If A, B, C are angles of a triangles, then cos A + cos B + cos C is equal to

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

The correct option is A

$c o s A + c o s B + c o s C = 2 c o s (\frac{A + B}{2}) c o s (\frac{A - B}{2}) + c o s C$
= $2 s i n \frac{C}{2} c o s (\frac{A - B}{2}) + 1 - 2 s i n^{2} \frac{C}{2}$
= $1 + 2 s i n \frac{C}{2} [c o s (\frac{A - B}{2}) - s i n \frac{C}{2}]$
= $1 + 2 s i n \frac{C}{2} [c o s (\frac{A - B}{2}) - c o s \frac{A + B}{2}]$

= $1 + 2 s i n \frac{A}{2} .2 s i n \frac{B}{2} . s i n \frac{C}{2}$
= $1 + 4 s i n \frac{A}{2} . s i n \frac{B}{2} . s i n \frac{C}{2}$
= $1 + \frac{r}{R}$

Suggest Corrections

Similar questions

A, B, C, D

cos A + cos B + cos C + cos D

∣ ∣ ∣ ∣ \begin{matrix} - 1 & c o s C & c o s B c o s C & - 1 & c o s A c o s B & c o s A & - 1 \end{matrix} ∣ ∣ ∣ ∣

\frac{c o s A}{a} = \frac{c o s B}{b} = \frac{c o s C}{c}

a = 2

A, B, C & D

A B C D,

cos A + cos B + cos C + cos D + 1

A B C

cos A, 1 - cos B, cos C

A . P .

sin A + sin C = 1

∠ B

Join BYJU'S Learning Program