三角恒等变换问题.化减：cos2a+6sin^2(a/2)-8sin^4(a/2)=

Question

三角恒等变换问题.
化减：cos2a+6sin^2(a/2)-8sin^4(a/2)=

Answer 1

cosa=2cos^2(a/2)-1=1-2sin^2(a/2).所以：sin^2(a/2)=(1-cosa)/2
所以：sin^4(a/2)=(1-cosa)^2/4=(1-2cosa+cos^2(a))/4
而：cos^2(a)=(cos2a+1)/2.
所以：sin^4(a/2)=(1-2cosa+(cos2a+1)/2)/4
所以原式=cos2a+(3-3cosa)-(2-4cosa+cos2a+1)
=cos2a+3-3cosa-2+4cosa-cos2a-1
=cosa+2.