Trigonometrical Ratios — Exercise 20(B)

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°

Answer

(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° = $\dfrac{1}{4}$.

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

Answer

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

Without using trigonometric tables, find the values of;

(i) $\dfrac{\sin 30^\circ - \sin 90^\circ +2\times \cos 0^\circ}{\tan 30^\circ \times \tan 60^\circ}$

(ii) $\dfrac{5 \sin^2 30^\circ + \cos^2 45^\circ- 4 \tan^2 30^\circ}{2 \sin 30^\circ \cos 30^\circ+ \tan 45^\circ}$

(iii) $\dfrac{\tan^2 60^\circ + 4 \cos^2 45^\circ + \sec^2 30^\circ + 5 \cos^2 90^\circ}{\cosec 30^\circ + \sec 60^\circ - \cot^2 30^\circ}$

(iv) 4(sin⁴ 30° + cos⁴ 60°) - 3 (cos² 45° - sin² 90°)

Answer

(i) Solving,

$\dfrac{\sin 30^\circ - \sin 90^\circ +2\times \cos 0^\circ}{\tan 30^\circ \times \tan 60^\circ}\\[1em] = \dfrac{\dfrac{1}{2} - 1 +2\times1}{\dfrac{1}{\sqrt{3}}\times {\sqrt{3}}}\\[1em] = \dfrac{\dfrac{1}{2} - 1 + 2}{\dfrac{1}{\sqrt{3}}\times {\sqrt{3}}}\\[1em] = \dfrac{\dfrac{1}{2} + 1}{1}\\[1em] = \dfrac{3}{2}.$

Hence, the required value = $\dfrac{3}{2}$.

(ii) As,

sin² 30° = (sin 30°)² = $\Big(\dfrac{1}{2}\Big)^2 = \dfrac{1}{4}$

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

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

Therefore,

$\dfrac{5 \sin^2 30^\circ + \cos^2 45^\circ- 4 \tan^2 30^\circ}{2 \sin 30^\circ \cos 30^\circ+ \tan 45^\circ}\\[1em] =\dfrac{5\times\dfrac{1}{4} + \dfrac{1}{2} - 4\times\dfrac{1}{3}}{2\times\dfrac{1}{2} \times\dfrac{\sqrt{3}}{2}+ 1 }\\[1em] = \dfrac{\dfrac{5}{4} + \dfrac{1}{2} -\dfrac{4}{3}}{\dfrac{\sqrt{3}}{2}+1}\\[1em] = \dfrac{\dfrac{15 + 6 - 16}{12}}{\dfrac{\sqrt{3}}{2}+1}\\[1em] = \dfrac{\dfrac{5}{12}}{\dfrac{\sqrt{3} + 2}{2}} \\[1em] = \dfrac{5}{6(2 + \sqrt{3})}.$

Hence, $\dfrac{5 \sin^2 30^\circ + \cos^2 45^\circ- 4 \tan^2 30^\circ}{2 \sin 30^\circ \cos 30^\circ+ \tan 45^\circ} = \dfrac{5}{6(2 + \sqrt{3})}$.

(iii) As,

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

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

sec² 30° = (sec 30°)² = $\Big(\dfrac{2}{\sqrt{3}}\Big)^2 = \dfrac{4}{3}$

cos² 90° = (cos 90°)² = 0

cot² 30° = (cot 30°)² = ($\sqrt{3}$)² = 3

Therefore,

$\dfrac{\tan^2 60^\circ + 4 \cos^2 45^\circ + \sec^2 30^\circ + 5 \cos^2 90^\circ}{\cosec 30^\circ + \sec 60^\circ - \cot^2 30^\circ}\\[1em] = \dfrac{3 + 4\times\dfrac{1}{2} + \dfrac{4}{3} + 0}{2 + 2 - 3}\\[1em] = \dfrac{3 + 2 + \dfrac{4}{3} + 0}{1}\\[1em] = 5+ \dfrac{4}{3}\\[1em] = \dfrac{19}{3} \\[1em] = 6\dfrac{1}{3}.$

Hence, $\dfrac{\tan^2 60^\circ + 4 \cos^2 45^\circ + \sec^2 30^\circ + 5 \cos^2 90^\circ}{\cosec 30^\circ + \sec 60^\circ - \cot^2 30^\circ} = \dfrac{19}{3}$

(iv) As,

sin⁴ 30° = (sin 30°)⁴ = $\Big(\dfrac{1}{2}\Big)^4 = \dfrac{1}{16}$

cos⁴ 60° = (cos 60°)⁴ = $\Big(\dfrac{1}{2}\Big)^4 = \dfrac{1}{16}$

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

sin² 90° = (sin 90°)² = 1

Therefore,

4 (sin⁴ 30° + cos⁴ 60°) - 3 (cos² 45° - sin² 90°)

= $4\Big(\dfrac{1}{16} + \dfrac{1}{16}\Big) - 3 \Big(\dfrac{1}{2} - 1\Big)$

= $4\Big(\dfrac{2}{16}\Big) - 3 \Big(\dfrac{-1}{2}\Big)$

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

= 2.

