Proving sin^2 + 2cos - 1 / 2 + cos - cos^2 = 1 / 1 + sec

[ihatemath]

ok ... here's my problem
i need to prove that the LHS = RHS


sin^2 + 2cos - 1 / 2 + cos - cos^2 = 1 / 1 + sec

i tried to use the pythagarean identies and substitute the 1 and 2
but tht isn't getting me anywhere .. please help

 

[Cyosis]

If you don't list the arguments of the functions we can't help you, also show the steps you've taken so far.

 

[ihatemath]

ummm ok .. here's what i got


sin^2 + 2cos - 1 / 2 + cos - cos^2 = 1 / 1 + sec

LHS :

= sin^2 + cos + cos - sin^2 - cos^2 / 1 + sin^2 + cos^2 + cos - cos^2
... here i canceled the "sin^2" and " - sin^2" on the top and in the bottom i canceled the "cos^2" and " - cos^2"

= cos + cos - (cos)(cos) / 1 + sin^2 + cos

= cos + cos - cos^2 / sin^2 + cos^2 + sin^2 + cos

= cos + cos / sin^2 + sin^2 + cos

... I don't noe what to do from here ... am i even tackling this problem the right way ??

RHS :

1 / 1 + sec

= 1 / 1 + 1/cos
= 1 + cos ... no problems here .. jus the LHS

 

[Cyosis]

You forgot the arguments. Sin does not exist, sin(x) does etc.

 

[ihatemath]

there all sin(x) or cos(x)

 

[Cyosis]

Okay then we're arriving at the next problem. I have a feeling that some brackets are missing, for example what is 1/1+sec(x) is it $$1+\sec(x)$$ or $$\frac{1}{1+\sec x}$$. Same goes for the right hand side. Use brackets to make divisions clear.

 

[ihatemath]

itz

$$\frac{1}{1+\sec x}$$

 

[Cyosis]

Yes, but what about the other side, similar ambiguities exist there. For me to assist you on this problem you will need to take all those ambiguities away.

 

[Mark44]

ok ... here's my problem
i need to prove that the LHS = RHS


sin^2 + 2cos - 1 / 2 + cos - cos^2 = 1 / 1 + sec

i tried to use the pythagarean identies and substitute the 1 and 2
but tht isn't getting me anywhere .. please help

Are you trying to prove that
$$sin^2(x) + 2cos(x) - \frac{1}{2} + cos(x) - cos^2(x) = \frac{1}{1 + sec(x)}?$$

That's actually the most reasonable way to interpret what you have written. (You've already told us what you meant on the right-hand side.