If tan ( x ) = 29 /2 (in Quadrant-I), find sin ( 2 x ) =

Answer 1

tanx = 29/2 = opposite side over adjacent side

hypotenuse squared = 29^2 + 2^2 = 841+4 = 845

hypotenuse = sqr845 = 13sqr5 = about or almost 29.1 rounded off to nearest tenth

sinx = 29/13sqr5

cosx = 2/13sqr5

use the double angle formula for sines: sin2x = 2sinxcosx

sin2x = 2sinxcosx = 2(29/13sqr5)(2/13sqr5) = 2(29)(2)/845 = 116/845

sin2x = 116/845 = about 0.137

another method is:

tanx = 29/2 = 14.5

use a calculator with an inverse trig function to get

x = 86.05

double it

2x = 86.05(2) = 172.1

sin172.1 = about 0.137

if you get the same answer with 2 different methods, there's great confidence you got it right

in this case down to 3 decimal places

or exactly 116/845

172.1 degrees is close to 180, so you expect sin172.1 to be close to zero

while x is in quadrant I, 2x is in quadrant II