Question

%2B%202cos%20%5C%3A%208a%7D%20%7D%20%7D%20" id="TexFormula1" title="m. \: 2cos \: a = \sqrt{2 + \sqrt{2 + \sqrt{2 + 2cos \: 8a} } } " alt="m. \: 2cos \: a = \sqrt{2 + \sqrt{2 + \sqrt{2 + 2cos \: 8a} } } " align="absmiddle" class="latex-formula">
please help me.....​
I need full work!!

Answer 1

While this is with theta instead of A it still is the same thing. I hope this helps

Answer 2

Answer:  see proof below

Step-by-step explanation:

Use the following Double Angle Identity:

cos 2A = 2 cos²A - 1

Proof RHS → LHS

Given:                            $\sqrt{2+\sqrt{2+\sqrt{2+2cos8A}}}$

Factor:                           $\sqrt{2+\sqrt{2+\sqrt{2(1+cos8A)}}}$

Let α = 4A:                      $\sqrt{2+\sqrt{2+\sqrt{2(1+cos2\alpha)}}}$

Double Angle Identity:   $\sqrt{2+\sqrt{2+\sqrt{2(1+2cos^2\alpha-1)}}}$

Simplify:                            $\sqrt{2+\sqrt{2+\sqrt{2(2cos^2\alpha)}}}$

                                          $\sqrt{2+\sqrt{2+2cos\alpha}}}$

Substitute (α = 4A):           $\sqrt{2+\sqrt{2+2cos4A}}}$

Factor:                               $\sqrt{2+\sqrt{2(1+cos4A)}}}$

Let β = 2A                          $\sqrt{2+\sqrt{2(1+cos2\beta)}}}$

Double Angle Identity:     $\sqrt{2+\sqrt{2(1+2cos^2\beta-1)}}}$

Simplify:                             $\sqrt{2+\sqrt{2(2cos^2\beta)}}}$

                                          $\sqrt{2+2cos\beta}$

Substitute (β = 2A):            $\sqrt{2+2cos2\alpha}$

Factor:                                 $\sqrt{2(1+cos2\alpha)}$

Double Angle Identity:       $\sqrt{2(1+2cos^2\alpha-1)}$

Simplify:                               $\sqrt{2(2cos^2A)}$

                                             2 cos A

2cos A =  2cos A   $\checkmark$