Question

prove it (1+ cos π/8) (1+ cos 3π/8) (1+ cos 5π/8) (1+ cos 7π/8) = 1/8​

Answer 1

$\begin{gathered}\Large{\sf{{\underline{Formula \: Used - }}}}  \end{gathered}$

$\boxed{ \bf{ \: cos(\pi \: - x) = - cosx}}$

$\boxed{ \bf{ \: {sin}^{2} x + {cos}^{2} x = 1}}$

$\boxed{ \bf{ \: 1 - cos2x = {2sin}^{2} x}}$

$\large\underline{\sf{Solution-}}$

$\sf \: Consider, LHS$

$\rm :\longmapsto\: \: \bigg(1 + cos\dfrac{\pi}{8} \bigg) \bigg(1 + cos\dfrac{3\pi}{8} \bigg) \bigg(1 + cos\dfrac{5\pi}{8} \bigg) \bigg(1 + cos\dfrac{7\pi}{8} \bigg)$

Now,

$\boxed{\sf \: cos\dfrac{7\pi}{8} = cos\bigg( \pi - \dfrac{\pi}{8}\bigg) = - cos\dfrac{\pi}{8}}$

and

$\boxed{\sf \: cos\dfrac{5\pi}{8} = cos\bigg( \pi - \dfrac{3\pi}{8}\bigg) = - cos\dfrac{3\pi}{8}}$

So, LHS can be rewritten as

$\rm = \: \bigg(1 + cos\dfrac{\pi}{8} \bigg) \bigg(1 + cos\dfrac{3\pi}{8} \bigg) \bigg(1 - cos\dfrac{3\pi}{8} \bigg) \bigg(1 - cos\dfrac{\pi}{8} \bigg)$

$\rm = \: \bigg(1 + cos\dfrac{\pi}{8} \bigg) \bigg(1 - cos\dfrac{\pi}{8} \bigg) \bigg(1 + cos\dfrac{3\pi}{8} \bigg) \bigg(1 - cos\dfrac{ 3\pi}{8} \bigg)$

$\rm = \: \bigg(1 - cos ^{2} \dfrac{\pi}{8} \bigg) \bigg(1 - cos ^{2} \dfrac{3\pi}{8} \bigg)$

$\rm\: = \: {sin}^{2} \dfrac{\pi}{8} \times {sin}^{2} \dfrac{3\pi}{8}$

$\rm \: = \: \dfrac{1}{2}\bigg( {2sin}^{2}\dfrac{\pi}{8} \bigg) \times \dfrac{1}{2}\bigg( 2 {sin}^{2}\dfrac{3\pi}{8} \bigg)$

$\rm \: = \dfrac{1}{4}\bigg(1 - cos\dfrac{\pi}{4} \bigg) \bigg(1 - cos\dfrac{3\pi}{4} \bigg)$

$\rm \: = \dfrac{1}{4}\bigg( 1 - \dfrac{1}{ \sqrt{2} } \bigg) \bigg(1 - cos(\pi - \dfrac{\pi}{4} )\bigg)$

$\rm \: = \: \dfrac{1}{4}\bigg(1 - \dfrac{1}{ \sqrt{2} } \bigg) \bigg(1 + cos\dfrac{\pi}{4} \bigg)$

$\rm \: = \dfrac{1}{4}\bigg(1 - \dfrac{1}{ \sqrt{2} } \bigg)\bigg(1 + \dfrac{1}{ \sqrt{2} } \bigg)$

$\rm \: = \dfrac{1}{4}\bigg(1 - \dfrac{1}{2} \bigg)$

$\rm \: = \dfrac{1}{4} \times \dfrac{1}{2}$

$\rm \: = \: \dfrac{1}{8}$

$\bf \: = \: RHS$

${\boxed{\boxed{\bf{Hence, Proved}}}}$

Additional Information :-

Trigonometry Formulas

sin(−θ) = −sin θ

cos(−θ) = cos θ

tan(−θ) = −tan θ

cosec(−θ) = −cosecθ

sec(−θ) = sec θ

cot(−θ) = −cot θ

Product to Sum Formulas

sin x sin y = 1/2 [cos(x–y) − cos(x+y)]

cos x cos y = 1/2[cos(x–y) + cos(x+y)]

sin x cos y = 1/2[sin(x+y) + sin(x−y)]

cos x sin y = 1/2[sin(x+y) – sin(x−y)]

Sum to Product Formulas

sin x + sin y = 2 sin [(x+y)/2] cos [(x-y)/2]

sin x – sin y = 2 cos [(x+y)/2] sin [(x-y)/2]

cos x + cos y = 2 cos [(x+y)/2] cos [(x-y)/2]

cos x – cos y = -2 sin [(x+y)/2] sin [(x-y)/2]

Sum or Difference of angles

cos (A + B) = cos A cos B – sin A sin B

cos (A – B) = cos A cos B + sin A sin B

sin (A+B) = sin A cos B + cos A sin B

sin (A -B) = sin A cos B – cos A sin B

tan(A+B) = [(tan A + tan B)/(1 – tan A tan B)]

tan(A-B) = [(tan A – tan B)/(1 + tan A tan B)]

cot(A+B) = [(cot A cot B − 1)/(cot B + cot A)]

cot(A-B) = [(cot A cot B + 1)/(cot B – cot A)]

cos(A+B) cos(A–B)=cos^2A–sin^2B=cos^2B–sin^2A

sin(A+B) sin(A–B) = sin^2A–sin^2B=cos^2B–cos^2A

Multiple and Submultiple angles

sin2A = 2sinA cosA = [2tan A /(1+tan²A)]

cos2A = cos²A–sin²A = 1–2sin²A = 2cos²A–1= [(1-tan²A)/(1+tan²A)]

tan 2A = (2 tan A)/(1-tan²A)