Verified Answer :

pluralistic

sin2x+cos2x=1
2sinxcosx+2cos^2x-1=1
2cosx(cosx+sinx-1)=0
2cosx=0  cosx+sinx=1
True

I did this with double angle formulas. sin2x=2sinxcosx, and cos2x=2cos^x-1. With plugging this in, I used factoring. At the end, there are two factors. One of the factors is cosx+sinx-1=0, which is cosx+sinx=1. Therefore, the answer is true.

lublana

Answer with Step-by-step explanation:

We are given that it is true since

\(sin2x+ cos2x=1\) then\(sinx+cosx=1\)

We have to prove that  \(sinx+cosx=1\)

We know that

\(sin2x=2sinxcosx\)

\(cos2x=2cos^2x-1\)

Substituting  the values then we get

\(2sinxcosx+2cos^2x-1=1\)

\(2sinxcosx+2cos^2x=1+1\)

\(2(sinxcosx+cos^2x)=2\)

\(sinxcosx+cos^2x-1=0\)

\(cos^2x+sinxcosx=1\)

\(cosx(cosx+sinx)=1\)

\(cosx=1,cosx+sinx=1\)

Hence, \(sinx+cosx=1\)

Related Questions