MY SOLUTION
Given that A + B + C = M, we can write B + C = M – A. Therefore

Cos A = – Cos (M – A) and Sin A = Sin (M – A)

Using sum and difference formulas, we have

Cos A = –Cos(M)Cos(A) – Sin(M)Sin(A), and
Sin A = Sin(M)Cos(A) – Cos(M)Sin(A)

Adding both equations and grouping, we have

(1 + Cos M)(Sin A – Cos A) = (Sin A + Cos B)Sin M ——— (1)

Dividing the original trigonometric equations in the question

tan A = – tan (M – A)
Using the difference formula for tangent,
tan A = (tan A – tan M)/(1 + tan(M)tan(A)) –—– (2)

Solving equation (2) yields

(1 + tan² A) tan M = 0.

This means tan M = 0, therefore Sin M = 0. Substituting Sin M = 0 in equation (1) yields

(1 + Cos M)(Sin A – Cos A) = 0

This means Cos M = –1, solving this equation yields

M = (2n + 1)180°, where n is any interger

MrShape, guiddoti, danielistics, NDSMELODY

guiddoti:
CosA = -Cos(B+C)
SinA =Sin(B+C)

Let B+C=K

CosA=-CosK........1

SinA=SinK............2

Resolving using Quadrant

in (1) K= theta ; A=180- theta

180=A+k......3

in (2) A= theta ; K= 180-(-theta)
K-A=180.......4

2K=360........5
K=180
Put K into 3, A=0

0,180,-180.

i dnt agree putting k = theta = b+c.........why dnt u use trig id instead

NDSMELODY:
i dnt agree putting k = theta = b+c.........why dnt u use trig id instead

It will be too long, so I option for quadrant.

Martinez39s:
MY SOLUTION
Given that A + B + C = M, we can write B + C = M – A. Therefore

Using sum and difference formulas, we have


Adding both equations and grouping, we have


Dividing the original trigonometric equations in the question

Solving equation (2) yields This means tan M = 0, therefore Sin M = 0. Substituting Sin M = 0 in equation (1) yields

This means Cos M = –1, solving this equation yields


MrShape, guiddoti, danielistics, NDSMELODY

Mehn I've been playing too much chess. Nice problem.

Martinez39s:
MY SOLUTION
Given that A + B + C = M, we can write B + C = M – A. Therefore

Using sum and difference formulas, we have


Adding both equations and grouping, we have


Dividing the original trigonometric equations in the question

Solving equation (2) yields This means tan M = 0, therefore Sin M = 0. Substituting Sin M = 0 in equation (1) yields

This means Cos M = –1, solving this equation yields


MrShape, guiddoti, danielistics, NDSMELODY

Any idea

Horiolah:


Any idea

I don't understand what you are trying to say.

Horiolah:


Any idea

Is that equation of a circle?

Martinez39s:
I don't understand what you are trying to say.

What's the equation for?

Gold Circle

Martinez39s:
MY SOLUTION
Given that A + B + C = M, we can write B + C = M – A. Therefore

Using sum and difference formulas, we have


Adding both equations and grouping, we have


Dividing the original trigonometric equations in the question

Solving equation (2) yields This means tan M = 0, therefore Sin M = 0. Substituting Sin M = 0 in equation (1) yields

This means Cos M = –1, solving this equation yields


MrShape, guiddoti, danielistics, NDSMELODY

Brother abeg help me with this

Horiolah:


Brother abeg help me with this

to find the cube root of 2+i2√3 (i.e (2+i2√3)^1/3 )

first, u have to convert to polar form (i.e in the form of [r(cos@ +isin@) ] )
r =√(x^2+ y^2)
x=2. y=2√3
substituting u have r=4

@=tan^-1(y/x)
substituting @ =60° to rads =π/3

our polar form will be = 4[cos(π/3)+isin(π/3)]

secondly, u apply de'movires theorem
which is [r(cos@ + isin@)]^n = [r^n(cos(n@)+isin(n@)]
in this case n=1/3
so substituting we have [4^1/3{cos(π/9)+isin(π/9)}]

conver your finally answer to CARTESIAN FORMAT (i.e x+iy)

Horiolah:


Brother abeg help me with this

z = 2 + (2√3)i has three distinct cube roots. z = 2 + (2√3)i in polar form is

z = 4[cos 60° + i(sin 60°)]

The cube roots of z are

1) z₁ = 4^⅓[cos 20° + i(sin 20°)]
2) z₂ = 4^⅓[cos 140° + i(sin 140°)]
3) z₃ = 4^⅓[cos 260° + i(sin 260°)]

I wrote the angles in degrees. You can convert them to radians if you want.

Hamzasaid:

to find the cube root of 2+i2√3 (i.e (2+i2√3)^1/3 )

first, u have to convert to polar form (i.e in the form of [r(cos@ +isin@) ] )
r =√(x^2+ y^2)
x=2. y=2√3
substituting u have r=4

@=tan^-1(y/x)
substituting @ =60° to rads =π/3

our polar form will be = 4[cos(π/3)+isin(π/3)]

secondly, u apply de'movires theorem
which is [r(cos@ + isin@)]^n = [r^n(cos(n@)+isin(n@)]
in this case n=1/3
so substituting we have [4^1/3{cos(π/9)+isin(π/9)}]

conver your finally answer to CARTESIAN FORMAT (i.e x+iy)

Thank you sir

Martinez39s:
z = 2 + (2√3)i has three distinct cube roots. z = 2 + (2√3)i in polar form is


The cube roots of z are

See this

Horiolah:


See this

z × w = (2 × 3) [cos (π/4 + π/6) + i(sin (π/4 + π/6))]
z × w = 6[cos (5π/12) + i(sin (5π/12))]