Respuesta :

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[tex]0<\dfrac\pi{12}<\dfrac\pi6[/tex] and [tex]\sin x[/tex] is monotonic on [tex]0<x<\dfrac\pi2[/tex], so we know

[tex]\sin0<\sin\dfrac\pi{12}<\sin\dfrac\pi6\iff0<\sin\dfrac\pi{12}<\dfrac12[/tex]

The important thing here is that [tex]\sin\dfrac\pi{12}[/tex] must be positive.

Recall the half-angle identity:

[tex]\sin^2\dfrac\pi{12}=\dfrac{1-\cos\frac\pi6}2\implies\sin\dfrac\pi{12}=\sqrt{\dfrac{1-\cos\frac\pi6}2}[/tex]

(as opposed to the negative square root)

So we have

[tex]\sin\dfrac\pi{12}=\sqrt{\dfrac{1-\frac{\sqrt3}2}2}=\sqrt{\dfrac12-\dfrac{\sqrt3}4}=\dfrac{\sqrt3-1}{2\sqrt2}[/tex]

kudzordzifrancis

Answer:

[tex]\sin (\frac{\pi}{12})=\frac{\sqrt{6}-\sqrt{2}}{4}[/tex]

Step-by-step explanation:

We want to find the exact value of

[tex]\sin \frac{\pi}{12}[/tex]

We rewrite the given expression to obtain:

[tex]\sin \frac{\pi}{12}=\sin (\frac{\pi}{3}-\frac{\pi}{4})[/tex]

We use the identity:

[tex]\sin (A-B)=\sin A \cos B-\sin B \cos A[/tex]

[tex]\sin ( \frac{\pi}{3}- \frac{\pi}{4})=\sin \frac{\pi}{3} \cos \frac{\pi}{4}-\sin \frac{\pi}{4} \cos \frac{\pi}{3}[/tex]

[tex]\sin ( \frac{\pi}{3}- \frac{\pi}{4})=\frac{\sqrt{3}}{2} \times \frac{\sqrt{2}}{2}-\frac{\sqrt{2}}{2} \times \frac{1}{2}[/tex]

[tex]\sin ( \frac{\pi}{3}- \frac{\pi}{4})=\frac{\sqrt{6}-\sqrt{2}}{4}[/tex]

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