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sin(-19π/6)=;sin(-16π/3)=;cos(-79π/6)=;tan(-26π/3)=;sin(15π/4)=;
cos(-60°)-sin(-210°)=?
人气:151 ℃ 时间:2020-03-14 22:40:14
解答
sin(-19π/6)=-sin(2pai+pai+pai/6)=1/2 sin(-16π/3)=-sin(4pai+pai+1/3pai)=根号3/2cos(-79π/6)=cos(79pai/6)=cos(12pai+pai+pai/6)=-根号3/2 tan(-26π/3)=-tan(8pai+2pai/3)=-tan(pai-pai/3)=tan(pai/...
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