# Re: Pre-calc

[ Follow Ups ] [ Post Followup ] [ Main Message Board ] [ FAQ ]

Posted by T.Gracken on October 21, 2002 at 19:35:47:

In Reply to: Re: Pre-calc posted by Jacquie on October 21, 2002 at 18:23:27:

: the problem was written out as
: If cotx= 3pi/4 find sinx for x between pi and 3 pi/2.
: i dont know how to use the triangle for this bc it is cot. cot=x/y right? so how do i find x im really confused, this is way over my head.

If you have seen (right) triangle definitions, then you know that sin(x)= opposite/hypotenuse {where x is the angle), cos(x)= adjacent/hypotenuse, and tan(x)=opposite/adjacent.

since cot(x) = 1/tan(x), we get cot(x)=adjacent/opposite.

so for cot(x)=3pi/4, we have adjacent=3pi, opposite = 4. Then the pythagorean theorem gives hypotenuse as sqrt(9pi2+16) and we can now get sin(x) and sin(x) will be negative if x is between pi and 3pi/2.

but... another way without triangles would be to use identities. (I'm guessing this is the way your teacher wants since your previous post had an identity in it)

One of the Pythagorean identities is: cot2(x)+1=csc2(x).

since cot2(x)+1=csc2(x), we can substitute the given value of cot(x) to get the equation:

(3pi/4)2+1=csc2(x)

or (9pi2)/16 + 1 = csc2(x)

which gives (9pi2+16)/16 = csc2(x)

...but, csc(x) is the reciprocal of sin(x), [that is, 1/sin(x) = csc(x), so sin(x)=1/csc(x)]

so we could write the reciprocals of each side and have

16/(9pi2+16) = sin2(x)

take square root of each side to get

(plus or minus) sqrt[16/(9pi2+16)] = sin(x)

and since sin(x) is negative if x is between pi and 3pi/2, use the negative value.

I'll leave the algebra simplification to you.

Name:
E-Mail:

Subject: