first quadrant because sin and cos are +
triangle is 1, 2, sqrt 45
sin x = 2/sqrt 5
cos x = 1/sqrt 5
I think you can take it from there
find sin 2x
if tan x=2 cosx>0
2 answers
since both the cosine and the tangent are positive, x must be an angle in the first quadrant.
from tanx = 2/1, draw a right-angled triangle with angle x at the origin that has a height of 2, and a base of 1
(tangentx = opposite/adjacent, so the opposite is 2 and the adjacent is 1)
so by Pythagoras, the hypotenuse is √5
so sinx - 2/√5 and cosx = 1/√5
sin2x = 2sinxcosx (one of our general identities)
= 2(2/√5)(1/√5) = 4/5
you can check this on the calculator
take inverse tan of 2 to get the angle,
double that angle
take the sine
you should get .8 which is 4/5
from tanx = 2/1, draw a right-angled triangle with angle x at the origin that has a height of 2, and a base of 1
(tangentx = opposite/adjacent, so the opposite is 2 and the adjacent is 1)
so by Pythagoras, the hypotenuse is √5
so sinx - 2/√5 and cosx = 1/√5
sin2x = 2sinxcosx (one of our general identities)
= 2(2/√5)(1/√5) = 4/5
you can check this on the calculator
take inverse tan of 2 to get the angle,
double that angle
take the sine
you should get .8 which is 4/5