#### Trigonometric Ratios of Sub Multiple Angles

Trigonometric Ratios of sub - multiple Angles

If A is an angl, then $$\frac{A}{2}$$, $$\frac{A}{3}$$, $$\frac{A}{4}$$ etc. are called sub - multiple angles of A. In this section we wilol discuss about the trigonometric ratios of angle A in terms of $$\frac{A}{2}$$ and $$\frac{A}{3}$$ .

1. Trigonometric ratios of angle A in terms of $$\frac{A}{2}$$

(a) SinA = sin($$\frac{A}{2}$$ + $$\frac{A}{2}$$) = sin (2 .$$\frac{A}{2}$$) = 2 sin$$\frac{A}{2}$$ cos$$\frac{A}{2}$$

(b) sinA = 2sin$$\frac{A}{2}$$ cos$$\frac{A}{2}$$ =$$\frac{2 sin \frac{A}{2} cos \frac{A}{2}}{cos^2 \frac{A}{2} + sin^2 \frac{A}{2}}$$ =$$\frac{2tan \frac{A}{2}}{1 + tan^2 \frac{A}{2}}$$

(c) sinA = $$\frac{2 tan \frac{A}{2}}{1 + tan^2 \frac{A}{2}}$$ =$$\frac{\frac{2}{cot \frac{A}{2}}}{1 + \frac{1}{cot^2 \frac{A}{2}}}$$ = $$\frac{2 cot \frac{A}{2}}{1 + cot^2 \frac{A}{2}}$$

(d) cosA = cos(2. $$\frac{A}{2}$$) = cos$$^2$$$$\frac{A}{2}$$ - sin$$^2$$$$\frac{A}{2}$$

(e ) cosA = cos2$$\frac{A}{2}$$ - sin$$^2$$$$\frac{A}{2}$$ = 1 - sin2$$\frac{A}{2}$$ - sin$$^2$$$$\frac{A}{2}$$ \ = 1 - 2sin$$^2$$$$\frac{A}{2}$$

(f) cosA = cos2$$\frac{A}{2}$$ - sin$$^2$$$$\frac{A}{2}$$ = cos2$$\frac{A}{2}$$ - 1 + cos2$$\frac{A}{2}$$ = 2 cos2$$\frac{A}{2}$$ - 1

(g) cosA = cos2$$\frac{A}{2}$$ - sin$$^2$$$$\frac{A}{2}$$ =$$\frac{cos^2 \frac{A}{2} - sin^2 \frac{A}{2}}{cos^2 \frac{A}{2} + sin^2 \frac{A}{2}}$$

= $$\frac{1 - tan^2 \frac{A}{2}}{1 + tan^2 \frac{A}{2}}$$ (By dividing numerator and denominator by cos2 $$\frac{A}{2}$$)

(h) cosA = $$\frac{1 - tan^2 \frac{A}{2}}{1 + tan^2 \frac{A}{2}}$$ = $$\frac{1 - \frac{1}{cot^2 \frac{A}{2}}}{{1 + \frac{1}{cot^2 \frac{A}{2}}}}$$

= $$\frac{cot^2 \frac{A}{2} - 1}{cot^2 \frac{A}{2} + 1}$$

(i) tanA = tan(2. $$\frac{A}{2}$$) = $$\frac{2 tan\frac{A}{2}}{1 - tan^2 \frac{A}{2}}$$

(j) tanA = $$\frac{2 tan\frac{A}{2}}{1 - tan^2 \frac{A}{2}}$$ = $$\frac{\frac{2}{cot \frac{A}{2}}}{1 - \frac{1}{cot^2 \frac{A}{2}}}$$ = $$\frac{2 cot \frac{A} {2}}{cot^2 \frac{A}{2} - 1}$$

(k) cotA = cot(2 . $$\frac{A}{2}$$) = $$\frac{cot^2 \frac{A}{2} - 1}{2 cot \frac{A}{2}}$$

