Page 54 - mathsvol1ch1to3ans

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sin (4A − 2B) + sin (4B − 2A)
13. Prove that = tan (A + B).
cos (4A − 2B) + cos (4B − 2A)
Solution:

sin (4A − 2B) + sin (4B − 2A) 2 sin(A + B) cos(3A − 3B)
= = tan (A + B)
cos (4A − 2B) + cos (4B − 2A) 2 cos(A + B) cos(3A − 3B)
4 cos 2A
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14. Show that cot (A + 15 ) − tan (A − 15 ) = .
1 + 2 sin 2A
Solution:
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cos (A + 15 ) sin (A − 15 )
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cot (A + 15 ) − tan (A − 15 ) = −
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sin (A + 15 ) cos (A − 15 )
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cos (A + 15 ) cos (A − 15 ) − sin (A − 15 ) sin (A + 15 )
=
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sin (A + 15 ) cos (A − 15 )
cos 2A
=
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sin (A + 15 ) cos (A − 15 )
2 cos 2A
=
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2 sin (A + 15 ) cos (A − 15 )
2 cos 2A
=
sin 2A + sin 30 ◦
4 cos 2A
=
1 + 2 sin 2A
Exercise 3.7:
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1. If A + B + C = 180 , prove that
(i) sin 2A + sin 2B + sin 2C = 4 sin A sin B sin C
Solution:
sin 2A + sin 2B + sin 2C = sin 2A + 2 sin(B + C) cos(B − C)
= 2 sin A cos A + 2 sin A cos(B − C)
= 2 sin A (cos A + cos(B − C))

A + B − C A − B + C
= 4 sin A cos cos
2 2

π − 2C π − 2B
= 4 sin A cos cos
2 2
= 4 sin A sin C sin B
A B C
(ii) cos A + cos B − cos C = −1 + 4 cos cos sin
2 2 2
Solution: