Trigonometry> urgent please...

if sinA+sinB+sinC=0 and cosA+cosB+cosC=0 then find the value of Sin(sqr)A+Sin(sqr)B+Sin(sqr)C.please pleaseplease pleasepleaseplease

Alok Vishwakarma , 13 Years ago

Grade 10

5 Answers

Ashwin Muralidharan IIT Madras

Last Activity: 13 Years ago

Hi Alok,

SinA + SinB = -Sinc ---------(1)

CosA + CosB = -CosC ---------(2)

Sq and add (1),(2)

2+2Cos(A-B) = 1 ------------(3)

Sq and Sub (2),(1)

Cos2A + Cos2B + 2Cos(A+B) = Cos2C

2Cos(A+B)Cos(A-B) + 2Cos(A+B) = Cos2C

or Cos(A+B){2Cos(A-B) + 2} = Cos2C

or Cos(A+B) = Cos2C ----------- [Using (3)]

Now SinA+SinB+SinC = 0

Squaring, Sin²A+Sin²B+Sin²C = -2∑SinASinB ----------(4)

Now Cos(A+B) = Cos2C

Similarly Cos(B+C) = Cos2A

and Cos(C+A) = Cos2B

So adding these, ∑CosACosB - ∑SinASinB = ∑Cos2A

Also Cos(A-B) = -1/2 [from (3)]

similarly Cos(B-C) = -1/2

and Cos(C-A) = -1/2

Adding ∑CosACosB + ∑SinASinB = -3/2

So -2∑SinASinB = ∑Cos2A + 3/2

Hence from (4) Sin²A+Sin²B+Sin²C = -2∑SinASinB = ∑Cos2A + 3/2

Now Cos2A+Cos2B+Cos2C = 2Cos(A+B)Cos(A-B) + Cos2C = -Cos(A+B) + Cos2C = 0 ------ [because Cos(A+B) = Cos2C; and Cos(A-B) = -1/2 which was proved previously]

Hence ∑Cos2A = 0.

Hence Sin²A+Sin²B+Sin²C = 3/2.

Note that you can have so many results from the above conditions, like

Sin²A+Sin²B+Sin²C = 3/2

∑Cos2A = 0

∑SinASinB = -3/4

∑CosACosB = -3/4

Hope this helps.

Best Regards,

Ashwin (IIT Madras).

jatinder

Last Activity: 10 Years ago

Cosa+cosb=0=sina+sinb prove that cos2a+cos2b=-cos(a+b)

mycroft holmes

Last Activity: 10 Years ago

Let z1 = cos A + i sin A; z2 = cos B + i sin B; z3 = cos C + i sin C Then z1+z2+z3 = 0 Also if I denote conjugate of z by _z, then we have _z1+_z2+_z3 = 0 But _z1 = 1/z1 (as |z1|^2 = z1 _z1 = 1) Hence 1/z1 + 1/z2 + 1/z3 = 0 or z1z2+z2z3+z3z1 = 0 z1+z2+z3 = 0 so that (z1+z2+z3)^2 = 0 or z1^2+z2^2+z3^2 + 2(z1z2+z2z3+z3z1) = 0 which means z1^2+z2^2+z3^2= 0 Equating real parts, we get cos 2A + cos 2B + cos 2C = 0 or 3- 2(sin^2 A + sin^2 B + sin^2 C) = 0 or sin^2 A+sin^2 B+sin^2 C = 3/2

Parshwa Murdia

Last Activity: 9 Years ago

Let $a=cos\alpha +isin\alpha , b =cos\beta +isin\beta , c=cos\gamma +isin\gamma$

$a+b+c=0$ => $cos\alpha +cos\beta +cos\gamma =0$ and $sin\alpha +sin\beta +sin\gamma =0$

$\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=a^{-1}+b^{-1}+c^{-1}$ = $(cos\alpha +cos\beta +cos\gamma) - i (sin\alpha +sin\beta +sin\gamma)$ = 0

$a^{2}+b^{2}+c^{2}= (a+b+c)^{2}-2abc.(\frac{1}{a}+\frac{1}{b}+\frac{1}{c})= 0$

Thus, $a^{2}+b^{2}+c^{2}= (cos2\alpha +cos2\beta +cos2\gamma)+i(sin2\alpha +sin2\beta +sin2\gamma)=0$

$cos2\alpha +cos2\beta +cos2\gamma=0$ and $sin2\alpha +sin2\beta +sin2\gamma=0$ .

This can be done using complex numbers.

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Rishi Sharma

Last Activity: 4 Years ago

Dear Student,
Please find below the solution to your problem.

SinA + SinB = -Sinc ---------(1)
CosA + CosB = -CosC ---------(2)
Sq and add (1),(2)
2+2Cos(A-B) = 1 ------------(3)
Sq and Sub (2),(1)
Cos2A + Cos2B + 2Cos(A+B) = Cos2C
2Cos(A+B)Cos(A-B) + 2Cos(A+B) = Cos2C
or
Cos(A+B){2Cos(A-B) + 2} = Cos2C
or
Cos(A+B) = Cos2C ----------- [Using (3)]
Now SinA+SinB+SinC = 0
Squaring, Sin2A+Sin2B+Sin2C = -2∑SinASinB ----------(4)
Now Cos(A+B) = Cos2C
Similarly Cos(B+C) = Cos2A
and Cos(C+A) = Cos2B
So adding these, ∑CosACosB - ∑SinASinB = ∑Cos2A
Also Cos(A-B) = -1/2 [from (3)]
similarly Cos(B-C) = -1/2
and Cos(C-A) = -1/2
Adding ∑CosACosB + ∑SinASinB = -3/2
So -2∑SinASinB = ∑Cos2A + 3/2
Hence from (4) Sin2A+Sin2B+Sin2C = -2∑SinASinB = ∑Cos2A + 3/2
Now
Cos2A+Cos2B+Cos2C = 2Cos(A+B)Cos(A-B) + Cos2C = -Cos(A+B) + Cos2C = 0 [because Cos(A+B) = Cos2C; and Cos(A-B) = -1/2 which was proved previously]
Hence ∑Cos2A = 0.
Hence Sin2A+Sin2B+Sin2C = 3/2.
Note that you can have so many results from the above conditions, like
Sin2A+Sin2B+Sin2C = 3/2
∑Cos2A = 0
∑SinASinB = -3/4
∑CosACosB = -3/4

Thanks and Regards