Page 45 - mathsvol1ch1to3ans

45

Note that the three vectors in polar form: (1, θ), (1, θ + 120) and (1, θ + 240)
divide the complete angle into three equal parts and from symmetry arguments,
their sum must be equal to 0. Taking this to Cartesian coordinates, this means
that the sum of their x components must be equal to 0. Thus, we infer that,cos θ +
◦
◦
cos(θ + 120 ) + cos(θ + 240 ) = 0.
◦
◦
Let a = cos θ, b = cos (θ + 120 ). Then cos (θ + 240 ) = −(a + b).
a a
Hence y = x and z = − x.
b a + b
2
a a a
2
xy + yz + zx = − − x .
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b b(a + b) a + b
2
Calculating coefficient of x separately, we have
2
a a 2 a a(a + b) − a − ab
− − = = 0 Hence xy + yz + zx = 0.
b b(a + b) a + b b(a + b)
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17. Prove that
2
2
(i) sin(A + B) sin(A − B) = sin A − sin B
Solution:
1
sin(A + B) sin(A − B) = (cos(A + B − A + B) − cos(A + B + A − B))
2
1
= (cos(2B) − cos(2A))
2
1
2
2
= 1 − 2 sin (B) − 1 + 2 sin (A)
2
2
2
= sin (A) − sin (B)
2
2
2
2
(ii) cos(A + B) cos(A − B) = cos A − sin B = cos B − sin A
Solution:
cos(A + B) cos(A − B) = (cos A cos B − sin A sin B) (cos A cos B + sin A sin B)
2
2
2
2
= cos A cos B − sin A sin B
2

2
2
2
= cos A 1 − sin B − (1 − cos A) sin B
2
2
= cos A − sin B
2
2
= (1 − sin A) − (1 − cos B)
2
2
= cos B − sin A
2
2
(iii) sin (A + B) − sin (A − B) = sin 2A sin 2B
Solution:
1
2
2
sin (A + B) − sin (A − B) = {(1 − cos[2(A + B)]) − (1 − cos[2(A − B)])}
2
1
= (cos 2(A − B) − cos 2(A + B))
2
= sin 2A sin 2B
2
2
(iv) cos 8θ cos 2θ = cos 5θ − sin 3θ