Question

If \( 2 \cos^2 45^\circ - 1 = \cos \theta \), then find the value of \( \theta \).

Hint:

The cosine function equals zero at \( 90^\circ \) and \( 270^\circ \).

Answer 1

We are given that \[ 2 \cos^2 45^\circ - 1 = \cos \theta. \] First, we know that \( \cos 45^\circ = \frac{1}{\sqrt{2}} \), so \[ \cos^2 45^\circ = \left( \frac{1}{\sqrt{2}} \right)^2 = \frac{1}{2}. \] Now, substitute \( \cos^2 45^\circ \) into the equation: \[ 2 \times \frac{1}{2} - 1 = \cos \theta. \] Simplifying the left-hand side: \[ 1 - 1 = \cos \theta. \] This gives: \[ \cos \theta = 0. \] Now, we need to find the value of \( \theta \) such that \( \cos \theta = 0 \). The cosine function is zero at \[ \theta = 90^\circ \text{ or } \theta = 270^\circ. \] Thus, the possible values of \( \theta \) are \( 90^\circ \) and \( 270^\circ \).
Conclusion: The values of \( \theta \) are \( 90^\circ \) and \( 270^\circ \).