# Simplifying expressions

• Mar 27th 2007, 10:32 AM
Kim2425
Simplifying expressions
1)Tan0 + Cot0

2)1-Sec(2)0
------------
1-Cosec(2)0

3) Sin0
----------
Cosec0 - Cot0

4) 1+Cos2 0

0 = 0 with a line through the middle, not sure how you do it on the forums, and anything in brackets is power of.

Any help with this would be appreciated.

thank you!
• Mar 27th 2007, 05:38 PM
topsquark
Quote:

Originally Posted by Kim2425
1)Tan0 + Cot0

Let's call "0" = t for convenience.

tan(t) + cot(t) = sin(t)/cos(t) + cos(t)/sin(t) <-- Get a common denominator then add the fractions:

= [sin^2(t) + cos^2(t)]/[sin(t)cos(t)]

= 1/[sin(t)cos(t)]

We can work with this a bit more:

= 2/[2sin(t)cos(t)] = 2/sin(2t) = 2*csc(2t)

-Dan

TECHNICAL COMMENT: Tan(t) and tan(t) are technically two different functions. The correct usage for the common trig functions is sin(t), cos(t), tan(t), csc(t), sec(t), and cot(t).
• Mar 27th 2007, 05:41 PM
topsquark
Quote:

Originally Posted by Kim2425
2)1-Sec^2(t)
------------
1-csc^2(t)

First note that:
1 - sec^2(t) = tan^2(t)
and
1 - csc^2(t) = cot^2(t)

So:
[1 - sec^2(t)]/[1 - csc^2(t)] = tan^2(t)/cot^2(t) = tan^4(t)

-Dan
• Mar 27th 2007, 05:46 PM
topsquark
Quote:

Originally Posted by Kim2425
3) Sin(t)
----------
Csc(t) - Cot(t)

sin(t)/[csc(t) - cot(t)] = sin(t)/[1/sin(t) - cos(t)/sin(t)]

Multiply the numerator and denominator of the overall fraction by sin(t):

sin^2(t)/[1 - cos(t)] (We need to keep in the back of our minds that sin(t) cannot be 0 from here on, but it couldn't do that in the original expression anyway.)

Multiply the numerator and denominator by 1 + cos(t):

[sin^2(t)*(1 + cos(t))]/[1 - cos^2(t)] = [sin^2(t)*(1 + cos(t))]/[sin^2(t)]

= 1 + cos(t)

-Dan
• Mar 27th 2007, 05:57 PM
topsquark
Quote:

Originally Posted by Kim2425
4) 1+Cos^2(t)

Is there a typo in here? I can't think of anything to do with this one.

-Dan