(l) cotA = $$\frac{cot^2 \frac{A}{2} - 1}{2cot \frac{A}{2}}$$ = $$\frac{\frac{1}{tan^2 \frac{A}{2}} - 1}{tan \frac{A}{2}}$$ = $$\frac{1 - tan^2 A}{2 tan \frac{A}{2}}$$

2. Some useful results

(a) 1 + cosA = 1 + cos2 $$\frac{A}{2}$$ - sin2 $$\frac{A}{2}$$ = 1 - sin2 $$\frac{A}{2}$$ + cos2 $$\frac{A}{2}$$ = cos2 $$\frac{A}{2}$$ + cos2 $$\frac{A}{2}$$ = 2 cos2 $$\frac{A}{2}$$

(b) 1 - cosA = 1 - (cos2 $$\frac{A}{2}$$ - sin2 $$\frac{A}{2}$$) = 1 - cos2 $$\frac{A}{2}$$ + sin2$$\frac{A}{2}$$ = sin2 $$\frac{A}{2}$$ + sin2 $$\frac{A}{2}$$ = 2 sin2 $$\frac{A}{2}$$

(c) 1 + sinA = cos2 $$\frac{A}{2}$$ + sin2 $$\frac{A}{2}$$ + 2 sin $$\frac{A}{2}$$ cos $$\frac{A}{2}$$ = (cos $$\frac{A}{2}$$ + sin $$\frac{A}{2}$$)2

(d) 1 - sinA = cos2$$\frac{A}{2}$$ + sin2$$\frac{A}{2}$$ - 2 sin v cos$$\frac{A}{2}$$= (cos $$\frac{A}{2}$$ - sin $$\frac{A}{2}$$)2

3. Trigonometric ratios of A in terms of $$\frac{A}{3}$$

(a) cosA = cos (3 .$$\frac{A}{3}$$) = 3 cos $$\frac{A}{3}$$ - 4cos $$^3$$ $$\frac{A}{3}$$

(b) sinA = sin (3. $$\frac{A}{3}$$) = 3 sin $$\frac{A}{3}$$- 4 sin$$^3$$ $$\frac{A}{3}$$

(c) tanA = tan (3. $$\frac{A}{3}$$) = $$\frac{3 tan \frac{A}{3} - tan^3 \frac{A}{3}}{1 - 3 tan^2 \frac{A}{3}}$$

 S. N Multiple Angle formulae Sub - Multiple Angle formulae 1 sin2A = 2 sinA. cosA sinA = 2sin $$\frac{A}{2}$$. cos $$\frac{A}{2}$$ 2 sin2A = $$\frac{2 tanA}{1 + tan^2 A}$$ sinA =$$\frac{2 tan A \frac{A}{2}}{1 + tan^2 \frac{A}{2}}$$ 3 sin2A =$$\frac{2cot A}{1 + cot^2 A}$$ sinA =$$\frac{2 cot \frac{A}{2}}{1 + cot^2 \frac{A}{2}}$$ 4 cos2A = cos2A - sin2A cosA = cos2 $$\frac{A}{2}$$ - sin2 $$\frac{A}{2}$$ 5 cos2A = 2cos2A - 1 cosA = 2cos2 $$\frac{A}{2}$$ - 1 6 cos2A = 1 - 2 sin2A cosA = 1 - 2sin2$$\frac{A}{2}$$ 7 cos2A =$$\frac{1 - tan^2 A}{1 + tan^2 A}$$ cos A = (\frac{1 - tan^2 \frac{A}{2}}{1 + tan^2 \frac{A}{2}}\) 8 cos2A =$$\frac{cot^2 A - 1}{cot^2 A + 1}$$ cosA =$$\frac{cot^2 \frac{A}{2} - 1}{cot^2 \frac{A}{2} =+ 1}$$ 9 tan2a =$$\frac{2 tanA}{1 - tan^2A}$$ tanA =$$\frac{2tan \frac{A}{2}}{1 - tan^2 \frac{A}{2}}$$

