19) sin (cos^(-1) 2/3 + tan^(-1) 1)

Answer

Explanation:

Step1: Recall trig - inverse function values

Let $\alpha=\cos^{-1}\frac{2}{3}$, then $\cos\alpha = \frac{2}{3}$ and $\sin\alpha=\sqrt{1 - \cos^{2}\alpha}=\sqrt{1 - (\frac{2}{3})^{2}}=\frac{\sqrt{9 - 4}}{3}=\frac{\sqrt{5}}{3}$. Also, $\tan^{-1}1=\frac{\pi}{4}$, so $\sin(\tan^{-1}1)=\cos(\tan^{-1}1)=\frac{\sqrt{2}}{2}$.

Step2: Use the sum - formula for sine

The sum - formula for sine is $\sin(A + B)=\sin A\cos B+\cos A\sin B$. Here $A = \cos^{-1}\frac{2}{3}$ and $B=\tan^{-1}1$. $\sin\left(\cos^{-1}\frac{2}{3}+\tan^{-1}1\right)=\sin\left(\cos^{-1}\frac{2}{3}\right)\cos\left(\tan^{-1}1\right)+\cos\left(\cos^{-1}\frac{2}{3}\right)\sin\left(\tan^{-1}1\right)$.

Step3: Substitute the values

Substitute $\sin\left(\cos^{-1}\frac{2}{3}\right)=\frac{\sqrt{5}}{3}$, $\cos\left(\tan^{-1}1\right)=\frac{\sqrt{2}}{2}$, $\cos\left(\cos^{-1}\frac{2}{3}\right)=\frac{2}{3}$, and $\sin\left(\tan^{-1}1\right)=\frac{\sqrt{2}}{2}$ into the above formula. $\sin\left(\cos^{-1}\frac{2}{3}+\tan^{-1}1\right)=\frac{\sqrt{5}}{3}\times\frac{\sqrt{2}}{2}+\frac{2}{3}\times\frac{\sqrt{2}}{2}=\frac{\sqrt{10} + 2\sqrt{2}}{6}$.

Answer:

$\frac{\sqrt{10}+2\sqrt{2}}{6}$