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Answer:

It is 900^{\circ}/7

Explanation:

From alpha^{\circ}/alpha=360^{\circ}/(2*pi) we get
alpha^{\circ}=180^{\circ}/pi*5/7*pi
canceling the pi we get
alpha^{\circ}=900^{\circ}/7

Answer:

tan(x+y)=(tanx+tany)/(1-tanxtany)

Explanation:

This can be expanded through the tangent angle addition formula:

tan(alpha+beta)=(tanalpha+tanbeta)/(1-tanalphatanbeta)

Thus,

tan(x+y)=(tanx+tany)/(1-tanxtany)


The tangent addition formula can be found using the sine and cosine angle addition formulas.

sin(alpha+beta)=sinalphacosbeta+cosalphasinbeta
cos(alpha+beta)=cosalphacosbeta-sinalphasinbeta

Since tanx=sinx/cosx,

tan(alpha+beta)=sin(alpha+beta)/cos(alpha+beta)=(sinalphacosbeta+cosalphasinbeta)/(cosalphacosbeta-sinalphasinbeta)

This can be written in terms of tangent by dividing both the numerator and denominator by cosalphacosbeta.

tan(alpha+beta)=((sinalphacosbeta+cosalphasinbeta)/(cosalphacosbeta))/((cosalphacosbeta-sinalphasinbeta)/(cosalphacosbeta))=(sinalpha/cosalpha(cosbeta/cosbeta)+sinbeta/cosbeta(cosalpha/cosalpha))/(cosalpha/cosalpha(cosbeta/cosbeta)-sinalpha/cosalpha(sinbeta/cosbeta))

Final round of simplification yields:

tan(alpha+beta)=(tanalpha+tanbeta)/(1-tanalphatanbeta)

Answer:

2cos((13pi)/16){cos((7pi)/16)+isin((7pi)/16)}

Explanation:

e^((5pi)/4i) - e^((3pi)/8i)

Euler formula : e^(itheta) = cos(theta) + isin(theta)

e^((5pi)/4i) = cos((5pi)/4) + isin((5pi)/4)
e^((3pi)/8i) = cos((3pi)/8) + isin((3pi)/8)

e^((5pi)/4i)-e^((3pi)/8i)
= cos((5pi)/4) + isin((5pi)/4) - (cos((3pi)/8) + isin((3pi)/8))
=cos((5pi)/4)-cos((3pi)/8)+i(sin((5pi)/4)-sin((3pi)/8))

color(blue)( cos(C) -cos(D) = 2cos((C+D)/2)cos((C-D)/2)

color(blue)(sin(C) - sin(D) = 2cos((C+D)/2)sin((C-D)/2)

cos((5pi)/4)-cos((3pi)/8) = 2cos(((5pi)/4+(3pi)/8)/2)cos(((5pi)/4-(3pi)/8)/2)
cos((5pi)/4)-cos((3pi)/8) = 2cos(1/2((10pi)/8+(3pi)/8))cos((1/2((10pi)/8-(3pi)/8))
cos((5pi)/4)-cos((3pi)/8) = 2cos(1/2((13pi)/8)cos((1/2((7pi)/8))
color(green)(cos((5pi)/4)-cos((3pi)/8) = 2cos((13pi)/16)cos((7pi)/16)

sin((5pi)/4)-sin((3pi)/8) = 2cos(1/2((5pi)/4+(3pi)/8)sin(1/2((5pi)/4-(3pi)/8))
sin((5pi)/4)-sin((3pi)/8) = 2cos(1/2((10pi)/8+(3pi)/8)sin(1/2((10pi)/8-(3pi)/8))
sin((5pi)/4)-sin((3pi)/8) = 2cos(1/2((13pi)/8)sin(1/2((7pi)/8))
color(green)(sin((5pi)/4)-sin((3pi)/8) = 2cos((13pi)/16)sin((7pi)/16)

e^((5pi)/4i)-e^((3pi)/8i)
=cos((5pi)/4)-cos((3pi)/8)+i(sin((5pi)/4)-sin((3pi)/8))
=2cos((13pi)/16)cos((7pi)/16) + i(2cos((13pi)/16)sin((7pi)/16))
=2cos((13pi)/16){cos((7pi)/16)+isin((7pi)/16)}

Answer:

=(-cos^3(theta)-2cos^2(theta)+cos(theta)+1)/(cos(theta)sqrt(1-cos^2theta))

