Without using trigonometric table, find the values of:

(i) sin 60° cos 30° + cos 60° sin 30°

(ii) sin 45° cos 30° - cos 45° sin 30°

(iii) cos 60° cos 45° + sin 60° sin 45°

(iv) cos 90° + cos² 45° sin 30° tan 45°

(i) sin 60° cos 30° + cos 60° sin 30°

= $\dfrac{\sqrt{3}}{2}\times\dfrac{\sqrt{3}}{2} + \dfrac{1}{2}\times\dfrac{1}{2}$

= $\dfrac{3}{4} + \dfrac{1}{4}$

= $\dfrac{3+1}{4} = \dfrac{4}{4}$

= 1.

Hence, sin 60° cos 30° + cos 60° sin 30° = 1.

(ii) sin 45° cos 30° - cos 45° sin 30°

= $\dfrac{1}{\sqrt{2}}\times\dfrac{\sqrt{3}}{2} - \dfrac{1}{\sqrt{2}} \times\dfrac{1}{2}$

= $\dfrac{\sqrt{3}}{2\sqrt{2}} - \dfrac{1}{2\sqrt{2}}$

= $\dfrac{\sqrt{3}-1}{2\sqrt{2}}$.

Hence, sin 45° cos 30° - cos 45° sin 30° = $\dfrac{\sqrt{3}-1}{2\sqrt{2}}$.

(iii) cos 60° cos 45° + sin 60° sin 45°

= $\dfrac{1}{2}\times\dfrac{1}{\sqrt{2}} + \dfrac{\sqrt{3}}{2}\times\dfrac{1}{\sqrt2}$

= $\dfrac{1}{2\sqrt{2}} + \dfrac{\sqrt{3}}{2\sqrt{2}}$

= $\dfrac{1 + \sqrt{3}}{2\sqrt{2}}$.

Hence, cos 60° cos 45° + sin 60° sin 45° = $\dfrac{1 + \sqrt{3}}{2\sqrt{2}}$.

(iv) cos 90° + cos² 45° sin 30° tan 45°

As, cos² 45° = (cos 45°)² = $\Big(\dfrac{1}{\sqrt{2}}\Big)^2 = \dfrac{1}{2}$

Therefore,

cos 90° + cos² 45° sin 30° tan 45°

= 0 + $\dfrac{1}{2}\times\dfrac{1}{2}\times1$

= $\dfrac{1}{4}$.

Hence, cos 90° + cos² 45° sin 30° tan 45° = $\dfrac{1}{4}$.