Hence, 4 (sin⁴ 30° + cos⁴ 60°) - 3 (cos² 45° - sin² 90°) = 2.

Verify each of the following :

(i) cos 60° cos 30° - sin 60° sin 30° = 0

(ii) cos 60° = (1 - 2 sin²30°) = (2 cos²30° - 1)

(iii) tan 30° = $\Big(\dfrac{\tan 60^\circ - \tan 30^\circ}{1 + \tan 60^\circ \tan 30^\circ}\Big)$

Answer

(i) cos 60° cos 30° - sin 60° sin 30°

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

= 0.

Hence, proved that cos 60° cos 30° - sin 60° sin 30° = 0.

(ii) As,

sin² 30° = (sin 30°)² = $\Big(\dfrac{1}{2}\Big)^2= \dfrac{1}{4}$

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

Therefore

Left Hand Side :

cos 60° = $\dfrac{1}{2}$

Right Hand Side :

(1 - 2 sin²30°)

=1 - 2 $\times \dfrac{1}{4} = \dfrac{1}{2}$

(2 cos²30° - 1)

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

Hence proved that cos 60° = (1 - 2 sin²30°) = (2 cos²30° - 1).

(iii) Left Hand Side :

tan 30° = $\dfrac{1}{\sqrt{3}}$

Right Hand Side

$\Big(\dfrac{\tan 60^\circ - \tan 30^\circ}{1 + \tan 60^\circ \tan 30^\circ}\Big) = \dfrac{\sqrt{3} - \dfrac{1}{\sqrt{3}}}{1 + \sqrt{3}\times\dfrac{1}{\sqrt{3}}}\\[1em] =\dfrac{\dfrac{3-1}{\sqrt{3}}}{1 + 1}\\[1em] =\dfrac{\dfrac{2}{\sqrt{3}}}{2}\\[1em] = \dfrac{1}{\sqrt{3}}.$

Hence proved that tan 30° = $\Big(\dfrac{\tan 60^\circ - \tan 30^\circ}{1 + \tan 60^\circ \tan 30^\circ}\Big)$

Verify each of the following :

(i) sin 60° cos 30° - cos 60° sin 30° = sin 30°

(ii) 2 sin 30° cos 30° = sin 60°

(iii) 2 sin 45° cos 45° = sin 90°

Answer

(i) Left Hand Side :

sin 60° cos 30° - cos 60° sin 30°

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

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

= $\dfrac{1}{2}$.

Right Hand Side :

sin 30° = $\dfrac{1}{2}$

Hence, proved that sin 60° cos 30° - cos 60° sin 30° = sin 30°.

(ii) Left Hand Side :

2 sin 30° cos 30°

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

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

Right Hand Side :

sin 60° = $\dfrac{\sqrt{3}}{2}$

Hence, proved that 2 sin 30° cos 30° = sin 60°.

(iii) Left Hand Side :

2 sin 45° cos 45°

= $2\times\dfrac{1}{\sqrt2} \times \dfrac{1}{\sqrt2}$

= 1

Right Hand Side :

sin 90° = 1

Hence, proved that 2 sin 45° cos 45° = sin 90°.

If A = 45°, verify that :

(i) sin 2A = 2 sin A cos A

(ii) cos 2A = (2 cos²A - 1) = (1 - 2 sin²A)

Answer

(i) L.H.S. :

sin 2A = sin 2(45°) = sin 90°

= 1

R.H.S. :

2 sin A cos A = 2 sin 45° cos 45°

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

= 1.

Hence, proved that sin 2A = 2 sin A cos A.

(ii) Substituting value of A = 45° in cos 2A, we get :

cos 2A = cos 2(45°)

= cos 90°

= 0.

Substituting value of A = 45° in 2 cos²A - 1, we get :

⇒ 2 cos²A - 1

= 2 cos²45° - 1

= 2 (cos 45°)² - 1

= $2\Big(\dfrac{1}{\sqrt{2}}\Big)^2$ - 1

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

= 1 - 1

= 0.

