

To get x-intercepts we can put y=0 and solve for #x in [0,4pi]#

So #2sin(x-pi/4)=0#

#=>x-pi/4=npi" where " n in ZZ#

#=>x=npi+pi/4" where " n in ZZ#

Putting #n=0# we get

#x=pi/4#

Putting #n=1# we get

#x=(5pi)/4#

Putting #n=2# we get

#x=(9pi)/4#

Putting #n=3# we get

#x=(13pi)/4#

(a) So points of x-intercepts over #[0,4pi]# are

#(pi/4,0);((5pi)/4,0);((9pi)/4,0);((13pi)/4,0)#

b) For maximum value of y the value of #sin(x-pi/4)=1#

So #x-pi/4=pi/2#
#=>x= pi/4+pi/2=(3pi)/4#

For #x=(3pi)/4" " y=2#

The point(in (0,2pi) where the graph of this function reaches
maximum is #((3pi)/4,2)#

enter image source here

1

Answer:

Start from a "basic cycle" for the #tan# function to obtain the results below.

Explanation:

Starting with the "basic cycle" for #tan(theta)# i.e. for #theta in [-pi/2,+pi/2]#
enter image source here

Then consider what values of #x# would place #(x+pi)# in this same range, i.e. #(x+pi) in [-pi/2,+pi/2]#
(Yes; I know: because #tan# has a cycle length of #pi# we could recognize that the cycle will repeat at exactly the same place, but let's do this for the more general case.)
#(x+pi)in[-pi/2,pi/2]color(white)("X"rarrcolor(white)("X")x in [-(3pi)/2,-pi/2]#
Giving us the "basic cycle" for #tan(x+pi)#:
enter image source here

Adding #5# to this to get #y=5+tan(x+pi)# simply shifts the points upward #5# units:

In the image below, I have added the "non-basic cycles" as well as the "basic cycle" used for analysis:
enter image source here

2

Answer:

See explanation...

Explanation:

Consider a right angled triangle with an internal angle #theta#:

enter image source here

Then:

#sin theta = a/c#

#cos theta = b/c#

So:

#sin^2 theta + cos^ theta = a^2/c^2+b^2/c^2 = (a^2+b^2)/c^2#

By Pythagoras #a^2+b^2 = c^2#, so #(a^2+b^2)/c^2 = 1#

So given Pythagoras, that proves the identity for #theta in (0, pi/2)#

For angles outside that range we can use:

#sin (theta + pi) = -sin (theta)#

#cos (theta + pi) = -cos (theta)#

#sin (- theta) = - sin(theta)#

#cos (- theta) = cos(theta)#

So for example:

#sin^2 (theta + pi) + cos^2 (theta + pi) = (-sin theta)^2 + (-cos theta)^2 = sin^2 theta + cos^2 theta = 1#

#color(white)()#
Pythagoras theorem

Given a right angled triangle with sides #a#, #b# and #c# consider the following diagram:

enter image source here

The area of the large square is #(a+b)^2#

The area of the small, tilted square is #c^2#

The area of each triangle is #1/2ab#

So we have:

#(a+b)^2 = c^2 + 4 * 1/2ab#

That is:

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

Subtract #2ab# from both sides to get:

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

2

Answer:

Use the formula for a circle #(x^2+y^2=r^2)#, and substitute #x=rcostheta# and #y=rsintheta#.

Explanation:

The formula for a circle centred at the origin is

#x^2+y^2=r^2#

That is, the distance from the origin to any point #(x,y)# on the circle is the radius #r# of the circle.

Picture a circle of radius #r# centred at the origin, and pick a point #(x,y)# on the circle:

graph{(x^2+y^2-1)((x-sqrt(3)/2)^2+(y-0.5)^2-0.003)=0 [-2.5, 2.5, -1.25, 1.25]}

If we draw a line from that point to the origin, its length is #r#. We can also draw a triangle for that point as follows:

graph{(x^2+y^2-1)(y-sqrt(3)x/3)((y-0.25)^4/0.18+(x-sqrt(3)/2)^4/0.000001-0.02)(y^4/0.00001+(x-sqrt(3)/4)^4/2.7-0.01)=0 [-2.5, 2.5, -1.25, 1.25]}

Let the angle at the origin be theta (#theta#).

Now for the trigonometry.

For an angle #theta# in a right triangle, the trig function #sin theta# is the ratio #"opposite side"/"hypotenuse"#. In our case, the length of the side opposite of #theta# is the #y#-coordinate of our point #(x,y)#, and the hypotenuse is our radius #r#. So:

#sin theta = "opp"/"hyp" = y/r" "<=>" "y=rsintheta#

Similarly, #cos theta# is the ratio of the #x#-coordinate in #(x,y)# to the radius #r#:

#cos theta = "adj"/"hyp"=x/r" "<=>" "x = rcostheta#

So we have #x=rcostheta# and #y=rsintheta#. Substituting these into the circle formula gives

#"      "x^2"     "+"      "y^2"     "=r^2#
#(rcostheta)^2+(rsintheta) ^2 = r^2#
#r^2cos^2theta + r^2 sin^2 theta = r^2#

The #r^2#'s all cancel, leaving us with

#cos^2 theta + sin^2 theta = 1#

This is often rewritten with the #sin^2# term in front, like this:

#sin^2 theta + cos^2 theta = 1#

And that's it. That's really all there is to it. Just as the distance between the origin and any point #(x,y)# on a circle must be the circle's radius, the sum of the squared values for #sin theta# and #cos theta# must be 1 for any angle #theta#.