If 0<β<α<π4,cos⁡(α+β)=35 and cos⁡(α−β)=45 then sin⁡2α is equals to

If 0

A
0
B
1
C
2
D
None of these

Solution:

$\begin{array}{l} \sin 2 α = \sin {(α + β) + (α - β)} \\ = \sin (α + β) \cos (α - β) + \cos (α + β) \sin (α - β) . . . . . . . . . . . . . . . . . . . . . . . \dots (i) \end{array}$

Now,

$\begin{array}{l} \sin (α + β) = \sqrt{1 - \cos^{2} (α + β)} \\ = \sqrt{1 - \frac{9}{25}} = \frac{4}{5} [∵ 0 < α + β < \frac{π}{2}] \end{array}$

and,

$\begin{array}{l} \sin (α - β) = \sqrt{1 - \cos^{2} (α - β)} \\ = \sqrt{1 - \frac{16}{25}} = \frac{3}{5} [∵ 0 < α - β < \frac{π}{4}] \end{array}$

From Eq. (i),

$\sin 2 α = \frac{4}{5} \cdot \frac{4}{5} + \frac{3}{5} \cdot \frac{3}{5} = \frac{16}{25} + \frac{9}{25} = \frac{25}{25} = 1$

If $0 < β < α < \frac{π}{4}, \cos (α + β) = \frac{3}{5}$ and $\cos (α - β) = \frac{4}{5}$ then $\sin 2 α$ is equals to

Solution:

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