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