Question

Write the given expression in terms of x and y only. sin(sin−1(x) + cos−1(y))

Answer 1

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Write the expression below in terms of x and y only:

(I'm going to call it "E")

$\mathsf{E=sin\!\left[sin^{-1}(x)+cos^{-1}(y)\right]\qquad\quad(i)}$


Let

$\begin{array}{lcl} \mathsf{\alpha=sin^{-1}(x)}&\qquad&\mathsf{then~-\,\dfrac{\pi}{2}\le \alpha\le \dfrac{\pi}{2}}\\\\\\ \mathsf{\beta=cos^{-1}(x)}&\qquad&\mathsf{then~0\le \beta\le \pi.} \end{array}$


so the expression becomes

$\mathsf{E=sin(\alpha+\beta)}\\\\ \mathsf{E=sin\,\alpha\,cos\,\beta+sin\,\beta\,cos\,\alpha\qquad\quad(ii)}$


•   Finding $\mathsf{sin\,\alpha:}$

$\mathsf{sin\,\alpha=sin\!\left[sin^{-1}(x)\right]}\\\\ \mathsf{sin\,\alpha=x\qquad\quad\checkmark}$


•   Finding $\mathsf{cos\,\alpha:}$

$\mathsf{sin^2\,\alpha=x^2}\\\\ \mathsf{1-cos^2\,\alpha=x^2}\\\\ \mathsf{cos^2\,\alpha=1-x^2}\\\\ \mathsf{cos\,\alpha=\sqrt{1-x^2}\qquad\quad\checkmark}$


because $\mathsf{cos\,\alpha}$ is positive for $\mathsf{\alpha\in \left[-\frac{\pi}{2},\,\frac{\pi}{2}\right].}$


•   Finding $\mathsf{cos\,\beta:}$

$\mathsf{cos\,\beta=cos\!\left[cos^{-1}(y)\right]}\\\\ \mathsf{cos\,\beta=y\qquad\quad\checkmark}$


•   Finding $\mathsf{sin\,\beta:}$

$\mathsf{cos^2\,\alpha=y^2}\\\\ \mathsf{1-sin^2\,\beta=y^2}\\\\ \mathsf{sin^2\,\beta=1-y^2}\\\\ \mathsf{sin\,\beta=\sqrt{1-y^2}\qquad\quad\checkmark}$


because $\mathsf{sin\,\beta}$ is positive for $\mathsf{\beta\in [0,\,\pi].}$


Finally, you get

$\mathsf{E=x\cdot y +\sqrt{1-y^2}\cdot \sqrt{1-x^2}}\\\\\\ \therefore~~\mathsf{sin\!\left[sin^{-1}(x)+cos^{-1}(y)\right]=x\cdot y +\sqrt{1-y^2}\cdot \sqrt{1-x^2}\qquad\quad\checkmark}$


I hope this helps. =)


Tags:   inverse trigonometric trig function sine cosine sin cos arcsin arccos sum angles trigonometry