Inverse Sine Function Definition and Range

Use the half-angle identities

$$\sin^2\frac{\theta}{2}=\frac{1-\cos\theta}{2}$$

and

$$\cos^2\frac{\phi}{2}=\frac{1+\cos\phi}{2}$$

Let

$$\theta=\arccos\left(\frac35\right)$$

Then

$$\cos\theta=\frac35$$

Hence

$$\sin^2\left(\frac{\theta}{2}\right)=\frac{1-\frac35}{2}$$

$$=\frac{\frac25}{2}$$

$$=\frac15$$

Now let

$$\phi=\arcsin\left(\frac{15}{17}\right)$$

Since

$$\phi\in\left[-\frac{\pi}{2},\frac{\pi}{2}\right]$$

we have

$$\cos\phi=\sqrt{1-\sin^2\phi}$$

$$=\sqrt{1-\left(\frac{15}{17}\right)^2}$$

$$=\sqrt{\frac{64}{289}}$$

$$=\frac{8}{17}$$

Therefore

$$\cos^2\left(\frac{\phi}{2}\right)=\frac{1+\frac{8}{17}}{2}$$

$$=\frac{\frac{25}{17}}{2}$$

$$=\frac{25}{34}$$

Thus

$$\sin^2\left(\frac12\arccos\left(\frac35\right)\right)+\cos^2\left(\frac12\arcsin\left(\frac{15}{17}\right)\right)$$

$$=\frac15+\frac{25}{34}$$

$$=\frac{34+125}{170}$$

$$=\frac{159}{170}$$

Hence

$$E=170\cdot\frac{159}{170}$$

$$=159$$

Final Answer :

$$159$$