 10 tan2A =$$\frac{2cot A}{cot^2 - 1}$$ tanA =$$\frac{2 cot \frac{A}{2}}{cot^2 \frac{}A{2} - 1}$$ 11 cot2A =$$\frac{cot^2 A - 1}{2 cot A}$$ cotA =$$\frac{cot^2 \frac{A}{2} - 1}{2 co \frac{A}{2}}$$ 12 cot2A =$$\frac{1 - tan^2 A}{2 tanA}$$ cotA =$$\frac{1 - tan^2 \frac{A}{2}}{2 tan \frac{A}{2}}$$ 13 sin3A = 3sinA - 4sin$$^3$$A sinA = 3sin $$\frac{A}{3}$$ - 4sin $$\frac{A}{3}$$ 14 cos3A = 4cos$$^3$$A - 3cosA cosA = 4cos $$^3$$ $$\frac{A}{3}$$ - 3cos $$\frac{A}{3}$$ 15 tan3A =$$\frac{3tanA - tan^3 A}{1 - tan^2 A}$$ tanA =$$\frac{3 tan \frac{A}{3} - tan^3\frac{A}{3}}{1 - 3tan^2 \frac{A}{3}}$$ 16 1 + cos2a = 2cos2A 1 + cosA = 2cos2 $$\frac{A}{2}$$ 17 1 - cos2A = 2sin2A 1 - cosA = 2sin2 $$\frac{A}{2}$$ 18 1 + sin2A = ( cosA + sinA )2 1 + sinA = (cos $$\frac{A}{2}$$$$\frac{A}{2}$$)2 19 1 - sin2A = (cosA - sinA)2 1 -sinA = (cos $$\frac{A}{2}$$$$\frac{A}{2}$$)2

Some properties of matrix multiplication:

(i) Multiplication of matrices is, in general, not commutative, i.e. AB not equal to BA, in general.

(ii) Multiplication of matrices in associative, i.e. if A, B and C are matrices conformable for multiplication, then (AB) C = A (BC).

(iii) Multiplication of matrices is distributive with respect to addition i.e. if A, B and C are matrices conformable for the requisite addition and multiplication, then A (B + C) = AB + AC and (A + B) C = AC + BC.

(iv) If A is a square matrix and I is a null matrix of the same order, then AI = IA = A.

 Sin A 2sin $$\frac{A}{2}$$ Cos $$\frac{A}{2}$$ Cos A Cos2  $$\frac{A}{2}$$ - Sin2 $$\frac{A}{2}$$ Cos A 2cos2  $$\frac{A}{2}$$ - 1 Cos A 1 - 2 sin2 $$\frac{A}{2}$$ 1 + Cos A 2cos2 $$\frac{A}{2}$$ 1 - Cos A 2 sin2 $$\frac{A}{2}$$

Here,

tan$$\frac {\theta}{2}$$ = $$\frac 34$$

L.H.S.

= cos$$\theta$$

= $$\frac {1 - tan^2\frac {\theta}2}{1 + tan^2\frac {\theta}2}$$

= $$\frac {1 - (\frac 34)^2}{1 + (\frac 34)^2}$$

= $$\frac {1 - \frac 9{16}}{1 + \frac 9{16}}$$

= $$\frac {\frac {16 - 9}{16}}{\frac {16 + 9}{16}}$$

= $$\frac 7{16}$$× $$\frac {16}{25}$$

= $$\frac 7{25}$$

Hence, L.H.S. = R.H.S. Proved

Here,

cos$$\frac {\theta}{2}$$ = $$\frac 23$$

L.H.S.

=cos$$\theta$$

= 4 cos3$$\frac {\theta}3$$ - 3 cos$$\frac {\theta}3$$

= 4× ($$\frac 23$$)3 - 3× $$\frac 23$$

= 4× $$\frac 8{27}$$ - $$\frac 63$$

= $$\frac {32 - 54}{27}$$

= $$\frac {-22}{27}$$

Hence, L.H.S. = R.H.S. Proved

Here,

tan$$\frac {\alpha}3$$ = $$\frac 15$$

L.H.S.

