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Simplified calculation of trigonometric function.

analyse

SinA+2sinBcosC=0。 By using the triangle interior angle sum theorem and inductive formula, we can get: sin(B+C)+2sinBcosC=0, and expand it into: 3sinBcosC+cosBsinC=0, cosC≠0, cosb ≠ 0. So it can be judged that 3 tanb =-tanc. Tana =-tan (b+c) expansion substitution can be obtained by using the properties of basic inequalities.

explain

Solution: in triangle ABC, A+B+C = 180 Sina = SIN (B+C) is substituted into sinA+sinBcosC=0.

De: sin(B+C)+2sinBcosC=0。

∴:3sinBcosC+cosBsinC=0

∴:3sinBcosC=﹣cosBsinC

∴:3tanB=﹣tanC

sina+2sinbcosc=0,sina=﹣2sinbcosc>0

∴:sinBcosC 0

∴:cosC