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.