Question

What is the proof or explanation for the trigonometric identity sin(theta+3pi/2)?

Answer 1

To prove the trigonometric identity for \( \sin(\theta + \frac{3\pi}{2}) \), we can use the sine addition formula as well as our knowledge of the unit circle.

The sine addition formula states that:

\[
\sin(a + b) = \sin(a)\cos(b) + \cos(a)\sin(b)
\]

In this case, we can let \( a = \theta \) and \( b = \frac{3\pi}{2} \). Thus, we have:

\[
\sin\left(\theta + \frac{3\pi}{2}\right) = \sin(\theta)\cos\left(\frac{3\pi}{2}\right) + \cos(\theta)\sin\left(\frac{3\pi}{2}\right)
\]

Now we need to find the values of \( \cos\left(\frac{3\pi}{2}\right) \) and \( \sin\left(\frac{3\pi}{2}\right) \):

- The angle \( \frac{3\pi}{2} \) radians corresponds to \( 270^\circ \). At this angle, the coordinates on the unit circle are \( (0, -1) \).

So, we have:

\[
\cos\left(\frac{3\pi}{2}\right) = 0
\]
\[
\sin\left(\frac{3\pi}{2}\right) = -1
\]

Substituting these values back into the sine addition formula:

\[
\sin\left(\theta + \frac{3\pi}{2}\right) = \sin(\theta)(0) + \cos(\theta)(-1)
\]

This simplifies to:

\[
\sin\left(\theta + \frac{3\pi}{2}\right) = -\cos(\theta)
\]

Thus, the identity is proven:

\[
\sin\left(\theta + \frac{3\pi}{2}\right) = -\cos(\theta)
\]