在三角形ABC中 、角A,B,C所对的分别为a,b,c ,已知cos2C=-1/4 .求sinC .

网友回答

cos2C= -1/4
所以cos2c=1-2(sinc)^2=-1/4可得sinc=+—根号10/4
又因为角c为三角形内角所以正弦值是正数
所以sinc=根号10/4
======以下答案可供参考======
供参考答案1:
cos2C=-1/4
1-2sin^2C=-1/4
2sin^2C=1+1/4=5/4
sin^2C=5/8
sinC=√10/4
a=2,2sinA=sinc
a/sinA=c/sinC
c=asinC/sinA=2*2=4
c＞aA为锐角sinA=sinC/2=√10/8
cosA=√{1-10/64} = 3√6/8
a^2=b^2+c^2-2bccosA
2^2=b^2+4^2-2b*4*3√6/8
b^2-3√6b+12=0
(b-√6)(b-2√6)=0
b=√6，或2√6