sin A = 2 sin (A/2) cos (A/2)
cos A = cos2 (A/2) - sin2(A/2)
tan A = 2 tan (A/2)/[1-tan2(A/2)]
cos A = 1 - 2sin2(A/2)
cos A = 2cos²(A/2) - 1
sin A = 2 tan (A/2)/[1+tan2(A/2)]
cos A = [1-tan2(A/2)]/[1+tan2 (A/2)]
sin2(A/2) = (1-cos A)/2
cos2(A/2) = (1+cos A)/2
tan2(A/2) = (1-cos A)/(1+cos A)
Example 1 :
Using half angle find the value of sin 15°
Solution :
We may write, 15° = 30°/2
So,
sin 15° = sin (30°/2)
We know that,
sin2A/2 = (1-cosA)/2
sin (A/2) = √ (1-cosA)/2
sin (30°/2) = √[(1-cos 30)/2]
= √[(1-√3/2)/2]
= √[(2-√3)/4]
= √(2-√3)/2
Example 2 :
Using half angle formula find the value of tan 15°
Solution :
15° = 30°/2
So, tan 15° = tan 30°/2
tan²(A/2) = (1-cos A)/(1+cos A)
tan(A/2) = √(1-cos A)/(1+cos A)
tan 30°/2 = √[(1-cos 30°)/(1+cos 30°)]
= √[(1-√3/2)/(1+√3/2)]
= √[( 2-√3 )/2] x [2/(2+√3)]
= √(2-√3)/(2+√3)
Rationalizing the denominator,
= √[(2-√3)2/(22 - (√3)2]
= √[(2²+(√3)² - 2(2)(√3) /(2² - (√3)²]
= √[(2²+(√3)² - 2(2)(√3) /(4-3]
= √[(2-√3)² /1]
= √[(2-√3)2
= 2-√3
So, the value of tan 15° is 2-√3.
Example 3 :
Using half angle find the value of cos 15°
Solution :
15° = 30°/2
So, cos 15° = cos 30°/2
cos²A/2 = (1+cos A)/2
cos (A/2) = √(1+cos A)/2
cos 30°/2 = √[(1+cos 30)/2]
= √[(1+√3/2)/2]
= √[(2+√3)/4]
= √(2+√3)/2
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