aiven $$ \sin \beta=\frac{8}{c 7} $$ where $$ 90^{\circ} \leq \beta \leq 270^{\circ} $$ dotormine witrad using a calechator a) $$ \cos 2 \beta $$ b) $$ \sin 2 \beta $$

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We are given
$$
\sin \beta=\frac{8}{17}
$$
with
$$
90^\circ\le\beta\le270^\circ.
$$
Because $$\sin\beta>0$$ and the angle is between $$90^\circ$$ and $$270^\circ$$, it follows that $$\beta$$ is in the second quadrant. (In the third quadrant, sine is negative.) In the second quadrant the cosine is negative. We now use the following steps.

**Step 1. Find $$\cos\beta$$.**

Using the Pythagorean identity
$$
\sin^2\beta+\cos^2\beta=1,
$$
we have
$$
\cos^2\beta=1-\sin^2\beta=1-\left(\frac{8}{17}\right)^2=1-\frac{64}{289}=\frac{289-64}{289}=\frac{225}{289}.
$$
Taking the square root, and choosing the negative sign (since cosine is negative in the second quadrant),
$$
\cos\beta=-\frac{15}{17}.
$$

**Step 2. Compute $$\cos 2\beta$$.**

One version of the double-angle formula is
$$
\cos2\beta=1-2\sin^2\beta.
$$
Substitute $$\sin\beta=\frac{8}{17}$$:
$$
\cos2\beta=1-2\left(\frac{8}{17}\right)^2=1-2\frac{64}{289}=1-\frac{128}{289}.
$$
Express 1 as $$\frac{289}{289}$$:
$$
\cos2\beta=\frac{289-128}{289}=\frac{161}{289}.
$$

**Step 3. Compute $$\sin 2\beta$$.**

The double-angle formula for sine is
$$
\sin2\beta=2\sin\beta\cos\beta.
$$
Substitute the known values:
$$
\sin2\beta=2\cdot\frac{8}{17}\cdot\left(-\frac{15}{17}\right)= -\frac{240}{289}.
$$

Thus, the answers are:

a) $$\cos 2\beta=\frac{161}{289}$$

b) $$\sin 2\beta=-\frac{240}{289}$$
a) $$ \cos 2\beta = \frac{161}{289} $$ b) $$ \sin 2\beta = -\frac{240}{289} $$

aiven \( \sin \beta=\frac{8}{c 7} \) where \( 90^{\circ} \leq \beta \leq 270^{\circ} \) dotormine witrad using a calechator a) \( \cos 2 \beta \) b) \( \sin 2 \beta \)

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