math
posted by jessika on .
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

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 
i don't understand how to use the formula in the second problem. what would i plug into A and B?

cos(xB)cos(x+B) = 2sinxsinB
LS
= cosxcosB + sinxsinB  (cosxcosB  sinxsinB)
=2sinxsinB
= RS 
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