Given that tanè = -5 and è is in the 2nd quadrant, compute exactly:
A. sin(è)
B. cos(è)
C. tan(2è)
3 answers
The e is supposed to be theta.
In quad 2
sin is +
cos is -
of course we can look up the angle knowing the tangent and then find the sin etc but that is too easy.
tan T = -5 = sin T/cos T
cos T = sqrt (1 - sin^2 T)
so
-5 = sin T / sqrt(1-sin^2 T)
25 (1-sin^2 T ) = sin^2 T
25 = 26 sin^2 T
sin^2 T = 25/26
sin T = 5/sqrt(26)
then cos T = -sqrt (1 - 25/26)
now use
tan 2T = 2 tan T /(1-tan^2 T)
sin is +
cos is -
of course we can look up the angle knowing the tangent and then find the sin etc but that is too easy.
tan T = -5 = sin T/cos T
cos T = sqrt (1 - sin^2 T)
so
-5 = sin T / sqrt(1-sin^2 T)
25 (1-sin^2 T ) = sin^2 T
25 = 26 sin^2 T
sin^2 T = 25/26
sin T = 5/sqrt(26)
then cos T = -sqrt (1 - 25/26)
now use
tan 2T = 2 tan T /(1-tan^2 T)
or
Just sketch the triangle in II
given tanè = -5/1 ----> y/x ---> y = 5, x = -1
r^2 = 5^2+1^1=26
r = √26
sinè = y/r = 5/√26
cosè = -1/√26
tan 2è = 2tanè/(1 - tan^2 è)
= -10/(1 - 25)
= -10/24
= 5/12
Just sketch the triangle in II
given tanè = -5/1 ----> y/x ---> y = 5, x = -1
r^2 = 5^2+1^1=26
r = √26
sinè = y/r = 5/√26
cosè = -1/√26
tan 2è = 2tanè/(1 - tan^2 è)
= -10/(1 - 25)
= -10/24
= 5/12