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Contenu principal

Formulaire de trigonométrie

Toutes les formules (ou presque)

Les inverses du cosinus, du sinus et de la tangente

\sec, left parenthesis, theta, right parenthesis, equals, start fraction, 1, divided by, cosine, left parenthesis, theta, right parenthesis, end fraction

\csc, left parenthesis, theta, right parenthesis, equals, start fraction, 1, divided by, sine, left parenthesis, theta, right parenthesis, end fraction

cotangent, left parenthesis, theta, right parenthesis, equals, start fraction, 1, divided by, tangent, left parenthesis, theta, right parenthesis, end fraction

tangent, left parenthesis, theta, right parenthesis, equals, start fraction, sine, left parenthesis, theta, right parenthesis, divided by, cosine, left parenthesis, theta, right parenthesis, end fraction

cotangent, left parenthesis, theta, right parenthesis, equals, start fraction, cosine, left parenthesis, theta, right parenthesis, divided by, sine, left parenthesis, theta, right parenthesis, end fraction

Avec des carrés

sine, squared, left parenthesis, theta, right parenthesis, plus, cosine, squared, left parenthesis, theta, right parenthesis, equals, 1, squared
tangent, squared, left parenthesis, theta, right parenthesis, plus, 1, squared, equals, \sec, squared, left parenthesis, theta, right parenthesis
cotangent, squared, left parenthesis, theta, right parenthesis, plus, 1, squared, equals, \csc, squared, left parenthesis, theta, right parenthesis

Les autres formules

On déduit les formules de l'angle double des formules d'addition.
Les formules d'addition
sin(θ+ϕ)=sinθcosϕ+cosθsinϕsin(θϕ)=sinθcosϕcosθsinϕcos(θ+ϕ)=cosθcosϕsinθsinϕcos(θϕ)=cosθcosϕ+sinθsinϕ\begin{aligned} \sin(\theta+\phi)&=\sin\theta\cos\phi+\cos\theta\sin\phi\\\\ \sin(\theta-\phi)&=\sin\theta\cos\phi-\cos\theta\sin\phi\\\\ \cos(\theta+\phi)&=\cos\theta\cos\phi-\sin\theta\sin\phi\\\\ \cos(\theta-\phi)&=\cos\theta\cos\phi+\sin\theta\sin\phi \end{aligned}
tan(θ+ϕ)=tanθ+tanϕ1tanθtanϕtan(θϕ)=tanθtanϕ1+tanθtanϕ\begin{aligned} \tan(\theta+\phi)&=\dfrac{\tan\theta+\tan\phi}{1-\tan\theta\tan\phi}\\\\ \tan(\theta-\phi)&=\dfrac{\tan\theta-\tan\phi}{1+\tan\theta\tan\phi} \end{aligned}
Les formules de l'angle double
sine, left parenthesis, 2, theta, right parenthesis, equals, 2, sine, theta, cosine, theta
cosine, left parenthesis, 2, theta, right parenthesis, equals, 2, cosine, squared, theta, minus, 1
tangent, left parenthesis, 2, theta, right parenthesis, equals, start fraction, 2, tangent, theta, divided by, 1, minus, tangent, squared, theta, end fraction
Les formules de l'angle moitié
sinθ2=±1cosθ2cosθ2=±1+cosθ2tanθ2=±1cosθ1+cosθ=1cosθsinθ=sinθ1+cosθ\begin{aligned} \sin\dfrac\theta2&=\pm\sqrt{\dfrac{1-\cos\theta}{2}}\\\\ \cos\dfrac\theta2&=\pm\sqrt{\dfrac{1+\cos\theta}{2}}\\\\ \tan\dfrac{\theta}{2}&=\pm\sqrt{\dfrac{1-\cos\theta}{1+\cos\theta}}\\ \\ &=\dfrac{1-\cos\theta}{\sin\theta}\\ \\ &=\dfrac{\sin\theta}{1+\cos\theta}\end{aligned}

Angles associés

sine, left parenthesis, minus, theta, right parenthesis, equals, minus, sine, left parenthesis, theta, right parenthesis
cosine, left parenthesis, minus, theta, right parenthesis, equals, plus, cosine, left parenthesis, theta, right parenthesis
tangent, left parenthesis, minus, theta, right parenthesis, equals, minus, tangent, left parenthesis, theta, right parenthesis
sin(θ+2π)=sin(θ)cos(θ+2π)=cos(θ)tan(θ+π)=tan(θ)\begin{aligned} \sin(\theta+2\pi)&=\sin(\theta)\\\\ \cos(\theta+2\pi)&=\cos(\theta)\\\\ \tan(\theta+\pi)&=\tan(\theta) \end{aligned}

Formules faisant intervenir π, slash, 2

sinθ=cos(π2θ)cosθ=sin(π2θ)tanθ=cot(π2θ)cotθ=tan(π2θ)secθ=csc(π2θ)cscθ=sec(π2θ)\begin{aligned} \sin\theta&= \cos\left(\dfrac{\pi}{2}-\theta\right)\\\\ \cos\theta&= \sin\left(\dfrac{\pi}{2}-\theta\right)\\\\ \tan\theta&= \cot\left(\dfrac{\pi}{2}-\theta\right)\\\\ \cot\theta&= \tan\left(\dfrac{\pi}{2}-\theta\right)\\\\ \sec\theta&= \csc\left(\dfrac{\pi}{2}-\theta\right)\\\\ \csc\theta&= \sec\left(\dfrac{\pi}{2}-\theta\right) \end{aligned}

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  • piceratops seed style l'avatar de l’utilisateur raphaelangeles13
    Someone can help me, I can't find the way to find 2tan(2t) from tan((2π/4)+t) - tan((π/4)-t) . Thanks for the answer. :)
    (1 vote)
    Default Khan Academy avatar l'avatar de l’utilisateur
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