In any triangle A B C cos 2 (A/2)+ cos 2 (B/2)+ cos 2 (C/2)=

Question

In any triangle A B C cos 2 (A/2)+ cos 2 (B/2)+ cos 2 (C/2)= (A) 1+(2 r/R) (B) 2-(1/2 R) (C) 2(1+(r/4 R)) (D) 2(1+(r/2 R))

Tardigrade Admin · Accepted Answer

We have, cos 2 (A/2)+ cos 2 (B/2)+ cos 2 (C/2) =(1/2) (1+ cos A)+(1+ cos B)+(1+ cos C) =(1/2) 3+ cos A+ cos B+ cos C =(1/2) 3+1+(r/R) [∵ cos A+ cos B+ cos C=1+(r/R)] =(1/2)(4+(r/R))=2+(r/2 R) cos 2 (A/2)+ cos 2 (B/2)+ cos 2 (C/2) =(1/2) (1+ cos A)+(1+ cos B)+(1+ cos C) =(1/2) 3+ cos A+ cos B+ cos C =(1/2) 3+1+(r/R) [∵ cos A+ cos B+ cos C=1+(r/R)] =(1/2)(4+(r/R))=2+(r/2 R)

Q. In any triangle $A BC$ $cos^{2} \frac{A}{2} + cos^{2} \frac{B}{2} + cos^{2} \frac{C}{2} =$

$1 + \frac{2 r}{R}$

$2 - \frac{1}{2 R}$

$2 (1 + \frac{r}{4 R})$

$2 (1 + \frac{r}{2 R})$

Q. In any triangle ABC cos22A​+cos22B​+cos22C​=

1+R2r​

2−2R1​

2(1+4Rr​)

2(1+2Rr​)

Q. In any triangle $A BC$ $cos^{2} \frac{A}{2} + cos^{2} \frac{B}{2} + cos^{2} \frac{C}{2} =$

$1 + \frac{2 r}{R}$

$2 - \frac{1}{2 R}$

$2 (1 + \frac{r}{4 R})$

$2 (1 + \frac{r}{2 R})$