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(17)

sin (0+ a) sin a

=cos 0+cot a sin 0; use this transformation in Ex.

(8), expand cos + cot a sin 0 in ascending powers of 0 by Art. 41 and equate the coefficients of 0 on each side.

(19) This result may be deduced from Ex. (13).

(20) Put π-a for a in (19).

[blocks in formation]

(1) (i)

2 cos 40o, 2 cos 160o, 2 cos 80o.

(ii) √2 cos 15o, 2 cos 135°, √2 cos 105°.

(iii) 2 cos 40° +1, 2 cos 160° +1, 2 cos 80° +1.

(iv) 2/2 cos 45°-4, 2√2 cos 1650 - 4, 2/2 cos 75° - 4,

(v)

(2) (i)

4 cos 10°-√3, 4 cos 130°/3, 4 cos 110o-/3.
3a=1o 22'. (ii) 3a=65°41′53′′.

(iii) 3a=74°55′47′′.

(3) (i) Find a and ẞ such that a=tan a, b=tan ß, then

(ii)

a+b=sin (a±ẞ) sec a sec ß.

a cos 0±b sin 0=a cos (0 = a) sec a, where tan a==

(iii) 4 sin (B+ C) sin

(C+A) sin § (A+B).

(iv) 4 sin (0 – a) sin (m) − a) cos (0 – ma).

(v) 4R sin A sin B sin C.

XLI.

(2) Divide the first equation by a, the second by b, square both sides and add.

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(9) cos 40=3(2m2 - 5), sin (a−40) = 1 sin a, and so on.

(85) π3.

XLIII.

(91) sin is greater than 0 – 103, when 0 is less than π cos is less than 1-102+04. [Art. 44, 45.]

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VII.

4. {cos (a+B+y) + cos 1 (B+y− 3a) + cos § (y+ a −38)

+cos(a+ẞ-3y)}.

9. (1) ee-mx cos (e e-mx sin nx); (2) cos {a+(n+1)ẞ} sin nẞcosec ß.

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10. г+-{x ̄1- x ̄2 (sin a)−2 sin 2a + 3x ̃3 (sin a)−3 sin 3a — ...}.

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