Evaluate cos( 50 + theta) - sin( 40 - theta)

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Solution

$\cos (50 ° + θ) - \sin (40 ° - θ) = \cos (50 ° + θ) - \sin [90 ° - (50 ° + θ)] = \cos (50 ° + θ) - \cos (50 ° + θ) [as, \sin (90 ° - x) = \cos x] = 0$

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c o s (40^{\circ} + θ) - s i n (50^{\circ} - θ) + \frac{c o s^{2} 40^{\circ} + c o s^{2} 50^{\circ}}{s i n^{2} 40^{\circ} + s i n^{2} 50^{\circ}}

cos (40^{o} + θ) - sin (50^{o} - θ) + cos 240^{o} + cos 250^{o} sin 240^{o} + sin 250^{o} = ?

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cos (40 - θ) - sin (50 + θ) + \frac{{cos}^{2} 40 + {cos}^{2} 50}{{sin}^{2} 40 + {sin}^{2} 50}

s i n (50 + θ) - c o s (40 - θ) = ?

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