If 0<x<pie/2 and tanx=2, then sinx=?

Question

If 0<x<pie/2 and tanx=2, then sinx=?

Answer 1

0 &lt; x &lt; π/2

tan(x) = 2

arctan(2) = 63.4°

sin(63.4°) = 0.89

Answer 2

tan x = opposite / adjacent = 2 / 1

So...

opposite = 2

adjacent = 1

Use pythagorean theorem to find &quot;hypotenuse&quot;

(opp)^2 + (adj)^2 = (hyp)^2

4 + 1 = (hyp)^2

5 = hyp^2

Sqrt 5 = hyp

Sin x

= opposite / hypotenuse

= 2 / sqrt 5

= 2(sqrt 5) / (sqrt 5)^2

= 2(sqrt 5) / 5

Answer 3

If 0&lt;x&lt;pie/2 means x lies in the first quadrant where sine, cosine and tangent are all positive.

tan(x)=2

x=arctan(2)=1.107 radians or 63.435 degrees

sin(x)=0.8944

Answer 4

tan x = 2 = opposit/adjacent , draw right triangle whose legs 1 &amp; 2 then hypotenuse is sqrt5 , then sin x = 2/sqrt5

or use calc, tanx =2 ,then x =63.435degree, sinx=0.89442

Answer 5

If 0 &lt; x &lt; PI, both the sine and the cosine are positive, thus sin x = sqrt(1 - cos^2 x)

2 = tan x = sin x / cos x

4 = (sin^2 x)/(cos^2 x) = (1 - cos^2 x)/ cos^2 x

Multiply by cos^2 x:

4cos^2 x = 1 - cos^2 x

5cos^2 x = 1

cos^2 x = 0.2

sin^2 x = 1 - 0.2 = 0.8

sin x = sqrt(0.8) = sqrt(4/5) = 2/sqrt(5) = 2sqrt(5)/5

Answer 6

tan(x)=sin(x)/cos(x)=sin(x)/sqrt( 1-sin^2(x) )=2 &lt;=&gt;

4(1-sin^2(x))=sin^2(x) &lt;=&gt; 5sin^2(x)=4 &lt;=&gt; sin(x)=2/sqrt(5)

note: sin^2(x)=sin(x) * sin(x)

sqrt = square root