Substituting value of A = 45° in 1 - 2 sin²A, we get :

⇒ 1 - 2 sin²A

= 1 - 2 sin²45°

= 1 - 2 (sin 45°)²

= 1 - $2\Big(\dfrac{1}{\sqrt{2}}\Big)^2$

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

= 1 - 1

= 0.

Hence, proved that cos 2A = (2 cos²A - 1) = (1 - 2 sin²A).

If A = 30°, prove that :

(i) sin 2A = $\dfrac{2 \tan A}{1 + \tan^2A}$

(ii) cos 2A = $\Big(\dfrac{1 - \tan^2A}{1 + \tan^2A}\Big)$

Answer

(i) Left Hand Side:

sin 2A = sin 2(30°) = sin 60°

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

Right Hand Side:

$\dfrac{2 \tan A}{1 + \tan^2A} = \dfrac{2\times \dfrac{1}{\sqrt{3}}}{1 + \dfrac{1}{3}}\\[1em] = \dfrac{\dfrac{2}{\sqrt{3}}}{\dfrac{4}{3}}\\[1em] = \dfrac{2}{\sqrt{3}}\times \dfrac{3}{4}\\[1em] = \dfrac{\sqrt{3}}{2}.$

Hence, proved that sin 2A = $\dfrac{2 \tan A}{1 + \tan^2A}$

(ii) Left Hand Side :

cos 2A = cos 2(30°) = cos 60°

= $\dfrac{1}{2}$

Right Hand Side :

$\dfrac{1 - \tan^2A}{1 + \tan^2A} \\[1em] = \dfrac{1 - \tan^230°}{1 + \tan^230°} \\[1em] =\dfrac{1 - \Big(\dfrac{1}{\sqrt{3}}\Big)^2}{1 + \Big(\dfrac{1}{\sqrt{3}}\Big)^2}\\[1em] = \dfrac{1 - \dfrac{1}{3}}{1 + \dfrac{1}{3}} \\[1em] = \dfrac{\dfrac{2}{3}}{\dfrac{4}{3}}\\[1em] = \dfrac{1}{2}.$

Hence, proved that cos 2A = $\Big(\dfrac{1 - \tan^2A}{1 + \tan^2A}\Big)$.

If A = B = 45°, show that :

(i) sin(A - B) = sin A cos B - cos A sin B

(ii) cos(A + B) = cosA cosB - sin A sin B

Answer

(i) Left Hand Side :

sin(A - B) = sin (45° - 45°) = sin 0

= 0

Right Hand Side :

sin A cos B - cos A sin B

= sin 45° cos 45° - cos 45° sin 45°

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

= 0.

Hence, proved that sin(A - B) = sin A cos B - cos A sin B.

(ii) Left Hand Side :

cos(A + B) = cos (45° + 45°) = cos 90°

= 0.

Right Hand Side :

cos A cos B - sin A sin B

= cos 45° cos 45° - sin 45° sin 45°

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

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

= 0.

Hence, proved that cos(A + B) = cosA cosB - sin A sin B.

If A = 60° and B = 30°, show that :

(sin A cos B + cos A sin B)² + (cos A cos B - sin A sin B)² = 1

Answer

Left Hand Side :

(sin A cos B + cos A sin B)² + (cos A cos B - sin A sin B)²

= (sin 60° cos 30° + cos 60° sin 30°)² + (cos 60° cos 30° - sin 60° sin 30°)²

= $\Big(\dfrac{\sqrt{3}}{2}\times\dfrac{\sqrt{3}}{2} + \dfrac{1}{2}\times\dfrac{1}{2}\Big)^2 + \Big(\dfrac{1}{2}\times\dfrac{\sqrt{3}}{2} - \dfrac{\sqrt{3}}{2}\times\dfrac{1}{2}\Big)^2$

= $\Big(\dfrac{3}{4} + \dfrac{1}{4})^2 + \Big(\dfrac{\sqrt{3}}{4}- \dfrac{\sqrt{3}}{4}\Big)^2$

= 1 + 0 = 1.

Right Hand Side = 1

Hence, proved that (sin A cos B + cos A sin B)² + (cos A cos B - sin A sin B)² = 1.

If A = 60° and B = 30°, prove that :

(i) sin (A + B) = sin A cos B + cos A sin B

(ii) cos (A + B) = cos A cos B - sin A sin B

(iii) cos (A - B) = cos A cos B + sin A sin B

(iv) tan (A - B) = $\dfrac{\tan A - \tan B}{1 + \tan A \tan B}$

Answer

(i) Left Hand Side

sin (A + B) = sin (60° + 30°) = sin 90°

= 1

Right Hand Side

sin A cos B + cos A sin B

= sin 60° cos 30° + cos 60° sin 30°

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

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

= 1

Hence, proved that sin (A + B) = sin A cos B + cos A sin B.

