Question

Prove that

2 CosA = √2 + √2 + √2 + 2cos 8A ​

Answer 1

Hey mate...

here's the answer...

√{2+√(2+2cos4A)}

=√[2+√{2+2(2cos²2A-1)}]

=√{2+√(2+4cos²2A-2)}

=√(2+√4cos²2A)

=√(2+2cos2A)

=√{2+2(2cos²A-1)}

=√(2+4cos²A-2)

=√4cos²A

=2cosA .....Proved...

Hope it helps ❤️

Answer 2

Question-

$\bf \: 2cosA \: = \sqrt{2 + \sqrt{2 + \sqrt{2 + 2cos8A} } } \:$

Answer:

2 cosA

Step-by-step explanation:

RHS = $\sf\: \sqrt{2 + \sqrt{2 + \sqrt{2 + 2cos8A} } } \:$

$= \sf \: \sqrt{2 + \sqrt{2 + \sqrt{2 (1 + cos8A)} } } \: \\ \\ \sf \: = \sqrt{2 + \sqrt{2 + \sqrt{2 \{1 + cos(2 \times 4 A \: ) \}} } } \\ \\ = \sf \: \sqrt{2 + \sqrt{2 + \sqrt{2(2 {cos}^{2} 4A)} } } \\ \\ \bf \because \: 1 + cos2x = 2 {cos}^{2} x \\ \\ = \sf \: \sqrt{2 + \sqrt{2 + 2cos4A} } \\ \\ = \sf \: \sqrt{2 + \sqrt{2(1 + cos4A)} } \\ \\ = \sf \: \sqrt{2 + \sqrt{2 \{ 1 + cos(2 \times 2A\}} } \\ \\ = \sf \: \sqrt{2 + \sqrt{2( {2cos}^{2} 2A)} } \\ \\ = \sf \: \sqrt{2 + 2cos2A} \\ \\ = \sf \: \sqrt{2(1 + cos2A)} \\ \\ = \sf \: \: \sqrt{2(2 {cos}^{2}A) } \\ \\ = \sf \: 2cosA \:$