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.

 Submultiple Angles





11 and 12 Grade Math

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