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Posted by on Sunday, March 28, 2010 at 7:28pm.

csc(theta)= -3/2
3pi/2 < theta < 2pi
sin(theta)= _____ ?
cos(theta)= _____ ?
tan(theta)= _____ ?
cot(theta)= _____ ?
sec(theta)= _____ ?

  • Precalculus(NEED HELP ASAP PLEASE!!) - , Sunday, March 28, 2010 at 7:35pm

    draw the triangle. This is not an onerous problem. Notice it is in the fourth quadrant.

  • Precalculus(NEED HELP ASAP PLEASE!!) - , Monday, March 29, 2010 at 3:16pm

    Knowing that cosecant is the reciprocal for the sin function, sinć =-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, cosć= rad5/3 and secć= 3/rad5. The tangent of an angle is the ratio of opposite/adjacent and the
    cotangent is adjacent/opposite giving us tanć = -2/rad5 cotć =-rad5/2. Once you do these for a while the
    functions become almost instinctual. Good Luck!

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