Question

If they form an increasing arithmetic progression, then

• A. $\sin (\beta +\gamma )=\sin (\alpha +\delta )$
• B. $\sin (\beta -\gamma )=\sin (\alpha -\delta )$
• C. $\tan 2(\alpha -\beta )=\tan (\beta -\delta )$
• D. $\cos (\alpha +\gamma )=\cos 2\beta$

Answer 1

Answer: (A), (C), (D)

$\sin (\beta +\gamma )=\sin (\alpha +\delta )$
$\tan 2(\alpha -\beta )=\tan (\beta -\delta )$
$\cos (\alpha +\gamma )=\cos 2\beta$

Given:$\alpha ,\beta ,\gamma ,\delta$ are in AP

Let $\alpha ,\beta =\alpha +d,\gamma =\alpha +2d,\delta =\alpha +3d$

$\sin (\beta +\gamma )=\sin (\alpha +d+\alpha +2d)=\sin (2\alpha +3d)$

$\sin (\alpha +\delta )=\sin (\alpha +\alpha +3d)=\sin (2\alpha +3d)$

So, A is correct.

$\sin (\beta -\gamma )=\sin (\alpha +d-\alpha -2d)=-\sin d$

$\sin (\alpha -\delta )=\sin (\alpha -\alpha -3d)=-\sin 3d$

So, B is incorrect.

$\tan 2(\alpha -\beta )=\tan \left(2\right(\alpha -\alpha -d\left)\right)=-\tan 2d$

$\tan (\beta -\delta )=\tan (\alpha +d-\alpha -3d))=-\tan 2d$

So, C is correct.

$\cos (\alpha +\gamma )=\cos (\alpha +\alpha +2d)=\cos \left(2\right(\alpha +d\left)\right)$

$\cos 2\beta =\cos 2(\alpha +d)$

So, D is correct.