Question

if cos B = 1/2{a+1/a},prove that:[i]2cos2B ={ a^2+1/a^2 } [ii]2cos4B ={ a^4+1/a^4 }

Answer 1

EXPLANATION :

$\sf \cos \: B = \frac{1}{2} (a + \frac{1}{a} )$

$\sf \implies \: { \cos}^{2} B = \frac{1}{4} {(a + \frac{1}{a} )}^{2} \underline{ \: \: \: \: \: \: \: \: }(1)$

$\sf \implies \: 1 - { \sin}^{2} B = \frac{1}{4} {(a + \frac{1}{a} )}^{2}$

$\sf \implies { \sin}^{2} B = 1 - \frac{1}{4} {(a + \frac{1}{a} )}^{2} \underline{ \: \: \: \: \: \: \: \: \: }(2)$

SUBTRACTING EQN 1 AND 2

$\sf {\boxed {\sf { \cos \: 2B = { \cos}^{2} B - { \sin}^{2} B}}}$

1ST PART

APPLYING THIS FORMULA

$\sf {2 \cos \: 2B = {2( \cos}^{2} B - { \sin}^{2} B)}$

$\implies \sf {2 \cos \: 2B = {2( \frac{1}{4} {(a + \frac{1}{a}) }^{2} - 1 + \frac{1}{4} {(a + \frac{1}{a}) }^{2} )}}$

$\implies \sf {\boxed {\sf {2 \cos \: 2B = {a}^{2} + \frac{1}{ {a}^{2} } }}} \: \: \: \: proved$

2ND PART

Same process