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Hi, if Cos 12° =h, find Sin12°in terms of h...subsequently, write down the value of Cot 12°...please help

 Jun 2, 2019
 #1
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\(cos^2x+sin^2x=1<=> h^2 + sin^212=1<=>sin12= \sqrt{1-h^2}=\sqrt{(1-h)(1+h)}\)

 

\(cot12=\frac{cos12}{sin12}=\frac{h}{\sqrt{(1-h)(1+h)}}\)

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 Jun 2, 2019
edited by Dimitristhym  Jun 2, 2019
edited by Dimitristhym  Jun 2, 2019
 #2
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By the Pythagorean Identity,

 

\(\sin^2(12^\circ)+\cos^2(12^\circ)\ =\ 1\)

                                                      We are given that   cos(12°) = h   so we can substitute  h  in for  cos(12°)

\(\sin^2(12^\circ)+h^2\ =\ 1\)

                                                      Subtract  h2  from both sides of the equation.

\(\sin^2(12^\circ)\ =\ 1-h^2\)

                                                      Because  12°  is in Quadrant I,  sin(12°)  is positive. So take positive sqrt of both sides.

\(\sin(12^\circ)\ =\ \sqrt{1-h^2}\)

 

By definition of cotangent,

 

\(\cot(12^\circ)\ =\ \frac{\cos(12^\circ)}{\sin(12^\circ)}\)

                                          Substitute  h  in for  cos(12°)  and substitute  √[ 1 - h2 ]  in for  sin(12°)

\(\cot(12^\circ)\ =\ \frac{h}{\sqrt{1-h^2}}\)_

 Jun 2, 2019
 #3
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Hectictar,

 

thank you very much!!..may I ask one more thing please, how would we approach finding Sin78?..Thank you very much for your time..

Guest Jun 3, 2019
 #4
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Note that there is a rule which says  sin( 90° - θ )  =  cos( θ )  for any angle measure  θ . So...

 

sin( 78° )  =  sin( 90° - 12° )  =  cos( 12° )  =  h

hectictar  Jun 3, 2019
edited by hectictar  Jun 4, 2019

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