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Maths::Old::Trigo
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09/16/2024
\(\cos(2a)\)
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Avec \(\tan x\) , \(\cos^{2}(x)=\)
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\(\sin(a+b)=\)
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\(\sin(a-b)=\)
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\(\mathrm{cos}\big(\mathrm{a+b}\big)=\)
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\(\mathrm{cos}\big(\mathrm{a-b}\big)=\)
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\(\mathrm{tan}\bigl(a+b\bigr)=\)
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\(\tan (a-b)=\)
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\(\sin(a)\sin(b)=\)
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\(\cos(a)\cos(b)=\)
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\(\sin(a)\cos(b)=\)
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\(\sin(2a)=\)
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\(\tan(2a)=\)
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Avec \(\cos x\) , \(\sin^{2}({ a})=\)
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Avec \(\cos x\) ,\(\cos^{2}(a)=\)
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Avec \(\cos x\) ,\(\mathrm{tan}^{2}(\mathrm{a})\,=\)
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Euler : \(\sin(x)=\)
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Euler : \(\cos(x)=\)
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\(\cos(\theta + \pi) =\)
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\(\sin(\theta + \pi) =\)
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\(\cos(\pi - \theta) =\)
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\(\sin(\pi - \theta) =\)
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\(\cos\left(\theta + \frac{\pi}{2}\right) =\)
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\(\sin\left(\theta + \frac{\pi}{2}\right) =\)
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\(\cos\left(\frac{\pi}{2} - \theta\right) =\)
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\(\sin\left(\frac{\pi}{2} - \theta\right) =\)
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Soit \(a,b\in \mathbb{R}\) non tous les deux nuls. Alors il existe un unique couple\(\left( M, \varphi \right) \in \mathbb{R}^{+*} \times \left[ 0,2 \…
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\(\sin p- \sin q=\)
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\(\sin p+ \sin q=\)
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\(\cos p- \cos q=\)
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\(\cos p+ \cos q=\)
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Pour tout \(x \in \mathbb{R}\) on a \(\left| \sin x \right| \leq\)
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\(\tan \left( \theta+ \pi \right)=\)
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\(\tan \left(- \theta \right)=\)
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\(\tan \left( \pi- \theta \right)=\)
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\(\left \{ \begin{matrix} \cos \theta= \cos \theta^{ \prime} \\ \sin \theta= \sin \theta^{ \prime} \end{matrix} \right. \Leftrightarrow\)
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