Question

Given that Sinx=1/3, find tan2x.

Answer 1

@burgandy U knw the formula for tan2x in terms of cosx and sinx?

Answer 2

no

Answer 3

\[\tan(2x) =\frac{ 2tanx }{ 1-\tan^2x }\]
here sinx is given so determine cosx first using \[\sin^2x +\cos^2x = 1\]

Answer 4

remember tanx = sinx/cosx

Answer 5

7/9?

Answer 6

@312856MLP

Answer 7

what is value of tanx u get?

Answer 8

sinx = 3/5 that means opposite side of angle x is 3 and the hypotenuse is 5 so I don't think this is in a unit circle because unit circle has 1 as the radius or the hypotenuse.

Since they gave sinx = 3/5 and I stated above, hypotenuse is 5, opposite is 3 so the other leg should be 4. Right triangle with sides 3, 4, 5.
therefore
cosx = 4/5

tanx = sinx/cosx = 3/4

By double angle formula

tan(2x) = 2tanx/(1 - tan²u)

tan(2x) = 2*(3/4)/[1 - (3/4)²]

tan(2x) = (3/2)/(7/16)

tan(2x) = 24/7 = 3 3/7

Answer 9

apparently not the right one. :(
would it be \[4\sqrt{2} /9\] ?

Answer 10

do you get it

Answer 11

@burgandy do u get it

Answer 12

or /7?

Answer 13

i m getting
\[tanx = \frac{ 1 }{ 2\sqrt{2} }\]
\[\tan2x = \frac{ \frac{ 2 }{ 2\sqrt{2} } }{ 1- {\frac{ 1 }{ 2\sqrt{2} }}^{2} }\]

Answer 14

\7

Answer 15

\[\frac{ \frac{ 1 }{ \sqrt{2} } }{ 1 - \frac{ 1 }{ 8 } }\]
therefore \[\frac{ 8 }{ 7\sqrt{2} }\]

Answer 16

But you can't have a radical on the bottom.

Answer 17

it is 3 and 3/7

Answer 18

ok
\[\frac{ 8 }{ 7\sqrt{2} }*\frac{ \sqrt{2} }{ \sqrt{2} }\]
\[\frac{ 8\sqrt{2} }{ 7*2 }\]
\[\frac{ 4\sqrt{2} }{ 7 }\]

Answer 19

is that ok?

Answer 20

Yes! Thank you :)