Question

The average value of $\sin 2°,\sin 4°,\sin 6°,...,\sin 180°\mathrm{is}$

• A.
  $\frac{1}{90}\cot 1°$
• B.
  $\frac{1}{90}\tan 1°$
• C.
  $\frac{1}{90}\tan 89°$
• D.
  $\frac{1}{90}\cot 89°$

Answer 1

Answer: (A), (C)

$\frac{1}{90}\tan 89°$
Here, we represent the sum of sines as:
$S=\sin 2°+\sin 4°+..+\sin 180°$.
Now, using the expression for sum of sines
$S=\sin \left(\alpha \right)+\sin \left(\alpha +\beta \right)+..+\sin \left(\alpha +(n-1)\beta \right)\Rightarrow S=\frac{\sin \left(\frac{\mathrm{n\beta }}{2}\right)}{\sin \left(\frac{\beta }{2}\right)}\sin \left(\alpha +(n-1)\frac{\beta }{2}\right).$
Now, for our series, we have $\alpha =2°,\beta =2°\mathrm{and}n=90$.
So, we write sum as:
$S=\frac{\sin \left(\frac{90\times 2°}{2}\right)}{\sin \left(\frac{2°}{2}\right)}\sin \left(2°+(90-1)\frac{2°}{2}\right)\Rightarrow S=\frac{\sin \left(90°\right)}{\sin \left(1°\right)}\sin \left(\frac{180°+2°}{2}\right)\Rightarrow S=\frac{\sin \left(90°\right)}{\sin \left(1°\right)}\sin \left(91°\right)\Rightarrow S=\frac{\sin \left(90°\right)}{\sin \left(1°\right)}\sin \left(91°\right)\Rightarrow S=\frac{1}{\sin \left(1°\right)}\sin \left(90°+1°\right)\Rightarrow S=\frac{\cos \left(1°\right)}{\sin \left(1°\right)}=\cot \left(1°\right).$
Now, we have 90 terms so average of sum is given as:
$A=\frac{S}{90}=\frac{1}{90}\cot (1°)$
Also, $\cot (1°)=\cot (90°-89°)=\tan (89°)\Rightarrow A=\frac{1}{90}\cot 1°=\frac{1}{90}\tan 89°.$