是的，tan150度等于-tan30度。解释如下：在三角函数中，正切函数具有周期性，其周期为180度。这意味着对于任何角度α，其正切值tanα和与其相隔180度的角度的正切值之间存在一种特殊关系。具体来说，tan = -tanα。这是因为正弦和余弦函数在周期内的正负变化导致了正切值的正负变化。...
tan(α) = h/l 其中，α表示坡度，h表示汽车爬升的高度，l表示斜坡的长度。如果我们知道了坡度，就可以通过tan和ctg计算出斜坡的角度。例如，当tan(α)=0.5时，我们可以用以下公式计算出坡度：α = arctan(0.5) ≈ 26.57度 同样地，我们也可以用ctg来计算斜坡的角...
积化和差公式: sin\alpha*sin\beta=(1/2)[cos(\alpha-\beta)-cos(\alpha+\beta)] cos\alpha*cos\beta=(1/2)[cos(\alpha+\beta)+cos(\alpha-\beta)] sin\alpha*cos\beta=(1/2)[sin(\alpha+\beta)+sin(\alpha-\beta)] 和差化积公式: sin\alpha+sin\beta=2sin[(\alpha+\beta)/2]*cos...
tan(π−α)=−tanα\tan(\pi - \alpha) = -\tan\alphatan(π−α)=−tanα 这个公式说明,当角度减少 π\piπ(即180度)时,tan的值会取反。 tan(2π+α)=tanα\tan(2\pi + \alpha) = \tan\alphatan(2π+α)=tanα 这个公式说明,当角度增加 2π2\pi2π(即360度)...
2.1.1220 Part 1 Section 20.1.2.3.3, alphaOff (Alpha Offset) 2.1.1221 Part 1 Section 20.1.2.3.4, blue (Blue) 2.1.1222 Part 1 Section 20.1.2.3.5, blueMod (Blue Modification) 2.1.1223 Part 1 Section 20.1.2.3.6, blueOff (Blue Offset) 2.1.1224 Part 1 Section 20.1.2.3.1...
By scanning the temperature during a DMA experiment the glass transition, or alpha relaxation, is reached in which the polymer segments are given enough mobility through thermal activation to easily move past each other and transition from a glassy to a rubbery material. This is characterized by ...
\tan\alpha 原式可利用诱导公式化简: sin^2(-α) = sin^2α,tan(360°-α) = -tanα,sin(180°-α) = sinα,cos(360°-α) = cosα,tan(180° α) = tanα。 代入原式并化简得: sin^2α tanα - sinαcosαtanα = sin^2α tanα - sin^2α = tanα。 因此,原式的值...
The inverse lifetime to lowest order in the fine-structure constant α is displaystylefrac1tau=2fracpi²-99pimₜₑₓₜₑalpha⁶, where mₑ is the mass of the electron. π is present in some structural engineering formulae, such as the buckling formula derived by Euler, which ...
5.0.0-alpha.8402 years ago 4.29.25952 years ago 5.0.0-alpha.8112 years ago 4.29.2352 years ago 4.29.2202 years ago 4.29.2102 years ago 5.0.0-alpha.7902 years ago 4.29.202962 years ago 4.29.19522 years ago 5.0.0-alpha.7222 years ago ...
理由:如图$2$中,$\because \angle BDA=\angle BAC=\alpha $,$\therefore \angle DBA+\angle BAD=\angle BAD+\angle CAE=180^{\circ}-\alpha $,$\therefore \angle DBA=\angle CAE$,在$\triangle ADB$和$\triangle CEA$中,$\left\{\begin{array}{l}{∠BDA=∠AEC}\\{∠DBA=∠CAE}\\{AB=AC...