More Complex Than it Seems!

Algebra Level 5

∑ r = 0 4 ( 4 r ) b 4 − r c r cos ⁡ [ r B − ( 4 − r ) C ] \large\sum_{r=0}^4 \dbinom4r b^{4-r} c^r \cos [ rB - (4-r)C ]

Let a , b a,b and c c denote the sides of a triangle opposite to the angles A , B A,B and C C respectively.

Compute the value of the summation above, where a = 5 , b = 12 a=5,b=12 and c = 13 c=13 .


The answer is 625.

Somyaneel Sinha by Somyaneel Sinha , 21, India

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1 solution

Somyaneel Sinha
Feb 11, 2016

Let S= ∑ r = 0 4 ( 4 C r ) b 4 − r c r cos ⁡ ( r B − ( 4 − r ) C ) \sum_{r=0}^4 (^4C_r) b^{4-r} c^{r} \cos\ (rB-(4-r)C)

and T= ∑ r = 0 4 ( 4 C r ) b 4 − r c r sin ⁡ ( r B − ( 4 − r ) C ) \sum_{r=0}^4 (^4C_r) b^{4-r} c^{r} \sin\ (rB-(4-r)C)

then S + iT= ∑ r = 0 4 ( 4 C r ) b 4 − r c r e i ( r B − ( 4 − r ) C ) \sum_{r=0}^4 (^4C_r) b^{4-r} c^{r} e^{i(rB-(4-r)C)}

=> S+iT= ( c e i B + b e − i C ) 4 (ce^{iB} + be^{-iC})^{4}

=> S+iT= ( ( c cos ⁡ B + b cos ⁡ C ) + i ( c sin ⁡ B − b sin ⁡ C ) ) 4 ((c \cos\ B + b \cos\ C)+ i(c \sin\ B - b \sin\ C))^{4}

since according to sine rule c sin ⁡ C \frac{c}{\sin\ C} = b sin ⁡ B \frac{b}{\sin\ B} => ( c sin ⁡ B − b sin ⁡ C ) (c \sin\ B - b \sin\ C) =0

=> T=0 whereas S= a 4 a^{4} [ using Projection formula]

=> S= 5 4 5^{4} = 625

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