Page 46 - mathsvol1ch1to3ans

46

Solution:
1
cos 8θ cos 2θ = cos 10θ + cos 6θ
2
1
2
= (2 cos 5θ − 1 + cos 6θ)
2
1
2
2
= (2 cos 5θ − 2 sin 3θ)
2
2
2
= cos 5θ − sin 3θ
2
2
2
18. Show that cos A + cos B − 2 cos A cos B cos(A + B) = sin (A + B).
Not For Sale - Veeraragavan C S veeraa1729@gmail.com
Solution:
2
2
2
2
cos A + cos B − 2 cos A cos B cos(A + B) = cos A + cos B − (cos(A + B) + cos(A − B)) cos(A + B)
2
2
2
= cos A + cos B − cos (A + B) − cos(A + B) cos(A − B)
1
2
2
2
= cos A + cos B − cos (A + B) − (cos 2A + cos 2B)
2
1 + cos 2A 1 + cos 2B 1
2
= + − cos (A + B) − (cos 2A + cos 2B)
2 2 2
2
= 1 − cos (A + B)
2
= sin (A + B)
3
19. If cos(α − β) + cos(β − γ) + cos(γ − α) = − , then prove that
2
cos α + cos β + cos γ = sin α + sin β + sin γ = 0.
Solution:
2 cos(α − β) + 2 cos(β − γ) + 2 cos(γ − α) + 3 = 0
2 cos α cos β + 2 sin α sin β + 2 cos β cos γ + 2 sin β sin γ + 2 cos γ cos α + 2 sin γ sin α+
2
2
2
2
2
2
sin α + cos α + sin β + cos β + sin γ + cos γ = 0
2
(sin α + sin β + sin γ) + (cos α + cos β + cos γ) 2 = 0
Hence cos α + cos β + cos γ = sin α + sin β + sin γ = 0.
20. Show that
1 + tan A 1 − tan A
◦
◦
(i) tan(45 + A) = (ii) tan(45 − A) = .
1 − tan A 1 + tan A
Solution:
◦
tan(45 ) + tan A
◦
tan(45 + A) =
◦
1 − tan(45 ) tan A
1 + tan A
=
1 − tan A
◦
tan(45 ) − tan A
◦
tan(45 − A) =
◦
1 + tan(45 ) tan A
1 − tan A
=
1 + tan A
cot A cot B − 1
21. Prove that cot(A + B) = .
cot A + cot B