Question

Let csc t= sqrt (5) with t in Q2 and find the following.
cos2A and Sec2A

Answer 1

which formula do we use for both?

Answer 2

the missing side is 2

Answer 3

Then, you have \(\color{black}{ {\rm adj=\sqrt{hyp^2-opp^2}=\sqrt{(\sqrt{5})^2-1}=\sqrt{4}=2}}\).

Yes, you are correct.

Answer 4

Then, you know that

\(\color{black}{ \cos(t)={\rm adj/hyp}}\)
\(\color{black}{ \sec(t)={\rm hyp/adj}}\)

Answer 5

The adjacent side is -2.

Answer 6

Wait so I know that Cos2A is =cos^2A-sin^2A but Sec is the hardest one for me

Answer 7

(Due to the angle being in the second quadrant)

Answer 8

well, sec = 1 /cos

Answer 9

So, you got:
\(\color{black}{ \cos(2t)=2\cos^2(t)-1=2\left(\frac{-2}{\sqrt{5}}\right)^2-1 }\).

Answer 10

and you know that
\(\color{black}{ \displaystyle \sec(2t)=\frac{1}{\cos(2t)}=\frac{1}{2\cos^2(t)-1}=\frac{1}{2\left(\frac{-2}{\sqrt{5}}\right)^2-1 } }\).

Answer 11

thank you

Answer 12

\({\rm yw}{\tiny~}:\)\()\)