Question

Let $f\left(n , x\right)=\displaystyle \int ncos\left(n x\right)dx,$ with $f\left(n , 0\right)=0.$ If the expression $\displaystyle \sum _{x = 1}^{89} f \left(1 , x\right)$ simplifies to $\frac{sin a sin ⁡ b}{sin ⁡ c},$ then the value of $\frac{b}{a c}$ is (where $a>b$ )

• A. $45$
• B. $89$
• C. $\frac{89}{45}$
• D. $\frac{45}{89}$

Answer 1

Answer: (C)

$f\left(n , x\right)=\frac{n s i n \left(n x\right)}{n}+C$
As $f\left(n , 0\right)=C=0$
$\Rightarrow f\left(n , x\right)=sin\left(n x\right)$
Thus, $\displaystyle \sum _{x = 1}^{89} f \left(1 , x\right) = sin 1 + sin ⁡ 2 + . . . . . + sin ⁡ 89$
$=\frac{s i n \left(\frac{1 + 89}{2}\right) s i n \frac{89 \times 1}{2}}{s i n \left(\frac{1}{2}\right)}$
$=\frac{s i n \left(45\right) s i n \left(\frac{89}{2}\right)}{s i n \left(\frac{1}{2}\right)}$