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The connection should be examined by the tracing a beam from contour and ultizing Snell’s rules

Сообщение от CN Etagy вкл 23.09.2022
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The connection should be examined by the tracing a beam from contour and ultizing Snell’s rules

To read it, you can find around three triangles: the bigger (environmentally friendly having pink area) has hypotenuse $1$ (and you may adjacent and contrary sides one means brand new hypotenuses of the almost every other a couple); the second biggest (yellow) hypotenuse $\cos(\beta)$ , adjoining front (off angle $\alpha$ ) $\cos(\beta)\cdot \cos(\alpha)$ , and you can opposite side $\cos(\beta)\cdot\sin(\alpha)$ ; and minuscule (pink) hypotenuse $\sin(\beta)$ , adjoining front side (of direction $\alpha$ ) $\sin(\beta)\cdot \cos(\alpha)$ , and you can other side $\sin(\beta)\sin(\alpha)$ .

With the proven fact that $\sin$ are an odd means and you may $\cos$ an even form, related formulas for the variation $\alpha – \beta$ are derived.

Aforementioned works out the latest Pythagorean choose, however, provides a without signal. In reality, the fresh new Pythagorean pick is normally familiar with write which, like $\cos(dos\alpha) = 2\cos(\alpha)^2 – 1$ or $step one – 2\sin(\alpha)^2$ .

Using the significantly more than with $\alpha = \beta/2$ , we get you to definitely $\cos(\beta) = dos\cos(\beta/dos)^2 -1$ , and that rearranged output the fresh “half-angle” formula: $\cos(\beta/2)^2 = (step one + \cos(\beta))/2$ .

Analogy

\cos((n+1)\theta) &= \cos(n\theta + \theta) = \cos(n\theta) \cos(\theta) – \sin(n\theta)\sin(\theta), \text< and>\\ \cos((n-1)\theta) &= \cos(n\theta – \theta) = \cos(n\theta) \cos(-\theta) – \sin(n\theta)\sin(-\theta). \end

That is the perspective having a multiple out-of $n+1$ shall be expressed with regards to the position that have a simultaneous out of $n$ and you may $n-1$ . Continue Reading

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