sinAcosB=12[sin(A−B)+sin(A+B)]\sin A \cos B = \frac{1}{2}[\sin(A - B) + \sin(A + B)]sinAcosB=21[sin(A−B)+sin(A+B)] 这个公式表示,正弦函数与余弦函数的乘积可以转化为两个正弦函数之和与差的一半。 English Answer: The product-to-sum formula for sin...
sinAcosB - sinBcosA + sinAcosB - sinBcosA commute sin(A - B) + sin(A - B) difference formula 2sin(A - B) Report 12/01/16 Jonathan T. where does the SinAcosA and -SinBcosB go? Report 12/04/16 Jonathan T. or is (SinAcosB)-(SinBcosA) the same as (SinACosA-sinBcosB)...
In a triangle ABC , iftanA=2sin2Cand3cosA=2sinBsinC, then C= View Solution Exams IIT JEE NEET UP Board Bihar Board CBSE Free Textbook Solutions KC Sinha Solutions for Maths Cengage Solutions for Maths DC Pandey Solutions for Physics ...