Problems on Submultiple Angles

We will learn how to solve the problems on submultiple angles formula.

1. If sin x = 3/5 and 0 < x < $\frac{π}{2}$, find the value of tan $\frac{x}{2}$

Solution:

tan $\frac{x}{2}$

= $\sqrt{\frac{1 - cos x}{1 + cos x}}$

= $\sqrt{\frac{1 - \frac{4}{5}}{1 + \frac{4}{5}}}$

= $\sqrt{\frac{1}{9}}$

= $\frac{1}{3}$

2. Show that, (sin$^{2}$ 24° - sin$^{2}$ 6° ) (sin$^{2}$ 42° - sin$^{2}$ 12°) = $\frac{1}{16}$

Solution:

L.H.S. = 1/4 (2 sin$^{2}$ 24˚ - 2 sin$^{2}$ 6˚)(2 sin$^{2}$ 42˚ - 2 sin$^{2}$ 12˚)

= ¼ [(1- cos 48°) - (1 - cos 12°)] [(1 - cos 84° ) - (1 - cos 24°)]

= ¼ (cos 12° - cos 48°)(cos 24° - cos 84°)

= ¼ (2 sin 30° sin 18° ) (2 sin 54° sin 30°)

= ¼ [2 ∙ ½ ∙ sin 18°] [2 ∙ sin(90° - 36°) × ½]

= ¼ sin 18° ∙ cos 36°

= $\frac{1}{4}$ ∙ $\frac{√5 - 1}{4}$ ∙ $\frac{√5 + 1}{4}$

= $\frac{1}{4}$ × $\frac{4}{16}$

= $\frac{1}{16}$ = R.H.S. Proved.

3. If tan x = ¾ and x lies in the third quadrant, find the values of sin $\frac{x}{2}$, cos $\frac{x}{2}$ and tan $\frac{x}{2}$.

Solution:

As x lies in the third quadrant, cos x is negative

sec$^{2}$ x = 1 + tan$^{2}$ x = 1 + (3/4)$^{2}$ = 1 + $\frac{9}{16}$ = $\frac{25}{16}$

⇒ cos$^{2}$ x = $\frac{25}{16}$

⇒ cos x = ± $\frac{4}{5}$, but cos x is negative

Therefore, cos x = -$\frac{4}{5}$

Also π < x < $\frac{3π}{2}$

⇒ $\frac{π}{2}$ < $\frac{x}{2}$ < $\frac{3π}{4}$

⇒ $\frac{x}{2}$ lies in second quadrant

⇒ cos $\frac{x}{2}$ is –ve and sin $\frac{x}{2}$ is +ve.

Therefore, cos $\frac{x}{2}$ = -$\sqrt{\frac{1 + cos x}{2}}$ = -$\sqrt{\frac{1 - \frac{4}{5}}{2}}$ = - $\frac{1}{√10}$

sin $\frac{x}{2}$ = -$\sqrt{\frac{1 - cos x}{2}}$ = $\sqrt{\frac{1 - (-\frac{4}{5})}{2}}$ = $\sqrt{\frac{9}{10}}$ = $\frac{3}{√10}$

tan $\frac{x}{2}$ = $\frac{sin \frac{x}{2}}{cos \frac{x}{2}}$ = $\frac{3}{√10}$($\frac{√10}{1}$) = -3

4. Show that using the formula of submultiple angles tan 6˚ tan 42˚ tan 66˚ tan 78˚ = 1.

Solution:

L.H.S = tan 6˚ tan 42˚ tan 66˚ tan 78˚

= $\frac{(2 sin 6˚ sin 66˚) (2 sin 42˚ sin 78˚)}{(2 cos 6˚ cos 66˚) ( 2 cos 42˚ cos 78˚)}$

= $\frac{( cos 60˚ - cos 72˚)( cos 36˚ - cos 120˚)}{( cos 60˚ + cos 72˚)( cos 36˚ + cos 120˚)}$

= $\frac{(\frac{1}{2} - sin 18˚) ( cos 36˚ + \frac{1}{2})}{(\frac{1}{2} + sin 18˚) ( cos 36˚ - \frac{1}{2})}$, [Since, cos 72˚ = cos (90˚ - 18˚) = sin 18˚ and cos 120˚ = cos ( 180˚ - 60˚) = - cos 60˚ = -1/2]

= $\frac{(\frac{1}{2} - \frac{√5 - 1}{4}) (\frac{√5 + 1}{4} + \frac{1}{2})}{(\frac{1}{2} + \frac{√5 - 1}{4}) (\frac{√5 + 1}{4} - \frac{1}{2})}$, [ putting the values of sin 18˚ and cos 36˚]

= $\frac{(3 - √5) ( 3 + √5)}{(√5 + 1) (√5 - 1) }$

= $\frac{9 - 5}{5 - 1}$

= $\frac{4}{4}$

= 1 = R.H.S. Proved.

5. Without using table prove that, sin 12° sin 48° sin 54˚ = $\frac{1}{8}$

Solution:

L. H. S. = sin 12° sin 48° sin 54°

= $\frac{1}{2}$ (2 sin 12°sin 48°) sin (90°- 36°)

= $\frac{1}{2}$ [cos 36°- cos 60°] cos 36°

= $\frac{1}{2}$ [√$\frac{√5 + 1}{4}$ - $\frac{1}{2}$] $\frac{√5 + 1}{4}$, [Since, cos 36˚ = $\frac{√5 + 1}{4}$]

= $\frac{1}{2}$ ∙ $\frac{√5 - 1}{4}$ ∙ $\frac{√5 + 1}{4}$

= $\frac{4}{32}$