Posted by jessika on Thursday, November 15, 2007 at 7:31pm.
i have a few questions
cosx + cosx =2secx
 
1+sinx 1sinx
cos(xB)cos(xB) = 2sinxsinB
csc2x= 1 secx cscx

2
cotx= cos5x+cos3x

sin 5xsin 3x

math  Reiny, Thursday, November 15, 2007 at 8:09pm
for the first one:
LS = cosx/(1+sinx) + cosx/(1sinx)
getting a common denominator of (1+sinx)(1sinx)
= [cosx(1sinx) + cosx(1+sinx)]/(1+sinx)(1sinx)
= 2cosx/(1sin^2)
= 2cosx/cos^2x
= 2/cosx
= 2secx
= RS
for cos(xB)cos(xB) = 2sinxsinB
use the formula cos(AB) = 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

math  Reiny, Thursday, November 15, 2007 at 8:19pm
for cos(xB)cos(xB) = 2sinxsinB
you must have a typo, the LS is zero the way it stands
I am sure you meant
cos(xB)cos(x+B) = 2sinxsinB
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