Question

How do I find the exact value of 2cos(x) when they give me something for sine?

Answer 1

what do you mean *something for sine* ?

Answer 2

like fraction ? (opposite /hypotenuse ) values ?

Answer 3

like \[\frac{ \sqrt5 }{ 5 }\]

Answer 4

hmm
if \[\large \rm \sin(x) =\frac{\sqrt{5}}{5}\]
according to cos definition \[\rm sin \rm \theta = \frac{ opposite }{ hypotenuse }~~~~ \cos \theta = \frac{ adjacent }{ hypotenuse } ~~\tan \theta = \frac{ opposite }{ adjacent }\]

draw a right triangle
find the cos x and then multiply by 2
that's it i guess

Answer 5

It asks me and I quote "Find the exact value of cos2x given that sine = sqrt5/5

Answer 6

cos (2x) or 2 cos(x) or cos^2(x) which one is correct?

Answer 7

cos2x sorry.

Answer 8

\[2 \cos \left(\sin ^{-1}\left(\frac{1}{\sqrt{5}}\right)\right)=\frac{4}{\sqrt{5}} \]

Answer 9

\[\frac{ \sqrt5 }{ 5 }\]

How did you get that?

Answer 10

how did i get sqrt{5}/5 ??

Answer 11

\[\rm \cos (2x) ~~or~~ \cos^2(x)\] which one is correct ??
2 is square or is it cos(2x)
then you can apply double angle formula..

Answer 12

The sine of x is given as sqrt{5}/5 .

Answer 13

This is what it says -

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Answer 14

alright then sqrt{5}/5 is given. as mentioned above.

Answer 15

I'd use the Pythagorean Theorem.

Just suppose you were given sin x = a/h. h would be the hypotenuse, and a the opporite side. The adjacent side (which you need for writing the cosine of the angle x) would be \[\pm \sqrt{h^2-opp^2}.\]

Answer 16

then the cosine of x would be \[\pm \frac{ adj.side }{ hypotenuse}\]

Answer 17

Or, use the identity\[\sin^2\theta + \cos^2\theta =1 \] to find cos theta.

Answer 18

its ga ood idea to draw a right triangle. use the given information
at which quadrant should it located (this information needed to find is cos negative or positive )
and ofc definitely need Pythagorean theorem a^2+b^2=c^2 where c = hypotenuse(longest side of right triangle )