Correct Answer - Option 1 :
\(\frac{1}{{\sqrt {10} }}\)
Concept:
cos 2A = 2 cos2 A - 1
cos A = \(\rm 2 cos^{2}\frac{A}{2} - 1\)
Calculation:
Given: \(\rm \cos A = \pm \frac{4}{5}\)
Range of A: 450° < A < 540°
⇒ 360° + 90° < A < 360° + 180°
⇒ 90° < A < 180°
So A lies on 2nd quadrant
Hence, \(\rm \cos A = - \frac{4}{5}\) [cos θ is negative in 2nd quadrant]
As we know that,
\(\rm cos A = 2 cos^{2}\frac{A}{2} - 1\)
\(- \frac{4}{5} = \) \(\rm 2 cos^{2}\frac{A}{2} - 1\)
\(\rm- \frac{4}{5} + 1 = 2 cos^{2}\frac{A}{2} \)
\(\rm \frac{-4 + 5}{5} = \rm 2 cos^{2}\frac{A}{2} \)
\(\rm \frac{1}{5} = \rm 2 cos^{2}\frac{A}{2} \)
\(\rm \frac{1}{10} = cos^{2}\frac{A}{2} \)
\(\rm cos\frac{A}{2} = \frac{1}{\sqrt{10}} \)
The formula of cos 2A and cos A:
\(\rm cos 2A = cos^2 A - sin^ 2 A\)
\(\rm cos 2A = 1 - 2 sin^2 A\)
\(\rm cos 2A = \frac{1 - tan^2A}{1 + tan^2A}\)
\(\rm cos A = cos^2\frac{A}{2} - sin^ 2\frac{A}{2}\)
\(\rm cos A = 1 - 2 sin^2 \frac{A}{2}\)
\(\rm cos A = \frac{1 - tan^2\frac{A}{2}}{1 + tan^2\frac{A}{2}}\)