Explanation:

tan(theta)-cot(theta)+sin(theta)
We have to write in terms of cos(theta)

color(blue)"Let us start by using the identity"
tan(theta) = sin(theta)/cos(theta) and cot(theta) = cos(theta)/sin(theta)

We get

tan(theta)-cot(theta)+sin(theta)
=sin(theta)/cos(theta) - cos(theta)/sin(theta) + sin(theta)

color(blue)"In order to simplify we need to use Least Common Denominator for all the fractions"

=(sin(theta)sin(theta))/(cos(theta)sin(theta)) -(cos(theta)cos(theta))/(cos(theta)sin(theta)) + (sin(theta)cos(theta)sin(theta))/(cos(theta)sin(theta))

=(sin^2(theta)-cos^2(theta) + sin^2(theta)cos(theta))/(cos(theta)sin(theta))

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

=((1-2cos^2(theta))+cos(theta)-cos^3(theta))/(cos(theta)sin(theta))

=(-cos^3(theta)-2cos^2(theta)+cos(theta)+1)/(cos(theta)sin(theta))

=(-cos^3(theta)-2cos^2(theta)+cos(theta)+1)/(cos(theta)sqrt(1-cos^2theta))

First of all we have to convert these two numbers into trigonometric forms.
If (a+ib) is a complex number, u is its magnitude and alpha is its angle then (a+ib) in trigonometric form is written as u(cosalpha+isinalpha).
Magnitude of a complex number (a+ib) is given bysqrt(a^2+b^2) and its angle is given by tan^-1(b/a)

Let r be the magnitude of (1-2i) and theta be its angle.
Magnitude of (1-2i)=sqrt(1^2+(-2)^2)=sqrt(1+4)=sqrt5=r
Angle of (1-2i)=Tan^-1((-2)/1)=tan^-1(-2)=theta

implies (1-2i)=r(Costheta+isintheta)

Let s be the magnitude of (6-8i) and phi be its angle.
Magnitude of (6-8i)=sqrt(6^2+(-8)^2)=sqrt(36+64)=sqrt100=10=s
Angle of (6-8i)=Tan^-1((-8)/6)=Tan^-1(-4/3)=phi

implies (6-8i)=s(Cosphi+isinphi)

Now,
(1-2i)/(6-8i)

=(r(Costheta+isintheta))/(s(Cosphi+isinphi))

=r/s*(Costheta+isintheta)/(Cosphi+isinphi)*(Cosphi-isinphi)/(Cosphi-isinphi

=r/s*(costhetacosphi+isinthetacosphi-icosthetasinphi-i^2sinthetasinphi)/(cos^2phi-i^2sin^2phi)

=r/s*((costhetacosphi+sinthetasinphi)+i(sinthetacosphi-costhetasinphi))/(cos^2phi+sin^2phi)

=r/s*(cos(theta-phi)+isin(theta-phi))/(1)

=r/s(cos(theta-phi)+isin(theta-phi))

Here we have every thing present but if here directly substitute the values the word would be messy for find theta -phi so let's first find out theta-phi.

theta-phi=tan^-1(-2)-tan^-1(-4/3)
We know that:
tan^-1(a)-tan^-1(b)=tan^-1((a-b)/(1+ab))

implies tan^-1(-2)-tan^-1(-4/3)=tan^-1(((-2)-(-4/3))/(1+(-2)(-4/3)))

=tan^-1((-6+4)/(3+8))=tan^-1(-2/11)

implies theta -phi=tan^-1(-2/11)

r/s(cos(theta-phi)+isin(theta-phi))

=sqrt5/10(cos(tan^-1(-2/11))+isin(tan^-1(-2/11)))

=sqrt(5/100)(cos(tan^-1(-2/11))+isin(tan^-1(-2/11)))

=sqrt(1/20)(cos(tan^-1(-2/11))+isin(tan^-1(-2/11)))

=1/(2sqrt(5))(cos(tan^-1(-2/11))+isin(tan^-1(-2/11)))

This is your final answer.

You can also do it by another method.
By firstly dividing the complex numbers and then changing it to trigonometric form, which is much easier than this.

First of all let's simplify the given number
(1-2i)/(6-8i).

Multiply and divide by the conjugate of the complex number present in the denominator i.e 6+8i.

(1-2i)/(6-8i)=((1-2i)(6+8i))/((6-8i)(6+8i))=(6+8i-12i-16i^2)/(6^2-(8i)^2)
=(6-4i+16)/(36-(-64))=(22-4i)/(36+64)=(22-4i)/100=22/100-(4i)/100=11/50-i/25

(1-2i)/(6-8i)=11/50-(i)/25

Let t be the magnitude of (11/50-(i)/25) and beta be its angle.

