Prove that- bccos 2 1 2 A- cacos 2 1 2 B-ab cos 2 1 2 C - s 2 -

bccos ^ 2 1 2 #160; A=bc. s( s a ) #160; bc =s( s a ) . ∴ L.H.S=s[ ( s a ) +( s b ) +( s c ) #160; ] =s[ 3s s( a+b+c ) #160; ] =3 s ...

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Cosine Rule

Prove that: ...

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Prove that: $b c {cos}^{2} \frac{1}{2} A + c a {cos}^{2} \frac{1}{2} B + a b {cos}^{2} \frac{1}{2} C = s^{2} .$

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A B C

cot \frac{A}{2} - cot \frac{B}{2} = \frac{(b - a) s}{Δ}

(b + c) sin \frac{A}{2} = a cos \frac{B - C}{2}

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

\frac{b - c}{a} {cos}^{2} \frac{1}{2} A + \frac{c - a}{b} {cos}^{2} \frac{1}{2} B + \frac{a - b}{c} {cos}^{2} \frac{1}{2} C = 0.

{(\frac{2^{a}}{2^{b}})}^{(a + b)} . {(\frac{2^{b}}{2^{c}})}^{(b + c)} . {(\frac{2^{c}}{2^{d}})}^{(c + d)} . {(\frac{2^{d}}{2^{a}})}^{(d + a)} = 1

{(\frac{2^{a}}{2^{b}})}^{a + b} \times {(\frac{2^{b}}{2^{c}})}^{b + c} \times {(\frac{2^{c}}{2^{a}})}^{c + a} = 1

cos 2 A + cos 2 B - cos 2 C = 1 - 4 sin A sin B cos C

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