Question

Find the exact value of \(tan (arcsin (\dfrac{2}{5})).\)

Answer 1

@abhisar

Answer 2

draw a triangle

Answer 3

there is a picture of an angle whose sine is \(\frac{2}{5}\)
all you need is the third side, which you find via pythagoras

Answer 4

to which quadrant belongs the unknow arc, whose sin is 2/5?

Answer 5

it is
\[x=\sqrt{5^2-2^2}=\sqrt{25-4}=\sqrt{21}\]

Answer 6

you want the tangent, which is "opposite over adjacent' making your answer
\[\tan(\arcsin(\frac{2}{5}))=\frac{2}{\sqrt{21}}\]

Answer 7

Thank you so much!

Answer 8

yw

Answer 9

@Michele_Laino
it makes no difference what quadrant you are in

Answer 10

why? If you are in the first quadrant tan is positive, whereas if you are in second quadrant tan is negative

Answer 11

reason is that
\[\arcsin(\theta)\] is a number between \(-\frac{\pi}{2}\) and \(\frac{\pi}{2}\)

Answer 12

@satellite73 Sorry, equation sin x=2/5 have two mainly solutions

Answer 13

ok i said that wrong
the arcsine of a number cannot be in the second quadrant is what i meant to say

Answer 14

oh no, you do not have two solutions, you have one solution only
arcsine is a well defined function

Answer 15

you are right in that
\[\sin(x)=\frac{2}{5}\] has more than one solution
in fact it has an infinite number of solutions

Answer 16

but there is only one
\[\arcsin(\frac{2}{5})\]
it is the solution to
\[\sin(x)=\frac{2}{5}\] only for
\[-\frac{\pi}{2}\le x\le\frac{\pi}{2}\]

Answer 17

@satellite73 sorry, note that I said mainly solutions, If x is a solution, also 180-x is a solution

Answer 18

like i said you have an infinite number of solutions to
\[\sin(x)=\frac{2}{5}\] that is true
but only one number is
\[\arcsin(\frac{2}{2})\] since arcsine is a function (cannot have two answers)

Answer 19

for example
\[\arcsin(\frac{1}{2})=\frac{\pi}{6}\]
not
\(\frac{\pi}{6}\) or \(\frac{5\pi}{6}\)

Answer 20

@satellite73 Ok! you are right! sorry!

Answer 21

no need to be sorry, it is common concern
if the question was "find the tangent of a number whose sine is 2/5" then you would have been right, you would not know if it was positive or negative

Answer 22

@satellite73 thank you! :)