Precalculus(NEED HELP ASAP PLEASE!!)
posted by jh .
3pi/2 < theta < 2pi
sin(theta)= _____ ?
cos(theta)= _____ ?
tan(theta)= _____ ?
cot(theta)= _____ ?
sec(theta)= _____ ?
draw the triangle. This is not an onerous problem. Notice it is in the fourth quadrant.
Knowing that cosecant is the reciprocal for the sin function, sinƒá =-2/3 because that is the reciprocal of
-3/2. Notice that it is also negative because reciprocal functions always share the same sign. For the
other functions, you will need to make use of the Pythagorean theorem. Since sin=
opposite/hypotenuse and the Pythagorean theorem is C2 = a2 + b2, we can substitute what we know.
for ¡¥a¡¦ we put 2, and ¡¥c¡¦ we put 3. Signs do not matter since they are being squared. This gives you
9 = 4 + b2 , and through a bit of math we get our adjacent side of the triangle, b= radical5, this is a +/- radical5,
Depending on which quadrant you are in and which function you are using to solve this will determine
the sign. In the fourth quadrant sine and tangent are negative, cosine is positive. Let¡¦s get the cosine and
secant functions values. Cosine is adjacent/hypotenuse and the secant is the reciprocal of that,
hypotenuse/ adjacent or rather 1 over adjacent/hypotenuse. Since we found the adjacent value to be
+/- rad5, cosƒá= rad5/3 and secƒá= 3/rad5. The tangent of an angle is the ratio of opposite/adjacent and the
cotangent is adjacent/opposite giving us tanƒá = -2/rad5 cotƒá =-rad5/2. Once you do these for a while the
functions become almost instinctual. Good Luck!