How to prove 2.Sin Ó¨.Cos Ó¨ = Sin 2Ó¨ ?

Question

Help me guys! I don't know how to prove the trigonometrical identity :

2.Sin Ó¨.Cos Ó¨ = Sin 2Ó¨

I know it can be proved by putting values of Ó¨ , but I want a geometrical proof.

Answer 1

The geometrical proof is too elaborate to be given here. But it can be derived without any difficulty from the geometrical proof of sin(A+B)=sinAcosB+cosAsinB. I hope the said proof is provided in your maths book and you're also familiar with it. Now, put A=B (i.e. take two equal angles). You'll find the given identity very easy to prove. Try it.

Answer 2

2.Sinx.cosx= sin2x

rhs= sin2x

= sin(x+x)

=sinxcosx+cosxsinx

=2sinx.cosx

=LHS QED

Answer 3

this is simple

we know that sin(A+B) = sinAcosB+cosAsinB

in the above formula put A=B=Ó¨

sin(Ó¨+Ó¨) = sinÓ¨cosÓ¨+cosÓ¨sinÓ¨ = 2sinÓ¨cosÓ¨

Answer 4

wtf!!!

Answer 5

A simple proof using trigonometrical identities goes as follows:

sin(a+b) = sin(a)cos(b) + sin(b)cos(a)

Hence, if a=b we have a+b=a+a=2a, and also

sin(a+b) = sin(2a) = sin(a)cos(b)+sin(b)cos(a) = 2sin(a)cos(a)

There is a different proof using complex numbers that is quite cool. Here it goes just for fun.

e^(i a) = cos(a)+i sin(a). Hence im(e^(i a)) = sin(a) where im(z) is the imaginary part of the complex number z.

Now we have im(e^(i 2a)) = sin(2a). But also

e^(i 2a) = e^(i a)*e^(i a) = (cos(a)+i sin(a))(cos(a)+i sin(a))

            = cos(a)^2 - sin(a)^2 + i 2sin(a)cos(a)

hence

im(e^(i 2a)) = 2sin(a)cos(a)

which proves the result.