(ii) Left Hand Side

cos (A + B) = cos (60° + 30°) = cos 90°

= 0

Right Hand Side

cos A cos B - sin A sin B

= cos 60° cos 30° - sin 60° sin 30°

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

= $\dfrac{\sqrt{3}}{4} - \dfrac{\sqrt{3}}{4}$

= 0.

Hence, proved that cos (A + B) = cos A cos B - sin A sin B.

(iii) Left Hand Side

cos (A - B) = cos (60° - 30°) = cos 30°

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

Right Hand Side

cos A cos B + sin A sin B

= cos 60° cos 30° + sin 60° sin 30°

$\dfrac{1}{2}\times\dfrac{\sqrt{3}}{2} + \dfrac{\sqrt{3}}{2}\times\dfrac{1}{2}\\[1em] = \dfrac{\sqrt{3}}{4} + \dfrac{\sqrt{3}}{4}\\[1em] = \dfrac{2\sqrt{3}}{4}\\[1em] =\dfrac{\sqrt{3}}{2}$

Hence, proved that cos (A - B) = cos A cos B + sin A sin B.

(iv) Left Hand Side :

tan (A - B) = tan (60° - 30°) = tan 30°

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

Right Hand Side

$\Rightarrow \Big(\dfrac{\tan A - \tan B}{1 + \tan A \tan B}\Big)\\[1em] = \Big(\dfrac{\tan 60^\circ - \tan 30^\circ}{1 + \tan 60^\circ \tan 30^\circ}\Big)\\[1em] = \dfrac{\sqrt{3} - \dfrac{1}{\sqrt{3}}}{1 + \sqrt{3}\times\dfrac{1}{\sqrt{3}}}\\[1em] =\dfrac{\dfrac{3-1}{\sqrt{3}}}{1 + 1}\\[1em] =\dfrac{\dfrac{2}{\sqrt{3}}}{2}\\[1em] = \dfrac{1}{\sqrt{3}}.$

Hence, proved that tan (A - B) = $\dfrac{\tan A - \tan B}{1 + \tan A \tan B}$.

Evaluate : $\dfrac{\cos 3A + 2 \cos 4A}{\sin 3A + 2\sin 4A}$, when A = 15°.

Answer

Solving,

$\Rightarrow \dfrac{\cos 3A + 2 \cos 4A}{\sin 3A + 2\sin 4A}\\[1em] = \dfrac{\cos 3(15^\circ) + 2 \cos 4(15^\circ)}{\sin 3(15^\circ) + 2\sin 4(15^\circ)}\\[1em] = \dfrac{\cos 45^\circ + 2 \cos 60^\circ}{\sin 45^\circ + 2\sin 60^\circ}\\[1em] = \dfrac{\dfrac{1}{\sqrt{2}}+ 2\times\dfrac{1}{2}}{\dfrac{1}{\sqrt{2}}+ 2\times\dfrac{\sqrt{3}}{2}}\\[1em] = \dfrac{\dfrac{1}{\sqrt{2}}+ 1 }{\dfrac{1}{\sqrt{2}}+ {\sqrt{3}}}\\[1em] = \dfrac{\dfrac{1 + \sqrt{2}}{\sqrt{2}}}{\dfrac{1 + \sqrt{6}}{\sqrt{2}}}\\[1em] = \dfrac{1 + \sqrt{2}}{1 + \sqrt{6}}.$

Hence, $\dfrac{\cos 3A + 2 \cos 4A}{\sin 3A + 2\sin 4A} = \dfrac{1 + \sqrt{2}}{1 + \sqrt{6}}$.

Evaluate : $\dfrac{3\sin 3A + 2\cos (2A + 5^\circ)}{2\cos 3A - \sin (2A - 10^\circ)}$ , when A = 20°.

