## Trigonometry 7th Edition

1 + $\frac{\sqrt 3}{2}$ or $\frac{ 2 + \sqrt 3 }{2}$
Given expression = $( \sin 60^{\circ} + \cos 60^{\circ})^{2}$ = $\sin^{2} 60^{\circ}$ + 2 $\sin 60^{\circ} \cos 60^{\circ}$ + $\cos^{2} 60^{\circ}$ [ on expanding using identity $(a+b)^{2} = a^{2} + 2ab + b^{2}$] = $(\frac{\sqrt 3}{2})^{2}$ + 2$\times \frac{\sqrt 3}{2} \times\frac{1}{2}$ + $(\frac{1}{2})^{2}$ = $\frac{3}{4} + \frac{\sqrt 3}{2} + \frac{1}{4}$ = $\frac{3}{4} + \frac{1}{4} + \frac{\sqrt 3}{2}$ = $\frac{(3+1)}{4} + \frac{\sqrt 3}{2}$ = $\frac{4}{4} + \frac{\sqrt 3}{2}$ = 1 +$\frac{\sqrt 3}{2}$ or $\frac{2 + \sqrt 3}{2}$