答案1：按公式tan(A+B) = (tanA+tanB)/(1-tanAtanB)展开 tan[(a+b)/2]=tan(a/2+b/2)=(tana/2+tanb/2)/(1-tana/2tanb/2) 答案2：tan(A/2+B/2)=(sinA+sinB)/(cosA+cosB) sinA+sinB =sin((A+B)/2+(A-B)/2)+sin((A+B)/2-(A-B)/2) =sin(A+B)/2 *cos(A-B)/2 cosA+cosB =cos((A+B)/2+(A-B)/2)+cos((A+B)/2-(A-B)/2) =cos(A+B)/2 *cos(A-B)/2 (sinA+sinB)/(cosA+cosB ...

tan(A/2) = (1–cosA)/sinA=sinA/(1+cosA) 和差化积. sin(a)+sin(b) = 2sin[(a+b)/2]cos[(a-b)/2] sin(a)-sin(b) = 2cos[(a+b)/2]sin[(a-b)/2] cos(a)+cos(b) = 2cos[(a+b)/2]cos[(a-b)/2] cos(a)-cos(b) = -2sin[(a+b)/2]sin[(a-b)/2] tanA+tanB=sin(A+B)/cosAcosB . 积化和差. sin(a)sin(b) = -1/2*[cos(a+b)-cos(a-b)] cos(a)cos(b) = 1/2*[cos(a+b)+cos(a-b)]

tan(a/2+b/2)=(tana/2+tanb/2)/(1-tana/2tanb/2) tan(a/2+b/2)=(sina+sinb)/(cosa+cosb) 分析过程如下： （1）按公式tan(A+B) = (tanA+tanB)/(1-tanAtanB)展开： …

In trigonometry, trigonometric identities are equalities that involve trigonometric functions and are true for every value of the occurring variables for which both sides of the equality are defined. Geometrically, these are identities involving certain functions of one or more angles.

三角函数基本关系倒数关系 \sec x = \frac{1}{\cos{x}} \csc{x} = \frac{1}{\sin{x}} \cot{x} = \frac{1}{\tan{x}} 平方关系 \sin^{2}x + \cos^{2}x = 1 1 + \tan^{2} x = \sec^{2}x 1 + \cot^{2}x = \csc^{2}x 商…

2021年9月5日 · 令 t=\tan{\frac{x}{2}} 则 \tan{x}=\frac{2tan\frac{x}{2}}{1-tan^2\frac{x}{2}}=\frac{2t}{1-t^2} \because\left(2t\right)^2+\left(1-t^2\right)^2=\left(1+t^2\right)^2. 根据 三角函数 的 几何意义 绘出 三边关系图 \therefore \sin{x}=\frac{2t}{1+t^2}\quad\cos{x}=\frac{1-t^2}{1+t^2} 而 x=2\arctan{t} 则 dx=\frac{2dt}{1+t^2}

Given Triangle abc, with angles A,B,C; a is opposite to A, b opposite B, c opposite C: a/sin(A) = b/sin(B) = c/sin(C) (Law of Sines) c ^2 = a ^2 + b ^2 - 2ab cos(C)

Begin by entering your mathematical expression into the above input field, or scanning it with your camera. Once you've entered the function and selected the operation, click the 'Go' button to generate the result. The calculator will instantly provide the solution to your trigonometry problem, saving you time and effort.

2019年11月25日 · In a triangle, if $\tan(A/2)$, $\tan(B/2)$, $\tan(C/2)$ are in arithmetic progression, then so are $\cos A$, $\cos B$, $\cos C$

2024年11月11日 · tanb公式，全称为正切双角公式，它的基本形式是： [ \tan (2b) = \frac {2\tan (b)} {1 - \tan^2 (b)} ] 这个公式在解决涉及角度和三角函数的问题时显得尤为有用。 下面我们来详细探讨tanb公式的应用和意义。 首先，tanb公式可以帮助我们快速计算出任意角度b的正切值的两倍。 这对于那些需要将角度翻倍，但直接计算较为复杂的情况非常有帮助。 例如，在解决一些几何问题时，我们可能需要知道一个角的正切值的两倍，tanb公式就能派上用场。 其次，tanb公式 …