Question

Trigo question #1

Answer 1

E

Answer 2

\[\cos^2(\theta)+ \sin^2(\theta)=1\\
- k - k =1\\
-2 k=1\\
k = - \frac 12
\]

Answer 3

The answer is B...

Answer 4

Since \(\sin^2(x) +\cos^2(x)=1\), and both \(\sin^2(x)\) and \(\cos^2(x)\) are solutions if you add them together you get that \[\sin^2(x) +\cos^2(x)=-2k\]Thus, \(k=-1/2\)

So B is the answer.

Answer 5

Wow, I didn't know that..

Answer 6

should have thought ... since k goes to the other end and becomes negative ... i thought exactly opposite.

Answer 7

I didn't know that too!! Thanks @KingGeorge

Answer 8

More visually, \[\begin{matrix} \sin^2(\theta) +k=0 \\ \cos^2(\theta)+k=0 \\\text{______________}\\ \sin^2(\theta)+\cos^2(\theta)+2k=0\end{matrix}\]So \(1=-2k\) which implies \(k=-1/2\)

Answer 9

I thought
SinQ = CosQ
the the only value it could have was pi/2

Answer 10

*pi/4 ... old school trigonometry