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.
11 and 12 Grade Math
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