Proving Trigonometric Identity Involving Double Angle Formula

[aisha]

Proove trig identity help please

I think this problem involves the double angle formula which i am not quite familiar with.

1+cos2(x)=cot(x)sin2(x)

I know the cot=1/tan(x) or cos(x)/sin(x)

sin2(x)=2sin(x)cos(x)

cos2(x)=1-2sin^2 (x)

But I am not sure what to do can u please help me?

 

[aisha]

here is what I am getting so far

1+cos(2x)=cot(x) sin2(x)

1+cos(x+x)=[cos(x)/sin(x)] multiplied by sin(x+x)

1+cos^2(x)=cos(x)/sin(x) multiplied by sin^2(x)

Im doing something totally wrong please HELP ME

 

[erik05]

cos2x is equal to cos^2x-sin^2x therefore for the left side:

1-sin^2x+cos^2x
cos^2x+cos^2x
2cos^2x

for the right side:
\frac{1}{tanx} (2sinxcosx)

\frac {2sinxcosx}{\frac{sinx}{cosx}}

2sinxcosx (\frac {cosx}{sinx})

2cos^2x

 

[aisha]

cos2x is equal to cos^2x-sin^2x therefore for the left side:

1-sin^2x+cos^2x
cos^2x+cos^2x
2cos^2x

for the right side:
\frac{1}{tanx} (2sinxcosx)

\frac {2sinxcosx}{\frac{sinx}{cosx}}

2sinxcosx (\frac {cosx}{sinx})

2cos^2x

Hi thanks I didnt know cos2x is equal to cos^2x-sin^2x I am just trying to understand what you did on the right side can u explain why you multiplied 1/tan(x) by (2sinxcosx) thanks soo much

 

[dextercioby]

Okay.Initially,the RHS was:

\cot x \sin 2x(1)

Now:

\cot x=\frac{1}{\tan x}(2)

\sin 2x=2\sin x\cos x (3)

Multiply (2) and (3) and find:

\cot x\sin 2x=\frac{2\sin x\cos x}{\tan x} (4)

,which is just what u asked to prove.

Daniel.

 

[aisha]

ok i totally understand that I think I am over looking one of the trig identities because I didnt know that cos2x=cos^2(x)-sin^2(x) and I also didnt know sin2x=2sin(x)cos(x)

Hey how do u know that?

 

[dextercioby]

Simple,apply the formulas for addition (both for "sine" and for "cosine") and choose equal arguments
\cos 2x=\cos (x+x)=...?

Daniel.