Question

If \(\rm \cos A = \pm \frac{4}{5}\) where 450° < A < 540°, then \(\cos \frac{{\rm{A}}}{2}\) is equal to
1. \(\frac{1}{{\sqrt {10} }}\)
2. \(- \sqrt {\frac{3}{{10}}}\)
3. \(\frac{{\sqrt 3 }}{{\sqrt {10} }}\)
4. None of these

Answer 1

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:

\(\rm cos 2A = cos^2 A - sin^ 2 A\)

\(\rm cos 2A = 1 - 2 sin^2 A\)

\(\rm cos 2A = \frac{1 - tan^2A}{1 + tan^2A}\)

\(\rm cos A = cos^2\frac{A}{2} - sin^ 2\frac{A}{2}\)

\(\rm cos A = 1 - 2 sin^2 \frac{A}{2}\)

\(\rm cos A = \frac{1 - tan^2\frac{A}{2}}{1 + tan^2\frac{A}{2}}\)