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We will learn how to solve the problems on submultiple angles formula.
1. If sin x = 3/5 and 0 < x < π2, find the value of tan x2
Solution:
tan x2
= √1−cosx1+cosx
= √1−451+45
= √19
= 13
2. Show that, (sin2 24° - sin2 6° ) (sin2 42° - sin2
12°) = 116
Solution:
L.H.S. = 1/4 (2 sin2 24˚ - 2 sin2 6˚)(2 sin2 42˚ - 2 sin2 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°
= 14 ∙ √5−14 ∙ √5+14
= 14 × 416
= 116 = R.H.S. Proved.
3. If tan x = ¾ and x lies in the third quadrant, find the values of sin x2, cos x2 and tan x2.
Solution:
As x lies in the third quadrant, cos x is negative
sec2 x = 1 + tan2 x = 1 + (3/4)2 = 1 + 916 = 2516
⇒ cos2 x = 2516
⇒ cos x = ± 45, but cos x is negative
Therefore, cos x = -45
Also π < x < 3π2
⇒ π2 < x2 < 3π4
⇒ x2 lies in second quadrant
⇒ cos x2 is –ve and sin x2 is +ve.
Therefore, cos x2 = -√1+cosx2 = -√1−452 = - 1√10
sin x2 = -√1−cosx2 = √1−(−45)2 = √910 = 3√10
tan x2 = sinx2cosx2 = 3√10(√101) = -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˚
= (2sin6˚sin66˚)(2sin42˚sin78˚)(2cos6˚cos66˚)(2cos42˚cos78˚)
= (cos60˚−cos72˚)(cos36˚−cos120˚)(cos60˚+cos72˚)(cos36˚+cos120˚)
= (12−sin18˚)(cos36˚+12)(12+sin18˚)(cos36˚−12), [Since, cos 72˚ = cos (90˚ - 18˚) = sin 18˚ and cos 120˚ = cos ( 180˚ - 60˚) = - cos 60˚ = -1/2]
= (12−√5−14)(√5+14+12)(12+√5−14)(√5+14−12), [ putting the values of sin 18˚ and cos 36˚]
= (3−√5)(3+√5)(√5+1)(√5−1)
= 9−55−1
= 44
= 1 = R.H.S. Proved.
5. Without using table prove that, sin 12° sin 48° sin 54˚ = 18
Solution:
L. H. S. = sin 12° sin 48° sin 54°
= 12 (2 sin 12°sin 48°) sin (90°- 36°)
= 12 [cos 36°- cos 60°] cos 36°
= 12 [√√5+14 - 12] √5+14, [Since, cos 36˚ = √5+14]
= 12 ∙ √5−14 ∙ √5+14
= 432
= 18 = R.H.S. Proved.
11 and 12 Grade Math
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