题目

化简 sinα的三次方sin3α+cos αde 三次方cos3α
麻烦再帮忙找些关于二倍角、恒等式变换的题.
感激!

提问时间：2020-07-27

答案

sin³αsin3α+cos³αcos3α
=sin³α(3sinα-4sin³α)+cos³α(4cos³α-3cosα)
=3(sin^4α-cos^4α)-4(sin^6α-cos^6α)
=3(sin²α+cos²α)(sin²α-cos²α)-4(sin²α-cos²α)(sin^4α+cos^4α+sin²αcos²α)
=3(sin²α-cos²α)-4(sin²α-cos²α)[(sin²α+cos²α)²-sin²αcos²α]
=(sin²α-cos²α)[3-4(1-sin²αcos²α)]
=(sin²α-cos²α)(4sin²αcos²α-1)
=(cos²α-sin²α)(1-sin²2α)
=cos2α*cos²2α
=cos³2α