prove that root 2 + root 2 + 2 cos 4 x is equal to 2 cos x
Answers
Step-by-step explanation:
Thus, the result proves that the
\sqrt { 2 + \sqrt { 2 + 2 \cos 4 A } } = 2 \cos A
2+
2+2cos4A
=2cosA
To prove:
\sqrt { 2 + \sqrt { 2 + 2 \cos 4 A } } = 2 \cos A
2+
2+2cos4A
=2cosA
Solution:
Given that
\sqrt { 2 + \sqrt { 2 + 2 \cos 4 A } }
2+
2+2cos4A
We already knew that the value of \cos 2Acos2A using the double angle formula will be,
\cos 2 A = 2 \cos ^ { 2 } A - 1cos2A=2cos
2
A−1
Here 2A is also known as multiple angles.
Similarly,
\cos 4 A = 2 \cos ^ { 2 } ( 2 A ) - 1cos4A=2cos
2
(2A)−1
Substituting this value in the given term, we get,
\sqrt { 2 + \sqrt { 2 + 2 \cos 4 \mathrm { A } } }
2+
2+2cos4A
\begin{lgathered}\begin{array} { c } { = \sqrt { 2 + \sqrt { 2 + 2 \cos ( 4 A ) } } } \\\\ { = \sqrt { 2 + \sqrt { 2 + 2 \left( 2 \cos ^ { 2 } 2 A - 1 \right) } } } \end{array}\\\end{lgathered}
=
2+
2+2cos(4A)
=
2+
2+2(2cos
2
2A−1)
\begin{lgathered}\begin{array} { c } { = \sqrt { 2 + \sqrt { 2 + 4 \cos ^ { 2 } 2 A - 2 } } } \\\\ { = \sqrt { 2 + \sqrt { 4 \cos ^ { 2 } 2 A } } } \\\\ { = \sqrt { 2 + 2 \cos 2 A } } \end{array}\end{lgathered}
=
2+
2+4cos
2
2A−2
=
2+
4cos
2
2A
=
2+2cos2A
\begin{lgathered}\begin{aligned} = & \sqrt { 2 + 2 \left( 2 \cos ^ { 2 } A - 1 \right) } \\\\ = & \sqrt { 2 + 4 \cos ^ { 2 } A - 2 } \\\\ & = \sqrt { 4 \cos ^ { 2 } A } \\\\ & = 2 \cos A \end{aligned}\end{lgathered}
=
=
2+2(2cos
2
A−1)
2+4cos
2
A−2
=
4cos
2
A
=2cosA
Thus,
\sqrt { 2 + \sqrt { 2 + 2 \cos 4 A } } = 2 \cos A
2+
2+2cos4A
=2cosA
Hence proved.