解一道三角函数求值题

青青子衿

已知$\sin\alpha=\cos\beta,\cos\alpha=\sin3\beta$，则$\sin^2\beta+\cos^2\alpha=?$

战巡

回复 1# 青青子衿

\[\sin^2(b)+\cos^2(a)=\sin^2(b)+1-\sin^2(a)=\sin^2(b)+1-\cos^2(b)=2\sin^2(b)\]
然后
\[\sin(a)^2+\cos^2(a)=1=\cos^2(b)+\sin^2(3b)\]
\[\sin^2(b)-\sin^2(3b)=0\]
\[[\sin(b)-\sin(3b)][\sin(b)+\sin(3b)]=0\]
\[[4\sin^3(b)-2\sin(b)][4\sin^3(b)-4\sin(b)]=0\]
\[\sin^2(b)\cos^2(b)[2\sin^2(b)-1]=0\]
然后就出来三个解了
\[2\sin^2(b)=0,1,2\]

转化与化归

回复 1# 青青子衿