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# Cos 12

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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

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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
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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