Without using trigonometric tables, find the values of;

(i) (3sin² 45° + 2cos²60°)

(ii) (3cos² 30° + tan²60°)

(iii) (cos 0° + sin 45° + sin 30°)(sin 90° - cos 45° + cos 60°)

(iv) 2 $\sqrt{2}$ cos 45°cos 60° + 2 $\sqrt{3}$ sin 30° tan 60° - cos 0°

(v) $\dfrac{4}{3}$ tan²30°+ sin²60° - 3 cos²60°+ $\dfrac{3}{4}$ tan²60°- 2 tan²45°

(vi) $\dfrac{\sin^2 45^\circ + \cos^2 45^\circ}{\tan^2 60^\circ}$

(i) As,

sin² 45° = (sin 45°)² = $\Big(\dfrac{1}{\sqrt{2}}\Big)^2 = \dfrac{1}{2}$

and

cos² 60° = (cos 60°)² = $\Big(\dfrac{1}{2}\Big)^2 = \dfrac{1}{4}$

Substituting values we get :

(3sin² 45° + 2cos²60°)

= $3\times\dfrac{1}{2} + 2\times\dfrac{1}{4}$

= $\dfrac{3}{2} + \dfrac{2}{4} = \dfrac{3}{2} + \dfrac{1}{2}$

= $\dfrac{3+1}{2} = \dfrac{4}{2}$ = 2.

Hence, 3 sin² 45° + 2 cos²60° = 2.

(ii) As,

cos² 30° = (cos 30°)² = $\Big(\dfrac{\sqrt{3}}{2}\Big)^2 = \dfrac{3}{4}$

and

tan² 60° = (tan 60°)² = $(\sqrt{3})^2$ = 3

Therefore,

3cos² 30° + tan²60°

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

= $\dfrac{9}{4}$ + 3

= $\dfrac{9 + 12}{4} = \dfrac{21}{4} = 5\dfrac{1}{4}$.

Hence, 3cos² 30° + tan²60° = $5\dfrac{1}{4}$.

(iii) (cos 0° + sin 45° + sin 30°)(sin 90° - cos 45° + cos 60°)

$= \Big( 1 + \dfrac{1}{\sqrt{2}} + \dfrac{1}{2} \Big)\Big(1 - \dfrac{1}{\sqrt{2}} + \dfrac{1}{2}\Big)\\[1em] = \Big(\dfrac{2\sqrt{2}+ 2 + \sqrt{2}}{2\sqrt{2}}\Big)(\dfrac{2\sqrt{2} - 2 + \sqrt{2}}{2\sqrt{2}}\Big)\\[1em] = \Big(\dfrac{3\sqrt{2}+ 2}{2\sqrt{2}}\Big)\Big(\dfrac{3\sqrt{2}- 2}{2\sqrt{2}}\Big)\\[1em] = \Big(\dfrac{(3\sqrt{2})^2- (2)^2 }{2\sqrt{2}\times {2\sqrt{2}}}\Big)\\[1em] = \Big(\dfrac{18-4}{8}\Big) \\[1em] = \dfrac{14}{8} \\[1em] = \dfrac{7}{4} \\[1em] = 1\dfrac{3}{4}.$

Hence, (cos 0° + sin 45° + sin 30°)(sin 90° - cos 45° + cos 60°) = $1\dfrac{3}{4}$.

(iv) $2\sqrt{2}$ cos 45°cos 60° + 2 $\sqrt{3}$ sin 30° tan 60° - cos 0°

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

= 1 + 3 - 1

= 3.

Hence, 2 $\sqrt{2}$ cos 45°cos 60° + 2 $\sqrt{3}$ sin 30° tan 60° - cos 0° = 3.

(v) As,

cos² 60° = (cos 60°)² = $\Big(\dfrac{1}{2}\Big)^2 = \dfrac{1}{4}$

and

tan² 30° = (tan 30°)² = $\Big(\dfrac{1}{\sqrt{3}}\Big)^2 = \dfrac{1}{3}$

tan² 45° = (tan 45°)² = 1

tan² 60° = (tan 60°)² = ($\sqrt{3}$)² = 3

sin² 60° = (sin 60°)² = $\Big(\dfrac{\sqrt{3}}{2}\Big)^2 = \dfrac{3}{4}$

Therefore,

$\dfrac{4}{3}$ tan²30°+ sin²60° - 3 cos²60°+ $\dfrac{3}{4}$ tan²60°- 2 tan²45°

$\dfrac{4}{3}\times\dfrac{1}{3}+ \dfrac{3}{4} - 3 \times\dfrac{1}{4} + \dfrac{3}{4}\times 3- 2 \times 1 \\[1em] = \dfrac{4}{9}+ \dfrac{3}{4} -\dfrac{3}{4} + \dfrac{9}{4}- 2 \\[1em] = \dfrac{4}{9} + \dfrac{9}{4} - 2 \\[1em] = \dfrac{16 + 81 - 72}{36}\\[1em] = \dfrac{25}{36}.$

Hence, $\dfrac{4}{3}$ tan²30°+ sin²60° - 3 cos²60°+ $\dfrac{3}{4}$ tan²60°- 2 tan²45° = $\dfrac{25}{36}$.

(vi) As,

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

sin² 45° = (sin 45°)² = $\Big(\dfrac{1}{\sqrt{2}}\Big)^2 = \dfrac{1}{2}$

tan² 60° = (tan 60°)² = ($\sqrt{3}$)² = 3

Therefore,

$\dfrac{\sin^2 45^\circ + \cos^2 45^\circ}{\tan^2 60^\circ}$

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

= $\dfrac{1}{3}$

Hence, $\dfrac{\sin^2 45^\circ + \cos^2 45^\circ}{\tan^2 60^\circ} = \dfrac{1}{3}$.