Find the Exact Value

(a)

$\sin^2\theta+\cos^2\theta\,=\,1$       by the Pythagorean Identity.

$(\frac13)^2+\cos^2\theta\,=\,1$        because we are given that  $\sin\theta\,=\,\frac13$

$\frac19+\cos^2\theta\,=\,1\\~\\ \cos^2\theta\,=\,1-\frac19\\~\\ \cos^2\theta\,=\,\frac89\\~\\ \cos\theta\,=\,\pm\sqrt{\frac89}\\~\\ \cos\theta\,=\,\pm\frac{2\sqrt2}{3}$

Since  θ  is in Quadrant II,  cos θ  must be negative. So   $\cos\theta\,=\,-\frac{2\sqrt2}{3}$

(b)

$\sin(\theta+\frac{\pi}{6})\,=\,\sin\theta\,\cos\frac{\pi}{6}+\cos\theta\,\sin\frac{\pi}{6}$     by the angle sum formula for sin

$\sin(\theta+\frac{\pi}{6})\,=\,(\frac13)(\frac{\sqrt3}{2})+(-\frac{2\sqrt2}{3})(\frac12)\\~\\ \sin(\theta+\frac{\pi}{6})\,=\,\frac{\sqrt3}{6}-\frac{2\sqrt2}{6}\\~\\ \sin(\theta+\frac{\pi}{6})\,=\,\frac{\sqrt3-2\sqrt2}{6}$

(c)

$\cos(\theta-\frac{\pi}{3})\,=\,\cos\theta\,\cos\frac{\pi}{3}+\sin\theta\,\sin\frac{\pi}{3}$     by the angle difference formula for cos

Now plug in the values for  cos θ,  cos(pi/3) ,  sin θ , and  sin(pi/3)  and simplify. Can you finish this one?

(d)

To use the angle sum formula for tan, let's first find  tan θ.

$\tan\theta\,=\,\frac{\sin\theta}{\cos\theta}\,=\,\frac{(\frac13)}{(-\frac{2\sqrt2}{3})}\,=\,-\frac{1}{2\sqrt2}\,=\,-\frac{\sqrt2}{4}$

$\tan(\theta+\frac{\pi}{4})\,=\,\frac{\tan\theta+\tan\frac{\pi}{4}}{1-\tan\theta\,\tan\frac{\pi}{4}}$     by the angle sum formula for tan

$\tan(\theta+\frac{\pi}{4})\,=\,\frac{(-\frac{\sqrt2}{4})+(1)}{1-(-\frac{\sqrt2}{4})(1)}\\~\\ \tan(\theta+\frac{\pi}{4})\,=\,\frac{4-\sqrt2}{4+\sqrt2}$