Posted by Timothy on Monday, February 25, 2008 at 7:36pm.
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:
Could I please get a kick in the right direction?
- Trigonometry - Christiaan, Monday, February 25, 2008 at 8:21pm
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:
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
csc(x)=1/sin(x) and cot(x)=cos(x)/sin(x)
Now, we know that cos²(x)=1-sin²(x)
So, when we fill this in in the equation we get that:
We can split this fraction in two seperate 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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