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Let \(\mathsf{\theta=cos^{-1}\!\left(\dfrac{4}{5}\right).}\)
\(\mathsf{0\le \theta\le\pi,}\) because that is the range of the inverse cosine funcition.
Also,
\(\mathsf{cos\,\theta=cos\!\left[cos^{-1}\!\left(\dfrac{4}{5}\right)\right]}\\\\\\
\mathsf{cos\,\theta=\dfrac{4}{5}}\\\\\\ \mathsf{5\,cos\,\theta=4}\)
Square both sides and apply the fundamental trigonometric identity:
\(\mathsf{(5\,cos\,\theta)^2=4^2}\\\\
\mathsf{5^2\,cos^2\,\theta=4^2}\\\\
\mathsf{25\,cos^2\,\theta=16\qquad\qquad(but,~cos^2\,\theta=1-sin^2\,\theta)}\\\\
\mathsf{25\cdot (1-sin^2\,\theta)=16}\)
\(\mathsf{25-25\,sin^2\,\theta=16}\\\\
\mathsf{25-16=25\,sin^2\,\theta}\\\\
\mathsf{9=25\,sin^2\,\theta}\\\\
\mathsf{sin^2\,\theta=\dfrac{9}{25}}\)
\(\mathsf{sin\,\theta=\pm\,\sqrt{\dfrac{9}{25}}}\\\\\\
\mathsf{sin\,\theta=\pm\,\sqrt{\dfrac{3^2}{5^2}}}\\\\\\
\mathsf{sin\,\theta=\pm\,\dfrac{3}{5}}\)
But \(\mathsf{0\le \theta\le\pi,}\) which means \(\theta\) lies either in the 1st or the 2nd quadrant. So \(\mathsf{sin\,\theta}\) is a positive number:
\(\mathsf{sin\,\theta=\dfrac{3}{5}}\\\\\\
\therefore~~\mathsf{sin\!\left[cos^{-1}\!\left(\dfrac{4}{5}\right)\right]=\dfrac{3}{5}\qquad\quad\checkmark}\)
I hope this helps. =)
Tags: inverse trigonometric function cosine sine cos sin trig trigonometry