=tan$$\alpha$$

= $$\frac {3 tan{\frac {\alpha}3} - tan^3{\frac {\alpha}3}}{1 - 3 tan^2 {\frac {\alpha}3}}$$

= $$\frac {3 × {\frac 15} - (\frac 15)^3}{1 - 3 × (\frac 15)^2}$$

= $$\frac {\frac 35 - \frac 1{125}}{1 - \frac 3{25}}$$

= $$\frac {\frac {75 - 1}{125}}{\frac {25 - 3}{25}}$$

= $$\frac {74}{125}$$× $$\frac {25}{25}$$

= $$\frac {37}{55}$$

Hence, L.H.S. = R.H.S. Proved

L.H.S.

=$$\frac {1 + cos\theta}{1 - cos\theta}$$

= $$\frac {1 + 2 cos^2{\frac \theta2} - 1}{1 - (1 - 2sin^2\frac {\theta}2)}$$

=$$\frac {2 cos^2{\frac \theta2}}{1 - 1 + 2sin^2\frac {\theta}2}$$

= $$\frac {2 cos^2\frac \theta2}{2 sin^2\frac \theta2}$$

=$$\frac {cos^2\frac \theta2}{sin^2\frac \theta2}$$

= cot2$$\frac \theta2$$

Hence, L.H.S. = R.H.S. Proved

L.H.S.

=$$\frac {1 - tan^2({\frac \pi4} - {\frac \theta4})}{1 +tan^2({\frac \pi4} - {\frac \theta4})}$$

= cos 2($$\frac \pi4$$ - $$\frac \theta4$$)

= cos ($$\frac \pi2$$ - $$\frac \theta2$$)

= sin$$\frac \theta2$$

Hence, L.H.S. = R.H.S. Proved

L.H.S.

=$$\frac {1 + sin\theta - cos\theta}{1 + sin\theta + cos\theta}$$

= $$\frac {1 + 2 sin\frac \theta2 cos\frac \theta2 - 1 + 2 sin^2\frac \theta2}{1 + 2 sin\frac \theta2 cos\frac \theta2 + 2 cos^2\frac \theta2 - 1}$$

= $$\frac {2 sin\frac \theta2 (cos\frac \theta2 + sin\frac \theta2)}{2 cos\frac \theta2 (cos\frac \theta2 + sin\frac \theta2)}$$

= $$\frac {sin\frac \theta2}{cos\frac \theta2}$$

= tan$$\frac \theta2$$

Hence, L.H.S. = R.H.S. Proved

L.H.S.

=$$\frac {sin^3 \frac \theta2 + cos^3 \frac \theta2}{sin \frac \theta2 + cos \frac \theta2}$$

= $$\frac {(sin \frac \theta2 + cos \frac \theta2) (sin^2\frac \theta2 + cos^2\frac \theta2 - sin\frac \theta2 cos\frac \theta2)}{(sin\frac \theta2 + cos\frac \theta2)}$$

= 1 - $$\frac 22$$ sin$$\frac \theta2$$ cos$$\frac \theta2$$

= 1 - $$\frac 12$$sin$$\theta$$

Hence, L.H.S. = R.H.S. Proved

L.H.S.

=$$\frac {sin\frac A2 + sinA}{1 + cos\frac A2 + cosA}$$

= $$\frac {sin\frac A2 + 2 sin\frac A2 cos\frac A2}{1 + cos\frac A2 + 2 cos^2\frac A2 - 1}$$

= $$\frac {sin \frac A2 (1 + 2 cos\frac A2)}{cos \frac A2 (1 + 2 cos\frac A2)}$$

= tan$$\frac A2$$

Hence, L.H.S. = R.H.S. Proved

Here,

sin $$\frac \theta3$$ = $$\frac 12$$

sin$$\theta$$

= 3 sin $$\frac \theta3$$ - 4 sin3$$\frac \theta3$$

= 3× $$\frac 12$$ - 4× ($$\frac 12$$)3

= $$\frac 32$$ - $$\frac 12$$

= $$\frac {3 - 1}{2}$$

= $$\frac 22$$

= 1 Ans

L.H.S.

