Precalculus(NEED HELP ASAP PLEASE!!)
posted by jh on .
pi < theta < 3pi/2
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!!
tan theta= 1/3 which means sintheta 1/sqr10
draw the triangle Notice it is in the third quadrant, where cosine is negative.
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
= 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