If sin Ф=1/3>0 and tan Ф<0; then Ф belongs to the second quadrant and cos Ф will be negative (cos Ф<0).
sin²Ф+cos²Ф=1
(1/3)²+cos²Ф=1
1/9+cos²Ф=1
cos²Ф=1-1/9
cos²Ф=8/9
cos Ф=⁺₋√(8/9)=⁺₋2√2 / 3
We have two possible solutions.
cos Ф=2√2 / 3 This solutions is not possible, because in this case cos Ф has to be negative (cos Ф<0)
cos Ф=(-2√2) / 3 this solutions is right.
answer:B. -2 square root 2 /3 (or (-2√2) /3 )
Answer:
actually it is a simultaneous equation from linear combinations
All real numbers
7w-(2+w) = 2(3w-1)
Expand
7w-(2+w)= 6w-2
7w-2-w=6w-2
Group like terms
6w-2=6w-2
Add 2 to both sides
6w=6w
subtract 6w from both sides
0=0
True for all w
