Find $\sin(A/2)$ given $\sin(A) = 4/5$ and $0^\circ < A < 360^\circ$ in the first quadrant.

STEP 1

Assumptions1. The sine of angle A is4/5. Angle A is in the first quadrant (Q I)
3. We are asked to find the sine of A/We will use the half-angle formula for sine, which is given by$$\sin \left(\frac{A}{}\right)=\pm \sqrt{\frac{1-\cos(A)}{}}$$3The sign of the square root depends on the quadrant of A/. Since A is in the first quadrant (Q I), A/ will also be in the first quadrant. In the first quadrant, sine is positive. So, we will use the positive square root.
4We don't know the cosine of A directly, but we can find it using the Pythagorean identity$$\cos^(A) =1 - \sin^(A)$$5Substitute the given value of sin(A) into the Pythagorean identity to find cos^(A):
\cos^(A) =1 - \left(\frac{4}{5}\right)^6Calculate the value of cos^(A):
$$\cos^(A) =1 - \left(\frac{16}{25}\right) = \frac{9}{25}$$7Since A is in the first quadrant where cosine is positive, we take the positive square root to find cos(A):
$$\cos(A) = \sqrt{\frac{9}{25}} = \frac{3}{5}$$8Now substitute the value of cos(A) into the half-angle formula$$\sin \left(\frac{A}{}\right)=\sqrt{\frac{1-\frac{3}{5}}{}}$$9Calculate the value inside the square root$$\sin \left(\frac{A}{}\right)=\sqrt{\frac{}{5}}$$10implify the square root to find the final answer$$\sin \left(\frac{A}{}\right)=\frac{\sqrt{}}{\sqrt{5}}$$The sine of A/ is $\frac{\sqrt{}}{\sqrt{5}}$.