Posted by jessika on Thursday, November 15, 2007 at 7:31pm.
for the first one:
LS = cosx/(1+sinx) + cosx/(1-sinx)
getting a common denominator of (1+sinx)(1-sinx)
= [cosx(1-sinx) + cosx(1+sinx)]/(1+sinx)(1-sinx)
= 2cosx/(1-sin^2)
= 2cosx/cos^2x
= 2/cosx
= 2secx
= RS
for cos(x-B)-cos(x-B) = 2sinxsinB
use the formula cos(A-B) = cosAcosB + sinAsinB on the Left Side, it comes apart very nicely after that
For the last two, try changing all trig ratios into sines and cosines.
Show me what you get
i dont understand how to use the formula in the second problem. what would i plug into A and B?
cos(x-B)-cos(x+B) = 2sinxsinB
LS
= cosxcosB + sinxsinB - (cosxcosB - sinxsinB)
=2sinxsinB
= RS
i dont understand how to use the formula in the second problem. what would i plug into A and B?
for cos(x-B)-cos(x-B) = 2sinxsinB
you must have a typo, the LS is zero the way it stands
I am sure you meant
cos(x-B)-cos(x+B) = 2sinxsinB
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