Question

Show that (cos(x)) + sin(x))^2 = 1+ sin(2x) for every number x.

Answer 1

\[
\begin{align}
\big((cos(x)+sin(x)\big)^2&=1+sin(2x)\\
\big((cos(x)+sin(x)\big)\cdot\big((cos(x)+sin(x)\big)&=\\
cos^2(x)+2sin(x)cos(x)+sin^2(x)&=\\
2sin(x)cos(x)+1&=1+sin(2x)\\
2sin(x)cos(x)&=sin(2x)
\end{align}
\]

This proof uses two fundamental trig identities:
\[
sin^2(x)+cos^2(x)=1\\
2sin(x)cos(x)=sin(2x)
\]

Answer 2

can you re-write this, I don't understand what you are writing. Sorry.

Answer 3

this is (Cosx + sinx)^2 = 1+sin2x

sin^2x + cos^2x =1 ( this is an identity )

open the bracket using (a+b)^2 = a^2 +2ab +b^2

Cos^2x +2CosxSinx + Sin^2x

now we know that Cos^2x +Sin^2x=1 right?

this means 2CosxSinx +1

Now there is another double angle identity

which says that Sin2x=2CosxSinx

So Sin2x+1= Right hand side as in the question

and Hence its proved