# Thread: Find sin(theta + pi/4)

1. ## Find sin(theta + pi/4)

given that π/2 ≤ θ ≤ 3π/2 and sin θ = -2/7 find sin(θ+π/4)

2. ## Re: Find sin(theta + pi/4)

Hello, absolutjessy!

$\displaystyle \text{Given: }\:\tfrac{\pi}{2} \,\le\,\theta\,\le\, \tfrac{3\pi}{2}\,\text{ and }\,\sin\theta \,=\,\text{-}\frac{2}{7}$
$\displaystyle \text{Find: }\:\sin\left(\theta + \tfrac{\pi}{4}\right)$

We are told that $\displaystyle \theta$ is in Quad 2 or Quad 3.
. . Hence, $\displaystyle \theta$ is in Quad 3.

$\displaystyle \sin\theta \,=\,\frac{\text{-}2}{7} \,=\,\frac{opp}{hyp}$

$\displaystyle \theta$ is in a right triangle with: $\displaystyle opp = \text{-}2,\;hyp = 7$
. . Then: .$\displaystyle adj \,=\, \text{-}\sqrt{45} \,=\,\text{-}3\sqrt{5} \quad\Rightarrow\quad \cos\theta \:=\:\text{-}\frac{3\sqrt{5}}{7}$

We have: .$\displaystyle \sin\left(\theta + \tfrac{\pi}{4}\right) \;=\; \sin\theta\cos\tfrac{\pi}{4} + \cos\theta\sin\tfrac{\pi}{4}$

n . . . . . . . . . . . . . . . $\displaystyle =\;\left(\text{-}\tfrac{2}{7}\right)\left(\tfrac{1}{\sqrt{2}} \right) + \left(\text{-}\tfrac{3\sqrt{5}}{7}\right)\left(\tfrac{1}{\sqrt{ 2}}\right)$

n . . . . . . . . . . . . . . . $\displaystyle =\;\frac{\text{-}2}{7\sqrt{2}} + \frac{\text{-}3\sqrt{5}}{7\sqrt{2}}$

n . . . . . . . . . . . . . . . $\displaystyle =\;-\frac{2 + 3\sqrt{5}}{7\sqrt{2}}$