## HALF ANGLE IDENTITIES IN TRIGONOMETRY

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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