Question

Please help me with this problem

Answer 1

Answer:  $\bold{b)\quad \dfrac{17\sqrt2}{26}}$

Step-by-step explanation:

$\text{Given }sin\ x=-\dfrac{5}{13}\quad \text{and is in Quadrant IV}\quad \implies\quad cos\ x=\dfrac{12}{13}\\\\\text{Use the cosine sum formula: }cos(A + B) = cosA\cdot cosB-sinA\cdot sinB\\\\\implies cos\bigg(x+\dfrac{\pi}{4}\bigg)=cosx\cdot cos\dfrac{\pi}{4}-sinx\cdot sin\dfrac{\pi}{4}\\\\\\=\dfrac{12}{13}\cdot\dfrac{\sqrt2}{2}-\bigg(-\dfrac{5}{13}\bigg)\cdot \dfrac{\sqrt2}{2}\\\\\\=\dfrac{12\sqrt2}{26}+\dfrac{5\sqrt2}{26}\\\\\\=\large\boxed{\dfrac{17\sqrt2}{26}}$

Answer 2

Answer:

○ 17√2\26

Explanation:

According to the Unit Circle, sin -0,3947911197 is -5⁄13, and given that it falls in between 1½π and 2π [Quadrant IIIII], we can either pick the positive sum of it, like what the moderator said, or we can simply just plug it in. In this case, either way would work. Anyway, we plug this into the cosine function to get this:

cos (-0,3947911197 + π\4) → cos 0,39060704 >> 17√2\26 [Approximately]

I am joyous to assist you anytime.