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Trigonometric function in high school math problems
It is known that A=π-(B+C)

∴cos[π-(b+c)]+cosbcosc-√3c osbsinc = 0

-cos(B+C)+cosBcosC-√3c osbsinc = 0

cosBcosC - √3cosBsinC=cos(B+C)

cosBcosC-√3c OSB sinc = cosBcosC-sinb sinc

∴√3cosBsinC=sinBsinC

∫sinC≠0

∴√3cosB=sinB, then tanB=√3.

∴B=π/3

According to sine theorem: 2r = b/sinb =1/sin (π/3) = 2/√ 3.

Then a+c=2RsinA+2RsinC=2R(sinA+sinC).

=2R[sinA + sin(π - A - π/3)]

=2R[sinA + sin(2π/3 - A)]

= 2R[sinA+sin(2π/3)cosA-cos(2π/3)sinA]

= 2R[ Sina+(√3/2)cosA+(1/2) Sina]

=2R[(3/2)sinA + (√3/2)cosA]

=(2/√3)? (√3)[(√3/2) Sina+(1/2)cosA]

=2sin(A + π/6)

∫△ABC is an acute triangle

When b