Answer

Solving,

$\Rightarrow \dfrac{3\sin 3(20^\circ) + 2 \cos (2(20^\circ)+ 5^\circ)}{2 \cos 3(20^\circ) - \sin (2(20^\circ) - 10^\circ)}\\[1em] = \dfrac{3 \sin 60^\circ + 2 \cos 45^\circ}{2 \cos 60^\circ - \sin 30^\circ}\\[1em] = \dfrac{3\times\dfrac{\sqrt{3}}{2} + 2\times\dfrac{1}{\sqrt{2}}}{2\times\dfrac{1}{2} - \dfrac{1}{2}} \\[1em] = \dfrac{\dfrac{3\sqrt{3}}{2} + \dfrac{2}{\sqrt{2}}}{\dfrac{1}{{2}}}\\[1em] = \dfrac{\dfrac{3\sqrt{3}}{2} + {\sqrt{2}}}{\dfrac{1}{2}}\\[1em] = \dfrac{3\sqrt{3}+2\sqrt{2}}{2}\times\dfrac{2}{1}\\[1em] = {3\sqrt{3} + 2\sqrt{2}}.$

Hence, $\dfrac{3\sin 3A + 2\cos (2A + 5^\circ)}{2\cos 3A - \sin (2A - 10^\circ)} = {3\sqrt{3} + 2\sqrt{2}}.$

Show that 4(sin⁴30° + cos⁴60°) - 3(cos²45° - sin²90°) = 2.

Answer

sin 30° = $\dfrac{1}{2}$

sin⁴30° = (sin 30°)⁴ = $\dfrac{1}{16}$

cos 60° = $\dfrac{1}{2}$

cos⁴60° = (cos 60°)⁴ = $\dfrac{1}{16}$

sin 45° = $\dfrac{1}{\sqrt{2}}$

cos²45° = (cos 45°)² = $\dfrac{1}{2}$

sin²90° = (sin 90°)² = 1

Left Hand Side

4(sin⁴30° + cos⁴60°)- 3(cos²45°- sin²90°)

= $4\Big(\dfrac{1}{16}+\dfrac{1}{16}\Big) - 3 \Big(\dfrac{1}{2}- 1\Big)$

= $4\Big(\dfrac{2}{16}\Big) - 3 \Big(\dfrac{-1}{2}\Big)$

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

= 2.

Right Hand Side = 2

Hence, proved that 4(sin⁴30° + cos⁴60°) - 3(cos²45°- sin²90°) = 2.

Without using tables, verify that :

(i) cos 60° = (cos²30° - sin²30)

(ii) sin 60° = $\dfrac{2\tan 30^\circ}{1 + \tan^2 30^\circ} = \dfrac{\sqrt{3}}{2}$

(iii) cos 60° = $\dfrac{1 -\tan^2 30^\circ}{1 + \tan^2 30^\circ} = \dfrac{1}{2}$

Answer

(i) cos²30° = (cos 30°)² = $\Big(\dfrac{\sqrt{3}}{2}\Big)^2 = \dfrac{3}{4}$

sin²30° = (sin 30°)² = $\Big(\dfrac{1}{2}\Big)^2 = \dfrac{1}{4}$

Left Hand Side

cos 60° = $\dfrac{1}{2}$

Right Hand Side

cos²30° - sin²30°

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

Hence, proved that cos 60° = (cos²30° - sin²30).

(ii) Left Hand Side :

sin 60° = $\dfrac{\sqrt{3}}{2}$

Right Hand Side :

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

$\Rightarrow \dfrac{2\tan 30^\circ}{1 + \tan^2 30^\circ}\\[1em] =\dfrac{2\times\dfrac{1}{\sqrt{3}}}{1 + \dfrac{1}{3}}\\[1em] = \dfrac{\dfrac{2}{\sqrt{3}}}{\dfrac{4}{3}}\\[1em] = \dfrac{2}{\sqrt{3}}\times\dfrac{3}{4}\\[1em] = \dfrac{\sqrt{3}}{2}.$

Hence, proved that sin 60° = $\dfrac{2\tan 30^\circ}{1 + \tan^2 30^\circ} = \dfrac{\sqrt{3}}{2}$.

(iii) Left Hand Side

cos 60° = $\dfrac{1}{2}$

Right Hand Side

$\dfrac{1 -\tan^2 30^\circ}{1 + \tan^2 30^\circ}\\[1em] =\dfrac{1 -\dfrac{1}{3}}{1 + \dfrac{1}{{3}}}\\[1em] = \dfrac{\dfrac{{3} - 1}{{3}}}{\dfrac{{3} + 1}{{3}}}\\[1em] =\dfrac{2}{4} \\[1em] = \dfrac{1}{2}.$

Hence, proved that cos 60° = $\dfrac{1 -\tan^2 30^\circ}{1 + \tan^2 30^\circ} = \dfrac{1}{2}$.

If 0° ≤ x ≤ 90°, state the numerical value of x for which sin x° = cos x°.

Answer

sin 45° = cos 45° = $\dfrac{1}{\sqrt{2}}$.

x = 45 is the only value for which sin x° = cos x°.