Magnitude of (11/50-i/25)=sqrt((11/50)^2+(-1/25)^2)=sqrt(121/2500+1/625)=sqrt((121+4)/2500)=sqrt(125/2500)=sqrt(1/20)=1/(2sqrt5)=t
Angle of (11/50-(i)/25)=Tan^-1((-1/25)/(11/50))=tan^-1(-2/11)=beta

implies (11/50-(i)/25)=t(Cosbeta+isinbeta)

implies (11/50-(i)/25)=1/(2sqrt5)(Cos(tan^-1(-2/11))+isin(tan^-1(-2/11))).

(i+8)/(3i-1)

=(8+i)/(-1+3i)

First of all we have to convert these two numbers into trigonometric forms.
If (a+ib) is a complex number, u is its magnitude and alpha is its angle then (a+ib) in trigonometric form is written as u(cosalpha+isinalpha).
Magnitude of a complex number (a+ib) is given bysqrt(a^2+b^2) and its angle is given by tan^-1(b/a)

Let r be the magnitude of (8+i) and theta be its angle.
Magnitude of (8+i)=sqrt(8^2+1^2)=sqrt(64+1)=sqrt65=r
Angle of (8+i)=Tan^-1(1/8)=theta

implies (8+i)=r(Costheta+isintheta)

Let s be the magnitude of (-1+3i) and phi be its angle.
Magnitude of (-1+3i)=sqrt((-1)^2+3^2)=sqrt(1+9)=sqrt10=s
Angle of (-1+3i)=Tan^-1(3/-1)=Tan^-1(-3)=phi

implies (-1+3i)=s(Cosphi+isinphi)

Now,
(8+i)/(-1+3i)

=(r(Costheta+isintheta))/(s(Cosphi+isinphi))

=r/s*(Costheta+isintheta)/(Cosphi+isinphi)*(Cosphi-isinphi)/(Cosphi-isinphi

=r/s*(costhetacosphi+isinthetacosphi-icosthetasinphi-i^2sinthetasinphi)/(cos^2phi-i^2sin^2phi)

=r/s*((costhetacosphi+sinthetasinphi)+i(sinthetacosphi-costhetasinphi))/(cos^2phi+sin^2phi)

=r/s*(cos(theta-phi)+isin(theta-phi))/(1)

=r/s(cos(theta-phi)+isin(theta-phi))

Here we have every thing present but if here directly substitute the values the word would be messy for find theta -phi so let's first find out theta-phi.

theta-phi=tan^-1(1/8)-tan^-1(-3)
We know that:
tan^-1(a)-tan^-1(b)=tan^-1((a-b)/(1+ab))

implies tan^-1(1/8)-tan^-1(-3)=tan^-1(((1/8)-(-3))/(1+(1/8)(-3)))

=tan^-1((1+24)/(8-3))=tan^-1(25/5)=tan^-1(5)

implies theta -phi=tan^-1(5)

r/s(cos(theta-phi)+isin(theta-phi))

=sqrt65/sqrt10(cos(tan^-1(5))+isin(tan^-1(5)))

=sqrt(65/10)(cos(tan^-1(5))+isin(tan^-1(5)))

=sqrt(13/2)(cos(tan^-1(5))+isin(tan^-1(5)))

This is your final answer.

You can also do it by another method.
By firstly dividing the complex numbers and then changing it to trigonometric form, which is much easier than this.

First of all let's simplify the given number
(i+8)/(3i-1)

=(8+i)/(-1+3i)

Multiply and divide by the conjugate of the complex number present in the denominator i.e -1-3i.

(8+i)/(-1+3i)=((8+i)(-1-3i))/((-1+3i)(-1-3i))=(-8-24i-i-3i^2)/((-1)^2-(3i)^2)
=(-8-25i+3)/(1-(-9))=(-5-25i)/(1+9)=(-5-25i)/10=-5/10-(25i)/10=-1/2-(5i)/2

(8+i)/(-1+3i)=-1/2-(5i)/2

Let t be the magnitude of (1/10-(5i)/2) and beta be its angle.

Magnitude of (-1/2-(5i)/2)=sqrt((-1/2)^2+(-5/2)^2)=sqrt(1/4+25/4)=sqrt(26/4)=sqrt(13/2)=t
Angle of (-1/2-(5i)/2)=Tan^-1((-5/2)/(-1/2))=tan^-1(5)=beta

implies (-1/2-(5i)/2)=t(Cosbeta+isinbeta)

implies (-1/2-(5i)/2)=sqrt(13/2)(Cos(tan^-1(5))+isin(tan^-1(5))).

