Question

The value of $si{n}^2{\left(1\frac{1}{2}\right)}^{\circ }+si{n}^2(3{)}^{\circ }+si{n}^2{\left(4\frac{1}{2}\right)}^{\circ }+....+si{n}^2{\left(88\frac{1}{2}\right)}^{\circ }$ is equal to

• A.
  29
• B.
  $\frac{59}{2}$
• C.
  $30$
• D. $\frac{29\sqrt{2}+1}{\sqrt{2}}$

Answer 1

The correct option is B

$\frac{59}{2}$
Consider, $si{n}^2{\left(1\frac{1}{2}\right)}^0+si{n}^2{\left(4\frac{1}{2}\right)}^0+....+si{n}^2{\left(88\frac{1}{2}\right)}^{\circ }$

This can written as : $si{n}^2\left(\frac{3}{2}\right)+si{n}^2(3{)}^{\circ }+si{n}^2{\left(\frac{9}{2}\right)}^{\circ }+....+si{n}^2{\left(\frac{177}{2}\right)}^{\circ }....(i)$

The angles of ${\frac{3}{2}}^{\circ },3^{\circ },{\frac{9}{2}}^{\circ },...,{\frac{177}{2}}^{\circ }$ forms an AP, with first term, a as $\frac{3}{2}$ and common difference, d as

$3-\frac{3}{2}=\frac{3}{2}$

Now, $n^{th}$ term of an AP, $a_n=a+(n-1)d.\therefore \frac{177}{2}=\frac{3}{2}+(n+1)\frac{3}{2}\Rightarrow \frac{177}{2}-\frac{3}{2}=\frac{3(n-1)}{2}\Rightarrow \frac{174}{2}=\frac{3(n-1)}{2}\Rightarrow n-1=58\Rightarrow n=59$

Thus, there are in total 59 terms. Therefore, the middle term will be $\frac{59+1}{2}$ term, i.e., ${30}^{th}term.$

$\Rightarrow a_{30}=\frac{3}{2}+(30-1)\frac{3}{2}=\frac{3}{2}+(29\times \frac{3}{2})=45$

Thus, the middle term of the given series is $si{n}^2{45}^{\circ }={\left(\frac{1}{\sqrt{2}}\right)}^2=\frac{1}{2}.$

Now,
$si{n}^2\left(\frac{3}{2}\right)cos\theta =co{s}^2{\left(\frac{177}{2}\right)}^{\circ }.[\therefore sin({90}^{\circ })-=]si{n}^2(3{)}^{\circ }=co{s}^2(87{)}^{\circ }$

Thus ,(i) can be written as $co{s}^2{\left(\frac{177}{2}\right)}^{\circ }+co{s}^2(87{)}^{\circ }+...+\frac{1}{2}+..+si{n}^2(87{)}^{\circ }+{\sin }^2{\left(\frac{177}{2}\right)}^{\circ }$

Using the identity, $si{n}^2\theta +co{s}^2\theta =1,Weget(1+1+..+1)+\frac{1}{2}=29+\frac{1}{2}=\frac{59}{2}$

Hence, the correct answer option b.