Proove trig identity help

  • Thread starter aisha
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  • #1
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Proove trig identity help plz

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 Im not sure what to do can u please help me? :rolleyes:
 

Answers and Replies

  • #2
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here is what im 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
 
  • #3
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[tex] cos2x [/tex] is equal to [tex] cos^2x-sin^2x [/tex] therefore for the left side:

[tex] 1-sin^2x+cos^2x [/tex]
[tex] cos^2x+cos^2x [/tex]
[tex] 2cos^2x [/tex]

for the right side:
[tex] \frac{1}{tanx} (2sinxcosx) [/tex]

[tex] \frac {2sinxcosx}{\frac{sinx}{cosx}} [/tex]

[tex] 2sinxcosx (\frac {cosx}{sinx}) [/tex]

[tex] 2cos^2x [/tex]
 
  • #4
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erik05 said:
[tex] cos2x [/tex] is equal to [tex] cos^2x-sin^2x [/tex] therefore for the left side:

[tex] 1-sin^2x+cos^2x [/tex]
[tex] cos^2x+cos^2x [/tex]
[tex] 2cos^2x [/tex]

for the right side:
[tex] \frac{1}{tanx} (2sinxcosx) [/tex]

[tex] \frac {2sinxcosx}{\frac{sinx}{cosx}} [/tex]

[tex] 2sinxcosx (\frac {cosx}{sinx}) [/tex]

[tex] 2cos^2x [/tex]

Hi thanks I didnt know [tex] cos2x [/tex] is equal to [tex] cos^2x-sin^2x [/tex] Im 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 :smile:
 
  • #5
dextercioby
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Okay.Initially,the RHS was:

[tex] \cot x \sin 2x [/tex](1)

Now:

[tex] \cot x=\frac{1}{\tan x} [/tex](2)

[tex] \sin 2x=2\sin x\cos x [/tex] (3)

Multiply (2) and (3) and find:

[tex]\cot x\sin 2x=\frac{2\sin x\cos x}{\tan x} [/tex] (4)

,which is just what u asked to prove.

Daniel.
 
  • #6
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ok i totally understand that I think Im 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? :redface:
 
  • #7
dextercioby
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Simple,apply the formulas for addition (both for "sine" and for "cosine") and choose equal arguments
[tex] \cos 2x=\cos (x+x)=...? [/tex]

Daniel.
 

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