Question

Give the exact value of the expression without using a calculator.

Answer 1

Answer:

$\sin \left(\tan^{-1} \left(\frac{8}{15}\right)\right)=\frac{8}{17}$.

Step-by-step explanation:

To find the exact value of the expression $\sin(\tan^{-1}(\frac{8}{15} ))$

First, we need to simplify the expression $\sin(\tan^{-1}(x))$.

Draw a triangle in the plane with vertices $(1,x)$, $(1,0)$, and the origin. Then $\tan^{-1}(x)$ is the angle between the positive x-axis and the ray beginning at the origin and passing through $(1,x)$.

Therefore,

$\sin(\tan^{-1}(x))=\frac{x}{\sqrt{1+x^{2} } }$

$\mathrm{Multiply\:by\:the\:conjugate}\:\frac{\sqrt{1+x^2}}{\sqrt{1+x^2}}$

$\frac{x\sqrt{1+x^2}}{\sqrt{1+x^2}\sqrt{1+x^2}}$

$\sqrt{1+x^2}\sqrt{1+x^2}=1+x^2$

$\sin(\tan^{-1}(x))=\frac{x\sqrt{1+x^2}}{1+x^2}$

Now, use the identity $\sin(\tan^{-1}(x))=\frac{x\sqrt{1+x^2}}{1+x^2}$

$\sin(\tan^{-1}(\frac{8}{15} ))=\frac{\left(\frac{8}{15}\right)\sqrt{1+\left(\frac{8}{15}\right)^2}}{1+\left(\frac{8}{15}\right)^2}\\\\\frac{\frac{8}{15}\sqrt{\left(\frac{8}{15}\right)^2+1}}{1+\frac{8^2}{15^2}}\\\\\frac{\frac{136}{225}}{1+\frac{8^2}{15^2}}\\\\\frac{\frac{136}{225}}{1+\frac{64}{225}}\\\\\frac{136}{225\cdot \frac{289}{225}}\\\\\frac{136}{289}=\frac{8}{17}\\\\\sin \left(\tan^{-1} \left(\frac{8}{15}\right)\right)=\frac{8}{17}$