【三角函数(2023:41:22)已知2sina=cosa,求2sin2a+3sinacosa-cos2a的值】
<p>问题：【三角函数(2023:41:22)已知2sina=cosa,求2sin2a+3sinacosa-cos2a的值】<p>答案：↓↓↓<p class="nav-title mt10" style="border-top:1px solid #ccc;padding-top: 10px;">郭煜的回答：<div class="content-b">网友采纳 2sin2a+3sinacosa-cos2a =4sinacosa+3sinacosa-(cosa2-sina2) =sina2-cosa2+7sinacosa 因为2sina=cosa 所以原式就=sina2-4sina2+7sina*2sina =11sina2
页:
[1]