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Question 5 Exercise 8.2
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sqrt{1-\sin^2 2\theta}$$ Since $2\theta$ in QII, therefore $\cos 2\theta$ is negative. \begin{align*}\c... \pm\sqrt{1-\sin\theta}$$ As $\theta$ lies in QI, therefore $\cos\theta >0$, thus \begin{align*} \cos\the... lies in QIII, we know that \(\sin 2\theta < 0\). Therefore: \begin{align*} \sin 2\theta & = -\sqrt{1 - \... lies in QIII, we know that \(\cos 2\theta < 0\). Therefore: \begin{align*} \cos 2\theta & = -\sqrt{1 - \
Question 1, 2 and 3 Exercise 8.2
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alpha$ lies in QII and $\cos$ is negative in QII, therefore \begin{align*} \cos \alpha & = - \sqrt{1-\sin
Question 4 Exercise 8.2
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\frac{13}{5}\end{align*} \(\theta\) lies in QIII, therefore $\sec \theta <o$\\ \begin{align*}\sec \theta