化简三角函数1-sin4Q-cos4Q1-sin6Q-sin6Q 化简注意,其中的4和6是次方
问题描述：
化简三角函数
1-sin4Q-cos4Q
1-sin6Q-sin6Q 化简
注意,其中的4和6是次方
答
1-(sinQ)^4-(cosQ)^4
=1-[(sinQ)^4+(cosQ)^4]
=1-[(sin²Q+cos²Q)²-2sin²Qcos²Q]
=1-[1²-1/2(2sinQcosQ)²]
=1-1+1/2sin²2Q
=1/2sin²2Q
1-(sinQ)^6-(cosQ)^6
=1-[(sinQ)^6+(cosQ)^6]
=1-(sin²Q+cos²Q)[(sin²Q)²-sin²Qcos²Q+(cos²Q)²]
=1-1[(sin²Q+cos²Q)²-3sin²Qcos²Q]
=1-[1-3/4(2sinQcosQ)²]
=1-1+3/4sin²2Q
=3/4sin²2Q
答
1-sinQ^4-cosQ^4
=(sinQ^2+cosQ^2)^2-sinQ^4-cosQ^4
=2sinQ^2cosQ^2
1-sinQ^6-cosQ^6
=(sinQ^2+cosQ^2)^3-sinQ^6-cosQ^6
=3sinQ^4cosQ^2+3sinQ^2cosQ^4
=3sinQ^2cosQ^2(sinQ^2+cosQ^2)
=3sinQ^2cosQ^2