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[vizman]

hi,

can someone tell me how to do this.

Prove that sin ( A + B + C) cos (B + C) - cos ( A + B + C) sin (B + C) = sin A

thanks in advance

 

[vizman]

solved. disregard post

 

[Riviet]

hi,

can someone tell me how to do this.

Prove that sin ( A + B + C) cos (B + C) - cos ( A + B + C) sin (B + C) = sin A

thanks in advance

Let S=sin, C=cos, A=a, B=b, C=c

LHS=C(b+c)[Sacos(b+c)+CaS(b+c)] - S(b+c)[CaC(b+c)-SaS(b+c)]

= SaC²(b+c) + CaS(b+c)C(b+c) - CaS(b+c)C(b+c) + SaS²(b+c)

= Sa[S²(b+c) + C²(b+c)]

=Sa(1)

=SinA

=RHS

EDIT:

 

[vizman]

Let S=sin, C=cos, A=a, B=b, C=c

LHS=C(b+c)[Sacos(b+c)+CaS(b+c)] - S(b+c)[CaC(b+c)-SaS(b+c)]

= SaC²(b+c) + CaS(b+c)C(b+c) - CaS(b+c)C(b+c) + SaS²(b+c)

= Sa[S²(b+c) + C²(b+c)]

=Sa(1)

=SinA

=RHS

EDIT:

sorry how does that equal to 1.

 

[Jono_2007]

it's the identity;
sin^2 θ+cos^2 θ=1
so when you consider that whats in the brackets is actualy an angle it directly refers to this identity.

hows that!

 

[vizman]

oh yeh. lol his substitution of sin for s confused me lol