In triangle ABC- prove that cosA2-cosB2-cosC2-4 cos-A4cos-B4cos-C4

In ABC, A+B+C=π cosA2+cosB2+cosC2=cosA2+2cosB+C4cosB C4=cosA2+2cosB C4cosπ A4 #160; #160; #160; #160; #160; #160; #160; #160; ...

You visited us 1 times! Enjoying our articles? Unlock Full Access!

Byju's Answer

Standard IX

Mathematics

Basics of Geometry

In triangle ...

Question

In triangle $A B C$ , prove that $cos \frac{A}{2} + cos \frac{B}{2} + cos \frac{C}{2} = 4 cos \frac{π - A}{4} cos \frac{π - B}{4} cos \frac{π - C}{4}$

Open in App

Solution

In $△ A B C, A + B + C = π$
$cos \frac{A}{2} + cos \frac{B}{2} + cos \frac{C}{2} = cos \frac{A}{2} + 2 cos \frac{B + C}{4} cos \frac{B - C}{4} = cos \frac{A}{2} + 2 cos \frac{B - C}{4} cos \frac{π - A}{4}$
$= sin \frac{π - A}{2} + 2 cos \frac{B - C}{4} cos \frac{π - A}{4}$
$= 2 cos \frac{π - A}{4} (sin \frac{π - A}{4} + cos \frac{B - C}{4})$
$= 2 cos \frac{π - A}{4} (cos (\frac{π - B}{4} + \frac{π - C}{4}) + cos (\frac{π - B}{4} - \frac{π - C}{4})) = 4 cos \frac{π - A}{4} cos \frac{π - B}{4} cos \frac{π - C}{4}$

Suggest Corrections

Similar questions

Q. If

A + B + C = π

, show that

cos (A / 2) + cos (B / 2) - cos (C / 2)

= 4 cos \frac{π + A}{4} cos \frac{π + B}{4} cos \frac{π - C}{4}

Q. If

A + B + C = π

show that

cos (A / 2) + cos (B / 2) cos (C / 2) = 4 cos {(π - A) / 4} cos {(π - B) / 4} cos {(π - C) / 4} = 4 cos {(B + C) / 4} cos {(C + A) / 4} cos {(A + B) / 4}

Q. In triangle ABC,

sin \frac{A}{2} + sin \frac{B}{2} + sin \frac{C}{2} \leq \frac{3}{2}

then show that

cos \frac{π + A}{4} cos \frac{π + B}{4} cos \frac{π + C}{4} \leq \frac{1}{8}

Q. In any triangle ABC,

\frac{1 - cos A + cos B + cos C}{1 - cos C + cos A + cos B} = \frac{tan (A / 2)}{tan (C / 2)}

Q. Prove that

$\frac{c o s A}{c c o s B + b c o s C} + \frac{c o s B}{a c o s C + c c o s A} + \frac{c o s C}{a c o s B + b c o s A} = \frac{a^{2} + b^{2} + c^{2}}{2 a b c}$

Join BYJU'S Learning Program

In triangle ABC, prove that cosA2+cosB2+cosC2=4cosπ−A4cosπ−B4cosπ−C4

In △ABC, A+B+C=πcosA2+cosB2+cosC2=cosA2+2cosB+C4cosB−C4=cosA2+2cosB−C4cosπ−A4 =sinπ−A2+2cosB−C4cosπ−A4 =2cosπ−A4(sinπ−A4+cosB−C4) =2cosπ−A4(cos(π−B4+π−C4)+cos(π−B4−π−C4))=4cosπ−A4cosπ−B4cosπ−C4

In triangle $A B C$ , prove that $cos \frac{A}{2} + cos \frac{B}{2} + cos \frac{C}{2} = 4 cos \frac{π - A}{4} cos \frac{π - B}{4} cos \frac{π - C}{4}$