Question
The function f is defined by f(x)= sin(x+4π/3). What is the value of f(3π/4)?
Answers
GPT-4o mini
To find the value of \( f(3\pi/4) \) for the function \( f(x) = \sin(x + 4\pi/3) \), we start by substituting \( x = 3\pi/4 \) into the function:
\[
f\left( \frac{3\pi}{4} \right) = \sin\left( \frac{3\pi}{4} + \frac{4\pi}{3} \right)
\]
Next, we need to calculate the argument of the sine function:
\[
\frac{3\pi}{4} + \frac{4\pi}{3}
\]
To add these fractions, we first find a common denominator. The denominators are 4 and 3, and the least common multiple is 12. We convert each fraction:
\[
\frac{3\pi}{4} = \frac{3\pi \cdot 3}{4 \cdot 3} = \frac{9\pi}{12}
\]
\[
\frac{4\pi}{3} = \frac{4\pi \cdot 4}{3 \cdot 4} = \frac{16\pi}{12}
\]
Now we can add the two fractions:
\[
\frac{9\pi}{12} + \frac{16\pi}{12} = \frac{25\pi}{12}
\]
Thus, we have:
\[
f\left( \frac{3\pi}{4} \right) = \sin\left( \frac{25\pi}{12} \right)
\]
Now, we simplify \( \frac{25\pi}{12} \). To find its equivalent angle within the principal range \( [0, 2\pi) \), we can subtract \( 2\pi \):
\[
2\pi = \frac{24\pi}{12}
\]
Now we do the subtraction:
\[
\frac{25\pi}{12} - \frac{24\pi}{12} = \frac{\pi}{12}
\]
Thus:
\[
\sin\left( \frac{25\pi}{12} \right) = \sin\left( \frac{\pi}{12} \right)
\]
Now we find \( \sin\left( \frac{\pi}{12} \right) \). We can use the sine half-angle identity to compute this:
\[
\sin\left( \frac{\pi}{12} \right) = \sin\left( 15^\circ \right) = \sin\left( 45^\circ - 30^\circ \right)
\]
Using the sine difference identity:
\[
\sin(a - b) = \sin a \cos b - \cos a \sin b
\]
where \( a = 45^\circ \) and \( b = 30^\circ \):
\[
\sin(15^\circ) = \sin(45^\circ) \cos(30^\circ) - \cos(45^\circ) \sin(30^\circ)
\]
Substituting the known sine and cosine values:
\[
\sin(45^\circ) = \frac{\sqrt{2}}{2}, \quad \cos(30^\circ) = \frac{\sqrt{3}}{2}, \quad \cos(45^\circ) = \frac{\sqrt{2}}{2}, \quad \sin(30^\circ) = \frac{1}{2}
\]
So we have:
\[
\sin(15^\circ) = \frac{\sqrt{2}}{2} \cdot \frac{\sqrt{3}}{2} - \frac{\sqrt{2}}{2} \cdot \frac{1}{2} = \frac{\sqrt{6}}{4} - \frac{\sqrt{2}}{4} = \frac{\sqrt{6} - \sqrt{2}}{4}
\]
Thus:
\[
f\left( \frac{3\pi}{4} \right) = \sin\left( \frac{\pi}{12} \right) = \frac{\sqrt{6} - \sqrt{2}}{4}
\]
Therefore, the value of \( f(3\pi/4) \) is
\[
\boxed{\frac{\sqrt{6} - \sqrt{2}}{4}}
\]
\[
f\left( \frac{3\pi}{4} \right) = \sin\left( \frac{3\pi}{4} + \frac{4\pi}{3} \right)
\]
Next, we need to calculate the argument of the sine function:
\[
\frac{3\pi}{4} + \frac{4\pi}{3}
\]
To add these fractions, we first find a common denominator. The denominators are 4 and 3, and the least common multiple is 12. We convert each fraction:
\[
\frac{3\pi}{4} = \frac{3\pi \cdot 3}{4 \cdot 3} = \frac{9\pi}{12}
\]
\[
\frac{4\pi}{3} = \frac{4\pi \cdot 4}{3 \cdot 4} = \frac{16\pi}{12}
\]
Now we can add the two fractions:
\[
\frac{9\pi}{12} + \frac{16\pi}{12} = \frac{25\pi}{12}
\]
Thus, we have:
\[
f\left( \frac{3\pi}{4} \right) = \sin\left( \frac{25\pi}{12} \right)
\]
Now, we simplify \( \frac{25\pi}{12} \). To find its equivalent angle within the principal range \( [0, 2\pi) \), we can subtract \( 2\pi \):
\[
2\pi = \frac{24\pi}{12}
\]
Now we do the subtraction:
\[
\frac{25\pi}{12} - \frac{24\pi}{12} = \frac{\pi}{12}
\]
Thus:
\[
\sin\left( \frac{25\pi}{12} \right) = \sin\left( \frac{\pi}{12} \right)
\]
Now we find \( \sin\left( \frac{\pi}{12} \right) \). We can use the sine half-angle identity to compute this:
\[
\sin\left( \frac{\pi}{12} \right) = \sin\left( 15^\circ \right) = \sin\left( 45^\circ - 30^\circ \right)
\]
Using the sine difference identity:
\[
\sin(a - b) = \sin a \cos b - \cos a \sin b
\]
where \( a = 45^\circ \) and \( b = 30^\circ \):
\[
\sin(15^\circ) = \sin(45^\circ) \cos(30^\circ) - \cos(45^\circ) \sin(30^\circ)
\]
Substituting the known sine and cosine values:
\[
\sin(45^\circ) = \frac{\sqrt{2}}{2}, \quad \cos(30^\circ) = \frac{\sqrt{3}}{2}, \quad \cos(45^\circ) = \frac{\sqrt{2}}{2}, \quad \sin(30^\circ) = \frac{1}{2}
\]
So we have:
\[
\sin(15^\circ) = \frac{\sqrt{2}}{2} \cdot \frac{\sqrt{3}}{2} - \frac{\sqrt{2}}{2} \cdot \frac{1}{2} = \frac{\sqrt{6}}{4} - \frac{\sqrt{2}}{4} = \frac{\sqrt{6} - \sqrt{2}}{4}
\]
Thus:
\[
f\left( \frac{3\pi}{4} \right) = \sin\left( \frac{\pi}{12} \right) = \frac{\sqrt{6} - \sqrt{2}}{4}
\]
Therefore, the value of \( f(3\pi/4) \) is
\[
\boxed{\frac{\sqrt{6} - \sqrt{2}}{4}}
\]