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square both side: (cosx+sinx)^x=(1/square root of 2) then solve: it will work.
It is like triangle. a^2+b^2=c^2 you know that one of the side and hyp. so, you can find the ohter by pluging in like this: 5^2+b^2= 13^2 25+b^2=169 b^2=144 b= 12
Can Anybody answer this question for me???? Please??? PlEASE help me with this!! I need it as soon as possible....... cos(x-(pi/4))+sin(x-(pi/4))=1 Find the answer from 0 to 2pi
Sorry. I misread it as y=-4 and x=-2 To solve this: y=x-2 y=2x Since (x-2) and (2x) both equal to y, then they are equal to each other. So,we can set equation like this: 2x=x-2 2x-x=-2 x= -2 To find y, plug in x value in the original equation. y=2(-2)=-4
Learn your identites well in order to prove these. 1. (1+sina) (1-sina) = 1-sina+sina-sin^2a = 1-sin^2a = cos^2a (according to identity) 2. cos^2b-sin^2b you know that sin^2b = 1-cos^2b, so: cos^2b-(1-cos^2b) =cos^2b-1+cos^2b =2cos^2b-1 3. sin^2a - sin^4a=
yes . It is right.

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