cutiecomittee123
  • cutiecomittee123
If sinx=4/5 and x is in the first quadrant, find each of the following (using double angle identities) cos2x sin2x tan2x
Mathematics
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SOLVED
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chestercat
  • chestercat
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anonymous
  • anonymous
\[\cos 2x=1-2\sin ^2x\] \[\sin 2x=2\sin x \cos x\] \[\tan 2x=\frac{ 2\tan x }{ 1-\tan ^2x }\] 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.
cutiecomittee123
  • cutiecomittee123
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
cutiecomittee123
  • cutiecomittee123
So to find tan, I just do (4/5)/(-7/25)?

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cutiecomittee123
  • cutiecomittee123
Well tan2x=(4/5)/(-7/25)?
anonymous
  • anonymous
you are given sin x ,not sin 2x
cutiecomittee123
  • cutiecomittee123
Yes, how does that effect my calculations thus far?
cutiecomittee123
  • cutiecomittee123
@surjithayer
Ac3
  • Ac3
you have to find sin2x your given sinx not sin2x
Ac3
  • Ac3
i found cosx=3/5 which means that sin2x=24/25
Ac3
  • Ac3
no you can find tan2x because it equals sin2x/cos2x so tan2x=-24/7
cutiecomittee123
  • cutiecomittee123
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
anonymous
  • anonymous
\[\cos x=\sqrt{1-\sin ^2x}=\sqrt{1-\left( \frac{ 4 }{ 5 } \right)^2}=\sqrt{1-\frac{ 16 }{ 25 }}\] \[=\sqrt{\frac{ 25-16 }{ 25 }}=\sqrt{\frac{ 9 }{ 25 }}=\frac{ 3 }{ 5 }\] \[\sin 2x=2\sin x \cos x=2*\frac{ 4 }{ 5 }*\frac{ 3 }{ 5 }=\frac{ 24 }{ 25 }\] \[\tan 2x=\frac{ \sin 2x }{ \cos 2x }=\frac{ \frac{ 24 }{ 25 } }{ \frac{ -7 }{ 25 } }=-\frac{ 24 }{ 7 }\]
cutiecomittee123
  • cutiecomittee123
But its cos2x that we need to find not cosx.
Ac3
  • Ac3
cos2x = 1-2(sinx)^2. <----- that's how you find cos2x. He found cosx so he could find sin2x.

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