If cos A = frac{7}{25} and frac{3 pi}{2}

Question

(A) Given $\cos A = \frac{7}{25}$ and $A$ is in the fourth quadrant,i.e.,$\frac{3 \pi}{2} < A < 2 \pi$. Since $\frac{3 \pi}{2} < A < 2 \pi$,we have $\

Vedclass · Accepted Answer

$\frac{1}{\sqrt{10}} + \frac{27}{625}$

Vedclass · Answer

(A) Given $\cos A = \frac{7}{25}$ and $A$ is in the fourth quadrant,i.e.,$\frac{3 \pi}{2} Since $\frac{3 \pi}{2} Using $\cos A = 2 \cos^2 \frac{A}{2} - 1$,we get $\cos^2 \frac{A}{2} = \frac{1 + \cos A}{2} = \frac{1 + 7/25}{2} = \frac{32/25}{2} = \frac{16}{25}$. Since $\frac{A}{2}$ is in the second quadrant,$\cos \frac{A}{2} = -\frac{4}{5}$. Using $\cos \frac{A}{2} = 2 \cos^2 \frac{A}{4} - 1$,we get $2 \cos^2 \frac{A}{4} = 1 + \cos \frac{A}{2} = 1 - \frac{4}{5} = \frac{1}{5}$,so $\cos^2 \frac{A}{4} = \frac{1}{10}$. Since $\frac{A}{4}$ is in the first quadrant,$\cos \frac{A}{4} = \frac{1}{\sqrt{10}}$. Also,$\cos 2A = 2 \cos^2 A - 1 = 2 \left(\frac{7}{25}\right)^2 - 1 = 2 \left(\frac{49}{625}\right) - 1 = \frac{98 - 625}{625} = -\frac{527}{625}$. Therefore,$\cos \frac{A}{4} + \cos \frac{A}{2} - \cos 2A = \frac{1}{\sqrt{10}} - \frac{4}{5} - \left(-\frac{527}{625}\right) = \frac{1}{\sqrt{10}} - \frac{500}{625} + \frac{527}{625} = \frac{1}{\sqrt{10}} + \frac{27}{625}$.

If $\cos A = \frac{7}{25}$ and $\frac{3 \pi}{2} < A < 2 \pi$,then $\cos \frac{A}{4} + \cos \frac{A}{2} - \cos 2A =$

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