Question #50171

In triangle ABC prove that a^2 *cot A + b^2 *cot B + c^2 *cot C =4*Delta. where Delta=area of the triangle.
1

Expert's answer

2014-12-29T13:03:09-0500

Answer on Question #50171 - Math - Trigonometry

In triangle ABCABC prove that a2cotA+b2cotB+c2cotC=4Δa^2 \cdot \cot A + b^2 \cdot \cot B + c^2 \cdot \cot C = 4 \cdot \Delta , where Δ\Delta - area of the triangle.



Solution:

OO is the center of the circumcircle. The area of the triangle AOBAOB is given by the following formula


SAOB=12AOOBsinAOB^S _ {A O B} = \frac {1}{2} A O \cdot O B \cdot \sin \widehat {A O B}


Since OO is the center of the circumcircle, the angle AOB^\widehat{AOB} two times larger than angle CC of the triangle ABCABC . Using this and the facts that AOAO and OBOB are the radii of the circumcircle (we'll denote its value by RR in the further) and that sin2α=2sinαcosα\sin 2\alpha = 2\sin \alpha \cdot \cos \alpha , the previous formula can be rewritten in the following way


SAOB=12AOOBsinAOB^=12R2sin2C=R2sinCcosCS _ {A O B} = \frac {1}{2} A O \cdot O B \cdot \sin \widehat {A O B} = \frac {1}{2} R ^ {2} \cdot \sin 2 C = R ^ {2} \cdot \sin C \cdot \cos C


The area of the triangle ABCABC can be given as a sum of areas of triangles AOBAOB , BOCBOC and COACOA :


Δ=SAOB+SBOC+SCOA=R2sinAcosA+R2sinBcosB+R2sinCcosC\Delta = S _ {A O B} + S _ {B O C} + S _ {C O A} = R ^ {2} \sin A \cdot \cos A + R ^ {2} \sin B \cdot \cos B + R ^ {2} \sin C \cdot \cos C


Now use the law of sines to express RR in terms of sines of angles of triangle ABCABC and values of its sides.


asinA=bsinB=csinC=2R\frac {a}{\sin A} = \frac {b}{\sin B} = \frac {c}{\sin C} = 2 RR2=14(asinA)2=14(bsinB)2=14(csinC)2R ^ {2} = \frac {1}{4} \left(\frac {a}{\sin A}\right) ^ {2} = \frac {1}{4} \left(\frac {b}{\sin B}\right) ^ {2} = \frac {1}{4} \left(\frac {c}{\sin C}\right) ^ {2}


Using this we obtain the following expression for Δ\Delta

Δ=14(a2cosAsinA+b2cosBsinB+c2cosCsinC)\Delta = \frac {1}{4} \left(a ^ {2} \frac {\cos A}{\sin A} + b ^ {2} \frac {\cos B}{\sin B} + c ^ {2} \frac {\cos C}{\sin C}\right)


Multiplying this expression by 4 we obtain


a2cotA+b2cotB+c2cotC=4Δ,a ^ {2} \cdot \cot A + b ^ {2} \cdot \cot B + c ^ {2} \cdot \cot C = 4 \cdot \Delta ,


as required.

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