L(x) = -sin(7pi/2)(x – 7pi/2) + cos(7pi/2)
The Correct Answer and Explanation is:
The function you’ve provided is:L(x)=−sin(7π2)(x−7π2)+cos(7π2)L(x) = -\sin\left(\frac{7\pi}{2}\right) \left( x – \frac{7\pi}{2} \right) + \cos\left(\frac{7\pi}{2}\right)L(x)=−sin(27π)(x−27π)+cos(27π)
To analyze and simplify this expression, let’s start by evaluating the sine and cosine functions at 7π2\frac{7\pi}{2}27π.
Step 1: Evaluate sin(7π2)\sin\left(\frac{7\pi}{2}\right)sin(27π) and cos(7π2)\cos\left(\frac{7\pi}{2}\right)cos(27π)
The angle 7π2\frac{7\pi}{2}27π is greater than 2π2\pi2π, so it corresponds to an angle that has gone around the unit circle more than once.
- sin(7π2)\sin\left(\frac{7\pi}{2}\right)sin(27π):
Since 7π2=2π+3π2\frac{7\pi}{2} = 2\pi + \frac{3\pi}{2}27π=2π+23π, this is effectively equivalent to the angle 3π2\frac{3\pi}{2}23π on the unit circle. The sine of 3π2\frac{3\pi}{2}23π is −1-1−1, so: sin(7π2)=−1\sin\left(\frac{7\pi}{2}\right) = -1sin(27π)=−1 - cos(7π2)\cos\left(\frac{7\pi}{2}\right)cos(27π):
Similarly, 7π2\frac{7\pi}{2}27π is coterminal with 3π2\frac{3\pi}{2}23π, so the cosine of 3π2\frac{3\pi}{2}23π is 000. Hence: cos(7π2)=0\cos\left(\frac{7\pi}{2}\right) = 0cos(27π)=0
Step 2: Simplify the Expression
Now, substitute these values into the original expression for L(x)L(x)L(x):L(x)=−(−1)(x−7π2)+0L(x) = -(-1) \left( x – \frac{7\pi}{2} \right) + 0L(x)=−(−1)(x−27π)+0L(x)=(x−7π2)L(x) = \left( x – \frac{7\pi}{2} \right)L(x)=(x−27π)
So the simplified function is:L(x)=x−7π2L(x) = x – \frac{7\pi}{2}L(x)=x−27π
Step 3: Interpretation
The function L(x)=x−7π2L(x) = x – \frac{7\pi}{2}L(x)=x−27π is a linear function, where the slope is 111 and the y-intercept is −7π2-\frac{7\pi}{2}−27π. This means that for every unit increase in xxx, L(x)L(x)L(x) increases by 1.
In summary:
- The function simplifies to L(x)=x−7π2L(x) = x – \frac{7\pi}{2}L(x)=x−27π.
- The original sine and cosine terms simplify due to the specific values of sin\sinsin and cos\coscos at 7π2\frac{7\pi}{2}27π, which ultimately result in a linear equation.
