Question

simplify the expression by using a double - angle formula or a half - angle formula. (a) $sqrt{\frac{1 - cos(80^{circ})}{2}}$ $sin(40^{circ})$ (b) $sqrt{\frac{1 - cos(4\theta)}{2}}$ $sin(2\theta)$ resources read it

Explanation:

Step1: Recall half - angle formula

The half - angle formula for sine is $\sin\frac{\alpha}{2}=\pm\sqrt{\frac{1 - \cos\alpha}{2}}$.

Step2: Solve part (a)

For $\sqrt{\frac{1-\cos(80^{\circ})}{2}}$, let $\alpha = 80^{\circ}$, then $\sqrt{\frac{1-\cos(80^{\circ})}{2}}=\sin\frac{80^{\circ}}{2}=\sin(40^{\circ})$.

Step3: Solve part (b)

For $\sqrt{\frac{1-\cos(4\theta)}{2}}$, let $\alpha = 4\theta$. By the half - angle formula $\sin\frac{\alpha}{2}=\pm\sqrt{\frac{1 - \cos\alpha}{2}}$, we have $\sqrt{\frac{1-\cos(4\theta)}{2}}=\sin\frac{4\theta}{2}=\sin(2\theta)$. But we need to consider the sign. In general, $\sqrt{\frac{1 - \cos(4\theta)}{2}}=\vert\sin(2\theta)\vert$. Since the original problem does not specify the range of $\theta$, the correct answer should be $\vert\sin(2\theta)\vert$.

Answer:

(a) $\sin(40^{\circ})$
(b) $\vert\sin(2\theta)\vert$