Question

prove that 2cosx=root2+root2+root2+2cos8x​

Answer 1

Answer:

Step-by-step explanation:

cos2 theta= 2cos^2 theta-1

$\sqrt{2\sqrt{2\sqrt{2(1+2cos^{2}4theta-1 } } }$

$\sqrt{2\sqrt{2+2cos4theta} }$

$\sqrt{2\sqrt{2(1+2cos^{2} theta-1} }$

$\sqrt{2+2costheta}$

$\sqrt{2(1+2cos^{2}theta-1 }$

$2costheta$ is the answer

Answer 2

Step-by-step explanation:

cos2 theta= 2cos^2 theta-1

\sqrt{2\sqrt{2\sqrt{2(1+2cos^{2}4theta-1 } } }

2

2

2(1+2cos

2

4theta−1

\sqrt{2\sqrt{2+2cos4theta} }

2

2+2cos4theta

\sqrt{2\sqrt{2(1+2cos^{2} theta-1} }

2

2(1+2cos

2

theta−1

\sqrt{2+2costheta}

2+2costheta

\sqrt{2(1+2cos^{2}theta-1 }

2(1+2cos

2

theta−1

2costheta2costheta is the answer