cos1290的知识精选

求值:sin(-1200°)·cos1290°+cos(-1020°)·sin(-1050°)+tan945°...
问题详情:求值:sin(-1200°)·cos1290°+cos(-1020°)·sin(-1050°)+tan945°.【回答】解:原式=-sin1200°·cos1290°+cos1020°·(-sin1050°)+tan945°=-sin120°·cos210°+cos300°·(-sin330°)+tan225°=(-sin60°)·(-cos30°)+cos6...
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