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Unit 2: Trigonometric Functions-II

Thus, it is veriﬁed that Notes
sin 2A = 2sin A cos A =

(ii) cos 2A =

2
2
cos A – sin A =

2
2cos A – 1 =

2
1 – 2sin A =

Thus, it is veriﬁed that
2
2
2
2
cos2A = cos A – sin A = 2cos A – 1 = 1 – 2sin A =
2.1.4 Trigonometric Function of 3A in terms of A
(a) sin 3A in terms of sin A

Substituting 2A for B in the formula
sin (A + B) = sin A cos B + cos A sin B, we get
sin(A + 2A) = sin A cos 2A + cos A sin 2A
2
= sin A(1 – 2sin A) + (cos A 2sin A cos A)
= sin A – 2sin A + 2sin A(1 – sin A)
2
3
3
3
= sin A – 2sin A + 2sin A – 2sin A
sin 3A = 3sin A – 4sin A ....(1)
3
(b) cos 3A in terms of cos A
Substituting 2A for B in the formula
cos(A + B) = cos A cos B – sin A sin B, we get
cos(A + 2A) = cos A cos 2A – sin A sin 2A
= cos A(2cos A – 1) – (sin A) 2sin A cos A
2

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