Question

sin(75°) + cos(-15°) = 
1. \({1+\sqrt3}\)
2. \(\frac{1}{\sqrt2}\)
3. 0
4. \(\frac{1+\sqrt3}{\sqrt2}\)

Answer 1

Correct Answer - Option 4 : \(\frac{1+\sqrt3}{\sqrt2}\)

Concept:

sin (x + y) = sin x cos y + cos x sin y

cos (x - y) = cos x cos y + sin x sin y

 

Calculation:

sin(75°) = sin(30° + 45)

= sin(30°) cos(45°) + cos 30° sin 45°                 (∵ sin(x + y) = sinx.cosy + cosx. siny)

= \(\frac{1}{2}\times \frac{1}{\sqrt2} +\frac{\sqrt3}{2}\times \frac{1}{\sqrt2}\)

= \(\frac{1+\sqrt3}{2\sqrt2}\)

Now, 

cos(-15°) = cos(30° - 45°)

= cos 30° cos 45° + sin 30° sin 45°               (∵ cos (x - y) = cosx.cosy + sinx sin y)

=  \(\frac{\sqrt3}{2}\times \frac{1}{\sqrt2}+\frac{1}{2}\times \frac{1}{\sqrt2} \)

= \(\frac{1+\sqrt3}{2\sqrt2}\)

∴ sin(75°) + cos(-15°) = \(\frac{1+\sqrt3}{2\sqrt2}\) + \(\frac{1+\sqrt3}{2\sqrt2}\)

= \(\frac{1+\sqrt3}{\sqrt2}\)

Hence, option (4) is correct.