# pre-cal/trig

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cos(x-(pi/4))+sin(x-(pi/4))=1
Find the answer from 0 to 2pi.
When I slove this, I got
cosx+sinx=(1/ square root of 2).

• pre-cal/trig -

if you get to cosx+sinx=1/square root of 2, well we are doing identities in my pre-cal class and cosx+sinx always equals 1, so it would be 1/square root of 2.

• pre-cal/trig -

Sorry, cosx+ sinx is not equal to 1 (from the identites). But, it was like this:
(sinx)^2+ (cosx)^2= 1

• pre-cal/trig -

cos(x-(pi/4))+sin(x-(pi/4))=1
Find the answer from 0 to 2pi

• pre-cal/trig -

square both side:
(cosx+sinx)^x=(1/square root of 2)
then solve: it will work.

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