- #1

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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

- Thread starter ihatemath
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- #1

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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

- #2

Cyosis

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

= 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 ??

1 / 1 + sec

= 1 / 1 + 1/cos

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

- #4

Cyosis

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

- #5

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

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Cyosis

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itz

[tex]\frac{1}{1+\sec x}[/tex]

[tex]\frac{1}{1+\sec x}[/tex]

- #8

Cyosis

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- #9

Mark44

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Are you trying to prove that

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

[tex]sin^2(x) + 2cos(x) - \frac{1}{2} + cos(x) - cos^2(x) = \frac{1}{1 + sec(x)}?[/tex]

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.

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