Question

`m. \: 2cos \: a = \sqrt{2 + \sqrt{2 + √(2 + 2cos \: 8a) } }`

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

Explanation:

Use the following Double Angle Identity:

cos 2A = 2 cos²A - 1

Proof RHS → LHS

Given:
`\sqrt{2+\sqrt{2+√(2+2cos8A)}}`

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

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

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

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

`\sqrt{2+√(2+2cos\alpha)}}`

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

Factor:
`\sqrt{2+√(2(1+cos4A))}}`

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

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

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

`√(2+2cos\beta)`

Substitute (β = 2A):
`√(2+2cos2\alpha)`

Factor:
`√(2(1+cos2\alpha))`

Double Angle Identity:
`√(2(1+2cos^2\alpha-1))`

Simplify:
`√(2(2cos^2A))`

2 cos A

2cos A = 2cos A
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