1.

If 0 < θ < π/8​

Answer»

Step-by-step explanation:

GIVEN:-

  • 0 < θ < π/8

To PROVE:-

  • \rm\sqrt{2 + \sqrt{2 + \sqrt{2 + 2cos4 \theta} } } = 2cos \theta/2

Formula Used:-

  • 1 + cosθ = 2 cos² θ/2

Proof:-

\rm <klux>LHS</klux> = \sqrt{2 + \sqrt{2 + \sqrt{2 + 2cos4 \theta} } }

\rm= \sqrt{2 + \sqrt{2 + \sqrt{2 (1+ cos4 \theta)} } }

\rm= \sqrt{2 + \sqrt{2 + \sqrt{2 (2 cos^{2} 2\theta)} } }

\rm= \sqrt{2 + \sqrt{2 + \sqrt{4 cos^{2} 2\theta} } }

\rm= \sqrt{2 + \sqrt{2 + 2cos2 \theta} }

\rm= \sqrt{2 + \sqrt{2 (1 + cos2 \theta)} }

\rm= \sqrt{2 + \sqrt{2 (2cos ^{2} \theta)} }

\rm= \sqrt{2 + \sqrt{4cos ^{2} \theta} }

\rm= \sqrt{2 + 2cos \theta }

\rm= \sqrt{2 (1 + cos \theta) }

\rm= \sqrt{2 (2cos^{2} ( \theta/2) }

\rm= \sqrt{4cos^{2} ( \theta/2) }

\rm= 2cos( \theta/2)

LHS = RHS

Hence Proved



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