tan theta= 1/3 which means sintheta 1/sqr10
draw the triangle Notice it is in the third quadrant, where cosine is negative.
cot(theta)= 3
pi < theta < 3pi/2
Find:
sin(theta)= -1 ?
cos(theta)= -3 ?
tan(theta)= 1/3 ?
sec(theta)= -1/3 ?
csc(theta)= -1 ?
That's what I came up with, but they are not correct, PLEASE help me where I went wrong!!
5 answers
You do not seem to have calculated the hypotenuse which you need for sin and cos and sec and csc
sqrt (1^2+3^2) = sqrt 10
sqrt (1^2+3^2) = sqrt 10
So,
tan(theta)=sin(theta)/cos(theta)
cos(theta)=sin(theta)/tan(theta)
=(1/sqrt(10))/(1/3)
= 3/sqrt(10) ??
But its incorrect!! What am I doing wrong?
tan(theta)=sin(theta)/cos(theta)
cos(theta)=sin(theta)/tan(theta)
=(1/sqrt(10))/(1/3)
= 3/sqrt(10) ??
But its incorrect!! What am I doing wrong?
If sin t=-1/4, find sin t + 6pi
tan theta= 5 divided by 12 and theta is in quadrant 3 what does sin 2 theta equal
2 answers