Menu Close

Question-185486

Tinku Tara 0 Comments
Pin




Question Number 185486 by sonukgindia last updated on 22/Jan/23
Answered by a.lgnaoui last updated on 23/Jan/23
A−(B+C)=0 A+B+C=2(B+C)=2A A+B−C=A−(B+C)+2B=2B A−B+C=A−(B+C)+2C=2C ((sin (A+B+C)+sin (A+B−C)+sin (A−B+C))/(2sin Acos Bcos C))= ((sin 2A+sin 2B+sin 2C)/(2sin Acos Bcos C))= ((2[sin Acos A+sin Bcos B+sin Ccos C])/(2sin Acos Bcos C))= ((cos A)/(cos Bcos C))+((tan B)/(sin Acos C))+((tan C)/(sin Acos B)) =((cos A)/(cos Bcos C))+(1/(sin A))(((tan B)/(cos C))+((tan C)/(cos B))) =((cos A)/(cos Bcos C))+(1/(sin A))(((sin B+sin C)/(cos Bcos C))) =(1/(cos Bcos C))[cos A+((sin B+sin C)/(sin A))] (1) sin A=sin (B+C) (1)=(1/(cos B cos C))[(cos (B+C)+((sin B+sin C)/(sin (B+C))))] =(1/(cos A+sinBsinC ))[cos A+((sin B+sin C)/(sin A))]
A−(B+C)=0A+B+C=2(B+C)=2AA+B−C=A−(B+C)+2B=2BA−B+C=A−(B+C)+2C=2Csin(A+B+C)+sin(A+B−C)+sin(A−B+C)2sinAcosBcosC=sin2A+sin2B+sin2C2sinAcosBcosC=2[sinAcosA+sinBcosB+sinCcosC]2sinAcosBcosC=cosAcosBcosC+tanBsinAcosC+tanCsinAcosB=cosAcosBcosC+1sinA(tanBcosC+tanCcosB)=cosAcosBcosC+1sinA(sinB+sinCcosBcosC)=1cosBcosC[cosA+sinB+sinCsinA](1)sinA=sin(B+C)(1)=1cosBcosC[(cos(B+C)+sinB+sinCsin(B+C))]=1cosA+sinBsinC[cosA+sinB+sinCsinA]
Commented by mr W last updated on 23/Jan/23
this is like when you try to simplify (1/2) to (4/(2+(3/(1−(1/2))))).
thisislikewhenyoutrytosimplify12to42+31−12.
Answered by Frix last updated on 23/Jan/23
a=b+c sin (a+b+c) +sin (a+b−c) +sin (a−b+c) = =sin (2b+2c) +sin 2b +sin 2c 2sin a cos b cos c = =2cos^2 b sin c cos c +2sin b cos b cos^2 c = =((sin (2b+2c) +sin 2b +sin 2c)/2) ⇒ answer is 2
a=b+csin(a+b+c)+sin(a+b−c)+sin(a−b+c)==sin(2b+2c)+sin2b+sin2c2sinacosbcosc==2cos2bsinccosc+2sinbcosbcos2c==sin(2b+2c)+sin2b+sin2c2⇒answeris2

Pin

Leave a Reply

Your email address will not be published. Required fields are marked *