Of course! Let's solve the problem step by step, making use of trigonometric identities.
Given:
[tex]\[
\sin(\theta) = \frac{3}{5} \quad \text{and} \quad \cos(\theta) = \frac{4}{5}
\][/tex]
### Finding [tex]\(\sin(2\theta)\)[/tex]
We will use the double-angle formula for sine:
[tex]\[
\sin(2\theta) = 2 \sin(\theta) \cos(\theta)
\][/tex]
Substitute the given values:
[tex]\[
\sin(2\theta) = 2 \left(\frac{3}{5}\right) \left(\frac{4}{5}\right)
\][/tex]
Calculate the product inside the parentheses first:
[tex]\[
\frac{3}{5} \times \frac{4}{5} = \frac{12}{25}
\][/tex]
Then multiply by 2:
[tex]\[
2 \times \frac{12}{25} = \frac{24}{25}
\][/tex]
Therefore,
[tex]\[
\sin(2\theta) = \frac{24}{25} = 0.96
\][/tex]
### Finding [tex]\(\cos(2\theta)\)[/tex]
We will use the double-angle formula for cosine:
[tex]\[
\cos(2\theta) = \cos^2(\theta) - \sin^2(\theta)
\][/tex]
First, compute [tex]\(\cos^2(\theta)\)[/tex] and [tex]\(\sin^2(\theta)\)[/tex]:
[tex]\[
\cos^2(\theta) = \left(\frac{4}{5}\right)^2 = \frac{16}{25}
\][/tex]
[tex]\[
\sin^2(\theta) = \left(\frac{3}{5}\right)^2 = \frac{9}{25}
\][/tex]
Then subtract [tex]\(\sin^2(\theta)\)[/tex] from [tex]\(\cos^2(\theta)\)[/tex]:
[tex]\[
\cos(2\theta) = \frac{16}{25} - \frac{9}{25} = \frac{7}{25}
\][/tex]
Therefore,
[tex]\[
\cos(2\theta) = \frac{7}{25} = 0.28
\][/tex]
Putting it all together, we find:
[tex]\[
\sin(2\theta) = 0.96 \quad \text{and} \quad \cos(2\theta) = 0.28
\][/tex]
So, the values are:
[tex]\[
\sin(2\theta) = 0.96 \quad \text{and} \quad \cos(2\theta) = 0.28000000000000014
\][/tex]
