Question

Prove that

\sin^{2} (α - \frac{π}{4}) + \sin^{2} (α + \frac{π}{12}) - \sin^{2} (α - \frac{π}{12}) = \frac{1}{2}

Accepted Answer

To Prove: $\sin^{2} (α - \frac{π}{4}) + \sin^{2} (α + \frac{π}{12}) - \sin^{2} (α - \frac{π}{12}) = \frac{1}{2}$

L.H.S. $= \sin^{2} (α - \frac{π}{4}) + \sin^{2} (α + \frac{π}{12}) - \sin^{2} (α - \frac{π}{12})$

Using the identity, $a^{2} - b^{2} = (a - b) (a + b)$ , we have

L.H.S. $= \sin^{2} (α - \frac{π}{4}) + (\sin (α + \frac{π}{12}) - \sin (α - \frac{π}{12})) (\sin (α + \frac{π}{12}) + \sin (α - \frac{π}{12}))$

Apply the formulae, $\sin C - \sin D = 2 \cos \frac{C + D}{2} \cdot \sin \frac{C - D}{2}$ and $\sin C + \sin D = 2 \sin \frac{C + D}{2} \cdot \cos \frac{C - D}{2}$

$= \sin^{2} (α - \frac{π}{4}) + ((2 \cos (\frac{2 α}{2})) \sin (\frac{π}{12})) \cdot (2 \sin (\frac{2 α}{2}) \cos \frac{π}{12})$

$= \sin^{2} (α - \frac{π}{4}) + (2 \sin α \cos α) (2 \sin \frac{π}{12} \cos \frac{π}{12})$

$= \sin^{2} (α - \frac{π}{4}) + (\sin 2 α) (\sin 2 (\frac{π}{12}))$

$= \sin^{2} (α - \frac{π}{4}) + (\sin 2 α) (\sin (\frac{π}{6}))$

L.H.S. $= \sin^{2} (α - \frac{π}{4}) + (\sin 2 α) (\frac{1}{2})$

We know that $\sin^{2} θ = \frac{1 - \cos 2 θ}{2}$

$= \frac{1 - \cos 2 (α - \frac{π}{4})}{2} + \frac{1}{2} \sin 2 α$

$= \frac{1 - \cos (2 α - \frac{π}{2})}{2} + \frac{1}{2} \sin 2 α$

$= \frac{1 - \cos (- (\frac{π}{2} - 2 α))}{2} + \frac{1}{2} \sin 2 α$

$= \frac{1 - \sin 2 α}{2} + \frac{1}{2} \sin 2 α$ $[∵ \cos (90 ° - θ) = \sin θ and \cos (- θ) = \cos θ]$

$= \frac{1}{2}$

Hence Proved.

Telangana Board Solutions for Exercise 5: Exercise 6 (e)

Telangana Board Mathematics Solutions for Exercise - Telangana Board Solutions for Exercise 5: Exercise 6 (e)

Questions from Telangana Board Solutions for Exercise 5: Exercise 6 (e) with Hints & Solutions

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