We are given that $\sin(x) = 0.6$ and $x$ is in the first quadrant ($x \in (0, \pi/2)$). We need to find the value of $\sin(2x) + \cos(2x)$.
Find Cosine Value
First, find $\cos(x)$ using the Pythagorean identity $\sin^2(x) + \cos^2(x) = 1$. Since $x$ is in the first quadrant, $\cos(x)$ is positive.
- Calculate $\cos^2(x)$: $ \cos^2(x) = 1 - \sin^2(x) $ $ \cos^2(x) = 1 - (0.6)^2 = 1 - 0.36 = 0.64 $
- Find $\cos(x)$: $ \cos(x) = \sqrt{0.64} = 0.8 $
Calculate Double Angle Values
Now, use the double angle formulas to find $\sin(2x)$ and $\cos(2x)$.
- Calculate $\sin(2x)$: $ \sin(2x) = 2 \sin(x) \cos(x) $ $ \sin(2x) = 2 \times (0.6) \times (0.8) = 2 \times 0.48 = 0.96 $
- Calculate $\cos(2x)$: $ \cos(2x) = \cos^2(x) - \sin^2(x) $ $ \cos(2x) = (0.8)^2 - (0.6)^2 = 0.64 - 0.36 = 0.28 $
Compute Final Expression
Finally, add the values of $\sin(2x)$ and $\cos(2x)$:
$ \sin(2x) + \cos(2x) = 0.96 + 0.28 = 1.24 $
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The given equation can be reduced to
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