\tan45\degree=1 \therefore\sin89\degree<\tan45\degree \because\cos1\degree \therefore\frac{1}{\cos1\degree}>1 \therefore\sin89\degree<\tan45\degree<\frac{1}{\cos1\degree} 即a 观察题目,可将1作为比较对象, 比较可得a<1,b=1,c>1, 故a 反馈 收藏 ...
\root \of {3} (m+1) 相关知识点: 试题来源: 解析 此题答案为:B. 解:∵\tan 60\degree=\tan (28\degree+32\degree)=, ∴\tan 28\degree+\tan 32\degree=\tan 60\degree·(1-\tan 28\degree·tan32\degree), ∵\tan 28\degree\tan 32\degree=m, ∴\tan 28\degree+\tan 32\degree=\...
tan89.9999=1/tan0.00001 tan0.00001\degree=tan(0.00001*\pi /180)≈0.00001*\pi /180 这就...
因为\tan60\degree={\tan10\degree+\tan50\degree\over 1-\tan10\degree\tan50\degree}=\sqrt{3}, 所以\tan10\degree+\tan50\degree+\sqrt{3} \tan10\degree\tan50\degree=\sqrt{3}(1-\tan10\degree\tan50\degree) +\sqrt{3}\tan10\degree\tan50\degree=\sqrt{3}。 故本题正确答案为\sqrt...
2.1.1658 Part 1 Section 22.1.2.26, deg (Degree) 2.1.1659 Part 1 Section 22.1.2.27, degHide (Hide Degree) 2.1.1660 Part 1 Section 22.1.2.28, den (Denominator) 2.1.1661 Part 1 Section 22.1.2.30, dispDef (Use Display Math Defaults) 2.1.1662 Part 1 Section 22.1.2.32, e (...
那天觀課,同事教 trigonometric graphs。談到 tangent graph 在 90 度處斷開,著同學試 tan(89∘), tan(89.9∘), tan(89.99∘) 之類,可見結果愈來愈大,去到 90 就無限大云云。 N 日後,有學生問我,何解相鄰結果似乎有 10 倍變化?見下: ...
因为\tan(20\degree+40\degree)=\tan60\degree=\root \of {3} 所以\tan(20\degree+40\degree)=\frac{\tan20\degree+\tan40\degree}{1-\tan20\degree\tan40\degree}=\root \of {3} \tan20\degree+\tan40\degree=\root \of {3} (1-\tan20\degree\tan40\degree)=\root \of {3} -\roo...
题目\tan60\degree的值等于( )。A: 1 B: \sqrt2 C: \sqrt3 D: 2相关知识点: 试题来源: 解析 C 正确率: 85%, 易错项: B 本题主要考查三角函数。 由特殊角的三角函数可得:\tan 60\degree=\sqrt{3}。 故本题正确答案为C。反馈 收藏
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已知锐角\alpha 满足\tan( {\alpha+20\degree} )=1,则锐角\alpha 的度数为( )。A: 10\degree B: 25\degre