# 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˚ = $$\frac{√5 + 1}{4}$$]

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

= $$\frac{4}{32}$$

= $$\frac{1}{8}$$ =  R.H.S.              Proved.