Answer:

Area =1/2*19*32.44 ~~308.13

Explanation:

Sketch
The area of a triangle =1/b*h where b = base and h=height

In this case tan((5pi)/8) =h/x and tan(pi/4) = h/y where

x+y = B =19
y=19-x

So xtan((5pi)/8) =ytan(pi/4)
xtan((5pi)/8) = (19-x)tan(pi/4)
x(tan((5pi)/8) +tan(pi/4)) = 19tan(pi/4)

:.x = 19tan(pi/4)/(tan((5pi)/8)+tan(pi/4))

h = (19tan(pi/4)/(tan((5pi)/8)+tan(pi/4)))tan((5pi)/8)
~~(19*1*2.414)/(2.414+1)
~~32.44

Area =1/2*19*32.44 ~~308.13

Answer:

(2sin^4theta-4sin^2theta+sinthetasin2theta+2)/(2sin^2theta)

Explanation:

Write in terms of sintheta and costheta.

=costheta-cos^2theta+cos^2theta/sin^2theta

Find a common denominator.

=(costhetasin^2theta)/sin^2theta-(cos^2thetasin^2theta)/sin^2theta+cos^2theta/sin^2theta

Combine.

=(costhetasin^2theta-cos^2thetasin^2theta+cos^2theta)/sin^2theta

The following simplification may seem unecessary, but is actually relevant. Its purpose will become clear in the following step.

=(sintheta(color(blue)(costhetasintheta))-color(green)(cos^2theta)sin^2theta+color(green)(cos^2theta))/sin^2theta

Use the following identities:

  • color(green)(cos^2theta=1-sin^2theta
  • 2costhetasintheta=sin2theta=>color(blue)(costhetasintheta=(sin2theta)/2

=(sintheta((sin2theta)/2)-(1-sin^2theta)sin^2theta+(1-sin^2theta))/sin^2theta

=((sinthetasin2theta)/2-sin^2theta+sin^4theta+1-sin^2theta)/sin^2theta

=((sinthetasin2theta)/2-2sin^2theta+sin^4theta+1)/sin^2theta

=(2sin^4theta-4sin^2theta+sinthetasin2theta+2)/(2sin^2theta)

Answer:

color(green)("area " = 1.125 " units"^2 -> 1 1/8 " units"^2)

Explanation:

color(blue)("Assumption: ")

As pi is used in the angular measure it is assumed that the unit is radians. (Not stated)

'~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
Tony B

color(blue)("Method Plane")

Determine /_cba

Using Sine Rule and /_cba determine length of side A
Determine h using h=Asin((5pi)/12)
Determine area hxxB/2

'~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
color(blue)("To determine" /_cba)

Sum internal angles of a triangle is 180^0 = pi" radians"

=>/_cba= pi-(5pi)/12-pi/12
color(brown)(/_cba = pi/2 -> 90^o)

'~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
color(blue)("Determine length of A")

Using B/(sin(b))=A/sin(a)

=> 3/(sin(pi/2)) =A/sin(pi/12)

=> A= (3xxsin(pi/12))/(sin(pi/2))

But sin(pi/2) = 1

color(blue)(=> A = 3xxsin(pi/12))

color(brown)("This is an exact value so keep it in this form for now to reduce error") color(brown)("on final calculation.")
'~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
color(blue)("To determine h")

h=Asin((5pi)/12)

=>h=3xxsin(pi/12)xxsin((5pi)/12)

'~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
color(blue)("To determine area")

"area "= B/2xxh

"area "= 3/2 xx3xx sin(pi/12)xxsin((5pi)/12)

but sin(pi/12)xxsin((5pi)/12)=1/4

"area "= 3/2 xx3xx1/4

color(green)("area " = 1.125 " units"^2 -> 1 1/8 " units"^2)

Answer:

The x component is: cos((5pi)/4)
The y component is: sin((5pi)/4)

Explanation:

Remembering our trigonometry, the vertical component of a vector is given by
r*sin(theta) where r is the length of the line,
and the horizontal component by
r*cos(theta)

https://en.wikipedia.org/wiki/Trigonometry

in the polar coordinate (1,(5pi)/4), r is 1, and the angle theta = (5pi)/4.

Hence:
The x component is: cos((5pi)/4)
The y component is: sin((5pi)/4)

In this case, (5pi)/4 is midway in the lower left quadrant, or 135^@, so both are equal to -1/sqrt2