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Prove that cosalpha+cosbeta+cosgamma(alp...

Prove that `cosalpha+cosbeta+cosgamma(alpha+beta+gamma)=4cos(alpha+beta)/(2)cos(beta+gamma)/2cos(gamma+alpha)/2`

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Prove that cosalpha+cosbeta+cosgamma+cos(alpha+beta+gamma)=4cos((alpha+beta)/2)cos((beta+gamma)/2)cos((gamma+alpha)/2)

Prove that cosalpha+cosbeta+cosgamma+cos(alpha+beta+gamma)=4cos((alpha+beta)/2)cos((beta+gamma)/2)cos((gamma+alpha)/2)

Prove that: cosalpha+cosbeta+cosgamma+cos(alpha+beta+gamma)=4cos((alpha+beta)/2)cos((beta+gamma)/2)cos((gamma+alpha)/2) .

Prove that: sinalpha+sinbeta+singamma-sin(alpha+beta+gamma)=4sin((alpha+beta)/2)sin((beta+gamma)/2)sin((gamma+alpha)/2)dot

Prove that : cos^2 (beta-gamma) + cos^2 (gamma-alpha) + cos^2 (alpha-beta) =1+2cos (beta-gamma) cos (gamma-alpha) cos (alpha-beta) .

If alpha,beta,gamma are the angles of a triangle and system of equations cos(alpha-beta)x+cos(beta-gamma)y+cos(gamma-alpha)z=0 cos(alpha+beta)x+cos(beta+gamma)y+cos(gamma+alpha)z=0 sin(alpha+beta)x+sin(beta+gamma)y+sin(gamma+alpha)z=0 has non-trivial solutions, then triangle is necessarily a. equilateral b. isosceles c. right angled "" d. acute angled

If cosalpha+cosbeta+cosgamma=0 a n d a l so sinalpha+sinbeta+singamma=0, then prove that cos2alpha+cos2beta+cos2gamma =sin2alpha+sin2beta+sin2gamma=0 sin3alpha+sin3beta+sin3gamma=3sin(alpha+beta+gamma) cos3alpha+cos3beta+cos3gamma=3cos(alpha+beta+gamma)

If cosalpha+cosbeta+cosgamma=0=sinalpha+sinbeta+singamma , then which of the following is/are true:- (a) cos(alpha-beta)+cos(beta-gamma)+cos(gamma-delta)=-3/2 (b) cos(alpha-beta)+cos(beta-gamma)+cos(gamma-delta)=-1/2 (c) sumcos2alpha+2cos(alpha+beta)+2cos(beta+gamma)+2cos(gamma+alpha)=0 (d) sumsin2alpha+2sin(alpha+beta)+2sin(beta+gamma)+2sin(gamma+alpha)=0

Prove that: |(sinalpha, cosalpha, 1),(sinbeta, cosbeta, 1),(singamma, cosgamma, 1)|=sin(alpha-beta)+sin(beta-gamma)+sin(gamma-alpha)

Let A and B denote the statements A : cos alpha + cos beta + cos gamma =0 B : sin alpha + siin beta + sin gamma = 0 If cos(beta - gamma) + cos (gamma -alpha) + cos (alpha -beta) = - (3)/(2) , then

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