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If we have one zero of +5i, we also have to have its conjugate which is -5i. Therefore, the factors that we will FOIL to get that polynomial are (x+5i)(x-5i). That distribution results in 
. I'm sure by now you know that i squared is equal to -1, so -25(-1) = 25. So our polynomial is 
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Let
(that is the range of the inverse sine function).
So,
![\mathsf{sin\,\theta=sin\!\left[sin^{-1}(x)\right]}\\\\ \mathsf{sin\,\theta=x\qquad\quad(i)}](https://tex.z-dn.net/?f=%5Cmathsf%7Bsin%5C%2C%5Ctheta%3Dsin%5C%21%5Cleft%5Bsin%5E%7B-1%7D%28x%29%5Cright%5D%7D%5C%5C%5C%5C%20%5Cmathsf%7Bsin%5C%2C%5Ctheta%3Dx%5Cqquad%5Cquad%28i%29%7D)
Square both sides:
Since
then 
is positive. So take the positive square root and you get
Then,
![\mathsf{tan\,\theta=\dfrac{sin\,\theta}{cos\,\theta}}\\\\\\ \mathsf{tan\,\theta=\dfrac{x}{\sqrt{1-x^2}}}\\\\\\\\ \therefore~~\mathsf{tan\!\left[sin^{-1}(x)\right]=\dfrac{x}{\sqrt{1-x^2}}\qquad\qquad -1\ \textless \ x\ \textless \ 1.}](https://tex.z-dn.net/?f=%5Cmathsf%7Btan%5C%2C%5Ctheta%3D%5Cdfrac%7Bsin%5C%2C%5Ctheta%7D%7Bcos%5C%2C%5Ctheta%7D%7D%5C%5C%5C%5C%5C%5C%20%5Cmathsf%7Btan%5C%2C%5Ctheta%3D%5Cdfrac%7Bx%7D%7B%5Csqrt%7B1-x%5E2%7D%7D%7D%5C%5C%5C%5C%5C%5C%5C%5C%20%5Ctherefore~~%5Cmathsf%7Btan%5C%21%5Cleft%5Bsin%5E%7B-1%7D%28x%29%5Cright%5D%3D%5Cdfrac%7Bx%7D%7B%5Csqrt%7B1-x%5E2%7D%7D%5Cqquad%5Cqquad%20-1%5C%20%5Ctextless%20%5C%20x%5C%20%5Ctextless%20%5C%201.%7D)
I hope this helps. =)
Tags: <em>inverse trigonometric function sin tan arcsin trigonometry</em>
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Let
(that is the range of the inverse sine function).
So,
Square both sides:
Since
Then,
I hope this helps. =)
Tags: <em>inverse trigonometric function sin tan arcsin trigonometry</em>