Questions: If sin θ=c and 0 ≤ θ<(π/2), which expression is equivalent to cos (π+θ)?

Transcript text: If $\sin \theta=c$ and $0 \leq \theta<\frac{\pi}{2}$, which expression is equivalent to $\cos (\pi+\theta)$ ?

Solution

Solution Steps

Step 1: Use the given information

Given $\sin \theta = c$ and $0 \leq \theta < \frac{\pi}{2}$, we know that $\theta$ is in the first quadrant, where both $\sin \theta$ and $\cos \theta$ are positive.

Step 2: Express $\cos \theta$ in terms of $c$

Using the Pythagorean identity: \[ \sin^2 \theta + \cos^2 \theta = 1 \] Substitute $\sin \theta = c$: \[ c^2 + \cos^2 \theta = 1 \] Solve for $\cos \theta$: \[ \cos \theta = \sqrt{1 - c^2} \]

Step 3: Use the cosine addition formula for $\cos (\pi + \theta)$

The cosine addition formula for $\cos (\pi + \theta)$ is: \[ \cos (\pi + \theta) = \cos \pi \cos \theta - \sin \pi \sin \theta \] We know that $\cos \pi = -1$ and $\sin \pi = 0$, so: \[ \cos (\pi + \theta) = (-1) \cdot \cos \theta - 0 \cdot \sin \theta \] Simplify: \[ \cos (\pi + \theta) = -\cos \theta \]

Step 4: Substitute $\cos \theta$ with $\sqrt{1 - c^2}$

From Step 2, we have $\cos \theta = \sqrt{1 - c^2}$, so: \[ \cos (\pi + \theta) = -\sqrt{1 - c^2} \]