Question

If $\alpha ,\beta ,\gamma ,\delta$ are in arithmetic progression.Then,$\cos \beta$ is equal to

• A. $\cos (\alpha +\gamma -\delta )$
• B. $\cos (\alpha +\delta -\gamma )$
• C. $\cos (2\gamma -\delta )$
• D. $\cos \left(\frac{\alpha +\gamma }{2}\right)$

Answer 1

Answer: (B), (C), (D)

The correct options are
B $\cos (\alpha +\delta -\gamma )$
C $\cos (2\gamma -\delta )$
D $\cos \left(\frac{\alpha +\gamma }{2}\right)$

$\alpha ,\beta ,\gamma ,\delta$ in A.P
Let common ratio is r, such that
$\beta =\alpha +r,\gamma =\alpha +2r,\delta =\alpha +3r$
A: $\alpha +\gamma -\delta =\alpha +\alpha +2r-\alpha -3r=\alpha -r\cos (\alpha +\gamma -\delta )=\cos (\alpha -r)\ne \cos \beta$

B: $\alpha +\delta -\gamma =\alpha +\alpha +3r-\alpha -2r=\alpha +r\cos (\alpha +\delta -\gamma )=\cos (\alpha +r)=\cos \beta$

C: $2\gamma -\delta =2\alpha +4r-\alpha -3r=\alpha +r\cos (2\gamma -\delta )=\cos (\alpha +r)=\cos \beta$

D: $\frac{\alpha +\gamma }{2}=\frac{\alpha +\alpha +2r}{2}=\alpha +r\cos \left(\frac{\alpha +\gamma }{2}\right)=\cos \frac{\alpha +\alpha +2r}{2}=\cos (\alpha +r)=\cos \beta$
Hence, option 'B', 'C' and 'D' are correct.