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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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