Answer:
[tex]\sin \left(\tan^{-1} \left(\frac{8}{15}\right)\right)=\frac{8}{17}[/tex].
Step-by-step explanation:
To find the exact value of the expression [tex]\sin(\tan^{-1}(\frac{8}{15} ))[/tex]
First, we need to simplify the expression [tex]\sin(\tan^{-1}(x))[/tex].
Draw a triangle in the plane with vertices [tex](1,x)[/tex], [tex](1,0)[/tex], and the origin. Then [tex]\tan^{-1}(x)[/tex] is the angle between the positive x-axis and the ray beginning at the origin and passing through [tex](1,x)[/tex].
Therefore,
[tex]\sin(\tan^{-1}(x))=\frac{x}{\sqrt{1+x^{2} } }[/tex]
[tex]\mathrm{Multiply\:by\:the\:conjugate}\:\frac{\sqrt{1+x^2}}{\sqrt{1+x^2}}[/tex]
[tex]\frac{x\sqrt{1+x^2}}{\sqrt{1+x^2}\sqrt{1+x^2}}[/tex]
[tex]\sqrt{1+x^2}\sqrt{1+x^2}=1+x^2[/tex]
[tex]\sin(\tan^{-1}(x))=\frac{x\sqrt{1+x^2}}{1+x^2}[/tex]
Now, use the identity [tex]\sin(\tan^{-1}(x))=\frac{x\sqrt{1+x^2}}{1+x^2}[/tex]
[tex]\sin(\tan^{-1}(\frac{8}{15} ))=\frac{\left(\frac{8}{15}\right)\sqrt{1+\left(\frac{8}{15}\right)^2}}{1+\left(\frac{8}{15}\right)^2}\\\\\frac{\frac{8}{15}\sqrt{\left(\frac{8}{15}\right)^2+1}}{1+\frac{8^2}{15^2}}\\\\\frac{\frac{136}{225}}{1+\frac{8^2}{15^2}}\\\\\frac{\frac{136}{225}}{1+\frac{64}{225}}\\\\\frac{136}{225\cdot \frac{289}{225}}\\\\\frac{136}{289}=\frac{8}{17}\\\\\sin \left(\tan^{-1} \left(\frac{8}{15}\right)\right)=\frac{8}{17}[/tex]
