2被根号3（Sin(B+π/3)+SinB)+3变形为4倍根号3Sin（B+6/π）Cos（π/6)+3 要求过程与用到的公式,_数学解答

2被根号3（Sin(B+π/3)+SinB)+3变形为4倍根号3Sin（B+6/π）Cos（π/6)+3 要求过程与用到的公式,

人气：445 ℃　时间：2024-11-23 19:41:00

解答

2 (根号3) [ sin (B +π/3) +sin B ]
=2 (根号3) *2 sin [ (B +π/3 +B) /2 ] cos [ (B +π/3 -B) /2 ]
=4 (根号3) *sin (B +π/6) cos (π/6) +3
=6 sin (B +π/6) +3.
= = = = = = = = =
和差化积：
sin A +sin B = 2 sin [ (A+B) /2 ] cos [ (A-B) /2 ].