Rewriting Trigonometric Functions

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Cegep1

Mathematics

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299 words

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2 minutes

$\cos (\arcsin (x + 1))$

\begin{aligned} Let θ & = \arcsin (x + 1) \\ \sin θ & = x + 1 \end{aligned}

\begin{aligned} Let c & = adjacent side of θ \\ 1^{2} & = c^{2} + (x + 1)^{2} \\ c^{2} & = 1 - (x + 1)^{2} \\ c & = \pm \sqrt{1 - (x + 1)^{2}} \end{aligned}

\begin{matrix} Note that \cos θ \geq 0 & \cos θ = \frac{c}{1} = c \\ ∴ c > 0 \end{matrix}

\begin{aligned} \cos (\arcsin (x + 1)) & = \cos θ \\ = c \\ = \sqrt{1 - (x + 1)^{2}} \end{aligned}

$\sin (2 \arccos (e^{x}))$

\begin{aligned} Let θ & = \arccos (e^{x}) \\ \cos θ & = e^{x} \end{aligned}

\begin{aligned} Let c & = opposite side of θ \\ 1^{2} & = c^{2} + (e^{x})^{2} \\ c^{2} & = 1 - e^{2 x} \\ c & = \pm \sqrt{1 - e^{2 x}} \end{aligned}

\begin{matrix} Note that \sin θ \geq 0 and \sin θ = \frac{c}{1} = c \\ ∴ c > 0 \end{matrix}

\begin{aligned} \sin (2 \arccos (e^{x})) & = \sin (2 θ) \\ = 2 \sin θ \cos θ \\ = 2 c e^{x} \\ = 2 \sqrt{1 - e^{2 x}} e^{x} \end{aligned}

$\sec (\arctan (\frac{1}{x}))$

\begin{aligned} Let θ & = \arctan (\frac{1}{x}) \\ \tan θ & = \frac{1}{x} \\ Thus 1 & = opposite side of θ \\ x & = adjacent side of θ \end{aligned}

\begin{matrix} Let c = hypotenuse of θ \\ Per Pythagorean Theorem, c = \sqrt{1^{2} + x^{2}} \end{matrix}

\begin{matrix} Since \sec θ > 0 and \sec θ = \frac{c}{x}, \\ \frac{c}{x} > 0 \\ Note that x \neq 0 because \frac{1}{x} is defined \\ Thus, x > 0 ⟹ c > 0 & x < 0 ⟹ c < 0 \end{matrix}

\begin{aligned} \sec (\arctan (\frac{1}{x})) & = \sec θ \\ = \frac{c}{x} \\ = {\begin{cases} \frac{\sqrt{1 + x^{2}}}{x} & x > 0 \\ - \frac{\sqrt{1 + x^{2}}}{x} & x < 0 \end{cases} \end{aligned}

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