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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
