Question

咋解这道三角函数

在三角形ABC中，设a＋c＝2b，A－C＝60度，求SinB的值？
2sin[(A+C)/2] × cos[(A-C)/2] = 2 × 2 ×sin(B/2) ×cos(B/2) 额，这个我看不懂.................谢谢

Answer 1

解：∵在三角形ABC中,

∴a/sinA = b/sinB = c/sinC

∵a + c = 2b

∴sinA + sinC = 2sinB

∵ sinA + sinC =sin[(A+C)/2+(A-C)/2]+sin[(A+C)/2-(A-C)/2]
=2sin[(A+C)/2] × cos[(A-C)/2]

sinB =2 ×sin(B/2) ×cos(B/2)

sinA + sinC = 2sinB

∴2sin[(A+C)/2] × cos[(A-C)/2] = 2 × 2 ×sin(B/2) ×cos(B/2)

sin[(A+C)/2] × cos30 = 2 × sin(B/2) ×cos(B/2)

cos[90 - (A+C)/2] × cos30 = 2 ×sin(B/2)× cos(B/2)

cos(B/2) × cos30 = 2×sin(B/2) ×cos(B/2)

cos30 = 2 ×sin(B/2)

∴sin(B/2) = √3 / 4

cos(B/2)=√13/4

∴sinB=2×√3/4×√13/4=√39/8

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