Question

If sinx=4/5 and x is in the first quadrant, find each of the following (using double angle identities)

cos2x

sin2x

tan2x

Answer 1

\[\cos 2x=1-2\sin ^2x\]
\[\sin 2x=2\sin x \cos x\]
\[\tan 2x=\frac{ 2\tan x }{ 1-\tan ^2x }\]
find cos x by the identity
\[\sin ^2x+\cos ^2x=1\]
then find tanx by \[\tan x=\frac{ \sin x }{ \cos x }\]
remember sin x ,cos x,tan x are positive in first quadrant.

Answer 2

Okay so I think I figured out how to solve for Cos2x
cos2x=1-2sin2x
cos2x=1-2(4/5)^2
cos2x=1-32/25
cos2x=-7/25

Answer 3

So to find tan, I just do (4/5)/(-7/25)?

Answer 4

Well tan2x=(4/5)/(-7/25)?

Answer 5

you are given sin x ,not sin 2x

Answer 6

Yes, how does that effect my calculations thus far?

Answer 7

@surjithayer

Answer 8

you have to find sin2x your given sinx not sin2x

Answer 9

i found cosx=3/5 which means that sin2x=24/25

Answer 10

no you can find tan2x because it equals sin2x/cos2x so tan2x=-24/7

Answer 11

huh? I think you are reading it wrong or something because I know the calculations have worked out for me so far using the sinx and plugging it into the other three

Answer 12

\[\cos x=\sqrt{1-\sin ^2x}=\sqrt{1-\left( \frac{ 4 }{ 5 } \right)^2}=\sqrt{1-\frac{ 16 }{ 25 }}\]
\[=\sqrt{\frac{ 25-16 }{ 25 }}=\sqrt{\frac{ 9 }{ 25 }}=\frac{ 3 }{ 5 }\]
\[\sin 2x=2\sin x \cos x=2*\frac{ 4 }{ 5 }*\frac{ 3 }{ 5 }=\frac{ 24 }{ 25 }\]
\[\tan 2x=\frac{ \sin 2x }{ \cos 2x }=\frac{ \frac{ 24 }{ 25 } }{ \frac{ -7 }{ 25 } }=-\frac{ 24 }{ 7 }\]

Answer 13

But its cos2x that we need to find not cosx.

Answer 14

cos2x = 1-2(sinx)^2. <----- that's how you find cos2x. He found cosx so he could find sin2x.