Precalc problem

[milt]

Precalc problem

Problem:

sin²ø( (csc²ø - 1 / cos²ø) - (cot²ø) ) = sec²ø - cos²ø - tan²ø

Helpful conversions:

tan²ø + 1 = sec²ø
sin²ø + cos²ø = 1
cot²ø + 1 = csc²ø
sin = 1/csc
cos = 1/sec
tan = sin/cos
csc = 1/sin
sec = 1/cos
cot = cos/sin

Basically I have to take the left side and make it look like the right side, and show the work on how to do it. This was given as an extra credit problem.

Any of you calc geniouses willing to help out with this would be appreciated. Tips/Hints/Suggestions/Ect. I'm stuck

 

[TMSTKSBK]

woohoo! proofs!

This reminded me how much I hated trig >_<

the idea in these problems is to manipulate the stuff so it looks the same.

So my step 1 is to convert it all to sines and cosines.

 

[khaine]

before i try to solve it ... in the 2nd set of parens do you mean:

csc²ø - (1 / cos²ø) or (csc²ø - 1) / cos²ø

 

[TheOutkastDev]

I've replaced theta with x for these equations, but this should be correct.

Split the csc²-1 so you can more easily factor the sin²x.

sin²x( csc²x/cos²x - 1/cos²x - cos²x/sin²x )

Use some substitutions to get everything in terms of sin and cos

sin²x( 1/sin²x * 1/cos²x - 1/cos²x - cos²x/sin²x )

Factor and Multply.

sin²x/sin²xcos²x - sin²x/cos²x - cos²xsin²x/sin²x

Reduce and replace common identities.

1/cos²x - tan²x - cos²x

sec²x - cos²x - tan²x

 

[milt]

Wow, splendid! Good job w/ that Outkast... Appreciate the help man.

 

[TheOutkastDev]

no problem.