Posted by **Timothy** on Monday, February 25, 2008 at 7:36pm.

Hello all,

In our math class, we are practicing the trigonometric identities (i.e., sin^2(x)+cos^2(x)=1 or cot(x)=cos(x)/sin(x). Now, we are working on proofs that two sides of an equation are equal (for example, sin(x)*csc(x)=1; sin(x)csc(x)=sin(x)/sin(x)=1; 1=1). I am not having much trouble with these problems, but two are proving rather difficult for me. They are:

csc^2(x)/cot(x)=csc(x)sec(x)

and

cot^3(x)/csc(x)=cos(x)*csc^2(x)-1

Could I please get a kick in the right direction?

Thanks,

Timothy

- Trigonometry -
**Christiaan**, Monday, February 25, 2008 at 8:21pm
1)

csc²(x)=1/sin²(x) and cot(x)=cos(x)/sin(x)

=> csc²(x)/cot(x) = sin(x)/(sin²(x)*cos(x))

We can scrap one of the sinus factors in both the numerator and denominator, so that the expression equals:

1/(sin(x)*cos(x))

Since csc(x)=1/sin(x) and sec(x)=1/cos(x)

the expression we found equals csc(x)*sec(x). So we have proven the equality.

- Trigonometry -
**Christiaan**, Monday, February 25, 2008 at 8:23pm
2)

csc(x)=1/sin(x) and cot(x)=cos(x)/sin(x)

=> cot^3(x)/csc(x)=cos^3(x)*sin(x)/sin^3(x)

=cos^3(x)/sin²(x)

Now, we know that cos²(x)=1-sin²(x)

=> cos^3(x)=cos(x)*(1-sin²(x))=cos(x)-cos(x)*sin²(x)

So, when we fill this in in the equation we get that:

cos^3(x)/sin²(x)=(cos(x)-cos(x)*sin²(x))/sin²(x)

We can split this fraction in two separate fractions, namely:

(cos(x)/sin²(x)) - (cos(x)*sin²(x)/sin²(x))

= cos(x)*csc²(x) - cos(x)

= cos(x) (csc²(x)-1)

It seems to me you forget a pair of brackets in your question.

- Trigonometry -
**Timothy**, Monday, February 25, 2008 at 8:44pm
Thank you very much for the assistance. I understand this perfectly. Have a good day. :)

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