Trigonometry> tan^2(π/16)+tan^(2π/16)+tan^2(3π/16)+.......

tan^2(π/16)+tan^(2π/16)+tan^2(3π/16)+......+tan^(7π/16)

Sanskar Mahato , 7 Years ago

Grade 11

1 Answers

Aarushi Ahlawat

Group terms (\pi/16) and (7\pi/16) and similarly (2\pi/16) and (6\pi/16) ….. It will be like given below:

$=tan^2(\pi/16)+tan^2(7\pi/16)+tan^2(2\pi/16)+tan^2(6\pi/16)+tan^2(3\pi/16)+tan^2(5\pi/16)+tan^2(4\pi/16)$

$=\left (tan(\pi/16)+tan(7\pi/16) \right )^2-2tan(\pi/16)tan(7\pi/16)+\left (tan(2\pi/16)+tan(6\pi/16) \right )^2-2tan(2\pi/16)tan(6\pi/16)+\left (tan(3\pi/16)+tan(5\pi/16) \right )^2-2tan(3\pi/16)tan(6\pi/16)+1$

$=\left (tan(\pi/16)+tan(7\pi/16) \right )^2-2tan(\pi/16)tan(\pi/2-\pi/16)+\left (tan(2\pi/16)+tan(6\pi/16) \right )^2-2tan(2\pi/16)tan(\pi/2-2\pi/16)+\left (tan(3\pi/16)+tan(5\pi/16) \right )^2-2tan(3\pi/16)tan(\pi/2-3\pi/16)+1$

$=\left (tan(\pi/16)+tan(7\pi/16) \right )^2-2tan(\pi/16)cot(\pi/16)+\left (tan(2\pi/16)+tan(6\pi/16) \right )^2-2tan(2\pi/16)cot(2\pi/16)+\left (tan(3\pi/16)+tan(5\pi/16) \right )^2-2tan(3\pi/16)cot(3\pi/16)+1$

$=\left (tan(\pi/16)+tan(7\pi/16) \right )^2-2+\left (tan(2\pi/16)+tan(6\pi/16) \right )^2-2+\left (tan(3\pi/16)+tan(5\pi/16) \right )^2-2+1$

$=\left (tan(\pi/16)+tan(7\pi/16) \right )^2+\left (tan(2\pi/16)+tan(6\pi/16) \right )^2+\left (tan(3\pi/16)+tan(5\pi/16) \right )^2-5$ $=\left (tan(\pi/16)+cot(\pi/16) \right )^2+\left (tan(2\pi/16)+cot(2\pi/16) \right )^2+\left (tan(3\pi/16)+cot(3\pi/16) \right )^2-5$

$=\left (\frac{sin^2(\pi/16)+cos^2(\pi/16)}{sin(\pi/16)cos(\pi/16)}\right )^2+\left (\frac{sin^2(2\pi/16)+cos^2(2\pi/16)}{sin(2\pi/16)cos(2\pi/16)}\right )^2+\left (\frac{sin^2(3\pi/16)+cos^2(3\pi/16)}{sin(3\pi/16)cos(3\pi/16)}\right )^2-5$

$=\left (\frac{1}{sin(\pi/16)cos(\pi/16)}\right )^2+\left (\frac{1}{sin(2\pi/16)cos(2\pi/16)}\right )^2+\left (\frac{1}{sin(3\pi/16)cos(3\pi/16)}\right )^2-5$

$=\left (\frac{2}{sin(2\pi/16))}\right )^2+\left (\frac{2}{sin(4\pi/16)}\right )^2+\left (\frac{2}{sin(6\pi/16)}\right )^2-5$

$=\left (\frac{2}{sin(2\pi/16))}\right )^2+\left (\frac{2}{\frac{1}{\sqrt{2}}}\right )^2+\left (\frac{2}{sin(6\pi/16)}\right )^2-5$

$=\left (\frac{2}{sin(2\pi/16))}\right )^2+8+\left (\frac{2}{sin(6\pi/16)}\right )^2-5$

$=\left (\frac{2}{sin(\pi/8))}\right )^2+\left (\frac{2}{sin(3\pi/8)}\right )^2+3$

$=\left (\frac{2}{sin(\pi/8))}\right )^2+\left (\frac{2}{cos(\pi/8)}\right )^2+3$

$=\left (\frac{2(cos^2(\pi/8)+sin^2(\pi/8))}{sin(\pi/8)cos(\pi/8))}\right )^2+3$

$=\left (\frac{2X2}{2sin(\pi/8)cos(\pi/8))}\right )^2+3$

$=\left (\frac{2X2}{sin(\pi/4)}\right )^2+3$

$=\left (\frac{2X2}{\frac{1}{\sqrt{2}}} \right )^2+3=32+3=35$

It looks lengthy by if you solve for only one group and use solution for remaining two groups with appropriate multiple for (\pi/16) term. That way it will be solved in few steps.

Last Activity: 7 Years ago