# Trig Identities

ok ... heres 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 isnt getting me anywhere .. plz help

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

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 cancelled the "sin^2" and " - sin^2" on the top and in the bottom i cancelled 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 dont 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
Homework Helper
You forgot the arguments. Sin does not exist, sin(x) does etc.

there all sin(x) or cos(x) :tongue2:

Cyosis
Homework Helper
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.

itz

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

Cyosis
Homework Helper
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
Mentor
ok ... heres 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 isnt getting me anywhere .. plz 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.