Question

Prove
$\sqrt{2 + \sqrt{2 + \sqrt{2 + cos8x = 2cosx} } }$
​

Answer 1

$\large\underline{\bold{Given \:Question - }}$

$\sf \: Prove \: that \: \sqrt{2 + \sqrt{2 + \sqrt{2 + 2cos8x} } } = 2cosx$

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

We know that

$\boxed{ \sf \: 1 + cosx = 2 {cos}^{2} \dfrac{x}{2} }$

Now, Consider,

$\rm :\longmapsto\: \sqrt{2 + \sqrt{2 + \sqrt{2 + 2cos8x} } }$

$\: \: = \sf \: \sqrt{2 + \sqrt{2 + \sqrt{2(1 + cos8x)} } }$

$\: \: = \sf \: \sqrt{2 + \sqrt{2 + \sqrt{2 \times 2 {cos}^{2}4x } } }$

$\: \: = \sf \: \sqrt{2 + \sqrt{2 + \sqrt{4 {cos}^{2}4x } } }$

$\: \: = \sf \: \sqrt{2 + \sqrt{2 + 2cos4x} }$

$\: \: = \sf \: \sqrt{2 + \sqrt{2(1 + cos4x)} }$

$\: \: = \sf \: \sqrt{2 + \sqrt{2 \times 2 {cos}^{2}2x } }$

$\: \: = \sf \: \sqrt{2 + \sqrt{4 {cos}^{2}2x } }$

$\: \: = \sf \: \sqrt{2 + 2cos2x}$

$\: \: = \sf \: \sqrt{2(1 + cos2x)}$

$\: \: = \sf \: \sqrt{2 \times 2 {cos}^{2}x }$

$\: \: = \sf \: \sqrt{4 {cos}^{2}x }$

$\: \: = \sf \: 2cosx$

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

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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)