Question

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！！！
如果 找得好，，再加分。。。

Answer 1

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

Answer 2

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

Answer 3

你的要求有点高，并且好象以前你关闭问题的比率太高，将近50%，
所以，没空理你

Answer 4

=（sin方阿尔法-cos方阿尔法）（3-4（1-sin方阿尔法cos方阿尔法））

Answer 5

才5分.................