Calculus: Early Transcendentals 8th Edition

$sin\frac{3\pi}{4}=sin\frac{\pi}{4}=\frac{\sqrt 2}{2}$ $csc\frac{3\pi}{4}=\frac{1}{sin\frac{3\pi}{4}}=\sqrt 2$ $cos\frac{3\pi}{4}=-cos\frac{\pi}{4}=-\frac{\sqrt 2}{2}$ $sec\frac{3\pi}{4}=\frac{1}{cos\frac{3\pi}{4}}=-\sqrt 2$ $tan\frac{3\pi}{4}=-tan\frac{\pi}{4}=-1$ $cot\frac{3\pi}{4}=\frac{1}{tan\frac{3\pi}{4}}=-1$
Since $\frac{3\pi}{4}$ lies in the second quadrant, its reference angle will be $\pi-\frac{3\pi}{4}=\frac{\pi}{4}$, which can be considered as a common angle to all trigonometric ratios. The trigonometric ratios and their inverse trigonometric ratios are given as follows: $sin\frac{3\pi}{4}=sin\frac{\pi}{4}=\frac{\sqrt 2}{2}$ $csc\frac{3\pi}{4}=\frac{1}{sin\frac{3\pi}{4}}=\sqrt 2$ $cos\frac{3\pi}{4}=-cos\frac{\pi}{4}=-\frac{\sqrt 2}{2}$ $sec\frac{3\pi}{4}=\frac{1}{cos\frac{3\pi}{4}}=-\sqrt 2$ $tan\frac{3\pi}{4}=-tan\frac{\pi}{4}=-1$ $cot\frac{3\pi}{4}=\frac{1}{tan\frac{3\pi}{4}}=-1$