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timofeeve [1]
3 years ago
11

Simplify the expression. tan(sin^−1 x)

Mathematics
2 answers:
Anna [14]3 years ago
8 0
Looking at that picture
we could use the pythagorean theorem to find the adjacent side,
or cosine value
\bf tan\left[ sin^{-1}(x) \right]\implies 
\begin{cases}
x=sin(\theta)\\
or\\
\cfrac{x}{1}=sin(\theta)\implies \cfrac{x=opposite=b}{1=radius=c}
\end{cases}\\
\\ \quad \\
c^2=a^2+b^2\implies \sqrt{c^2-b^2}=a
\\ \quad \\
\sqrt{1^2-x^2}=a=cos(\theta)\\
----------------------------\\
thus
\\ \quad \\
tan\left[ sin^{-1}(x) \right]\implies tan(\theta)\implies \cfrac{sin(\theta)}{cos(\theta)}\implies \cfrac{\frac{x}{1}}{\sqrt{1-x^2}}
\\ \quad \\
\cfrac{x}{\sqrt{1-x^2}}

Blizzard [7]3 years ago
3 0
If you're using the app, try seeing this answer through your browser:  brainly.com/question/2799412

_______________


Let  \mathsf{\theta=sin^{-1}(x)\qquad\qquad-\dfrac{\pi}{2}\ \textless \ \theta\ \textless \ \dfrac{\pi}{2}.}

(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)}


Square both sides:

\mathsf{sin^2\,\theta=x^2\qquad\qquad(but~sin^2\,\theta=1-cos^2\,\theta)}\\\\ \mathsf{1-cos^2\,\theta=x^2}\\\\ \mathsf{1-x^2=cos^2\,\theta}\\\\ \mathsf{cos^2\,\theta=1-x^2}


Since \mathsf{-\,\dfrac{\pi}{2}\ \textless \ \theta\ \textless \ \dfrac{\pi}{2},} then \mathsf{cos\,\theta} is positive. So take the positive square root and you get

\mathsf{cos\,\theta=\sqrt{1-x^2}\qquad\quad(ii)}


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.}


I hope this helps. =)


Tags:  <em>inverse trigonometric function sin tan arcsin trigonometry</em>

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