=$$\frac {cos\alpha}{1 - sin\alpha}$$

= $$\frac {\frac {1 - tan^2\frac \alpha2}{1 + tan^2\frac \alpha2}}{1 - \frac {2 tan\frac \alpha2}{1 + tan^2\frac \alpha2}}$$

=$$\frac {\frac {1 - tan^2\frac \alpha2}{1 + tan^2\frac \alpha2}}{\frac {1 + tan^2\frac \alpha2 - 2 tan\frac \alpha2}{1 + tan^2\frac \alpha2}}$$

= $$\frac {1 - tan^2\frac \alpha2}{(1 - tan\frac \alpha2)^2}$$

= $$\frac {(1 + tan\frac \alpha2)(1 - tan\frac \alpha2)}{(1 - tan\frac \alpha2)^2}$$

=$$\frac {1 + tan\frac \alpha2}{1 - tan\frac \alpha2}$$

Hence, L.H.S. = R.H.S. Proved

L.H.S.

=$$\frac {sin 2A}{1 + cos 2A}$$× $$\frac {cosA}{1 + cosA}$$

= $$\frac {2 sinA cosA}{2 cos^2A}$$× $$\frac {cosA}{1 + cosA}$$

= $$\frac {sinA}{1 + cosA}$$

= $$\frac {2 sin\frac A2 cos\frac A2}{2 cos^2\frac A2}$$

= tan$$\frac A2$$

Hence, L.H.S. = R.H.S. Proved

Let A = 30°

cos 30° = $$\frac {\sqrt 3}2$$

We know,

sin $$\frac A2$$ =± $$\sqrt {\frac {1 - cosA}{2}}$$

sin $$\frac {30°}2$$ =± $$\sqrt {\frac {1 - cos 30°}{2}}$$

sin 15° =± $$\sqrt {\frac {1 - \frac {\sqrt 3}2}{\frac 21}}$$

or, sin 15° = ± $$\sqrt {\frac {2 - \sqrt 3}{2} × \frac {1}{2}}$$

or, sin 15° =± $$\sqrt {\frac {2 - \sqrt 3}{4}}$$

or, sin 15° =± $$\sqrt {\frac {4 - 2\sqrt 3}{8}}$$

or, sin 15° =± $$\sqrt {\frac {3 - 2\sqrt 3 + 1}{8}}$$

or, sin 15° =± $$\sqrt {\frac {(\sqrt 3 -1)^2}{8}}$$

∴ sin 15° =± $$\frac {\sqrt 3 -1}{2\sqrt 2}$$ Ans

L.H.S.

=$$\frac {2 sin\beta + sin 2\beta}{2 sin\beta - sin 2\beta}$$

=$$\frac {2 sin\beta + 2 sin\beta cos\beta}{2 sin\beta - 2 sin\beta cos\beta}$$

= $$\frac {2 sin\beta (1 + cos\beta)}{2 sin\beta (1 - cos\beta)}$$

= $$\frac {2 cos^2\frac \beta2}{2 sin^2\frac \beta2}$$

[$$\because$$ 1 + cos$$\theta$$ = 2 cos2$$\frac \theta2$$,1 - cos$$\theta$$ = 2 sin2$$\frac \theta2$$]

= cot2$$\frac \beta2$$

Hence, L.H.S. = R.H.S. Proved

Here,

sin $$\frac \theta3$$ = $$\frac 45$$

sin$$\theta$$

= 3 sin $$\frac \theta3$$ - 4 sin3$$\frac \theta3$$

= 3× $$\frac 45$$ - 4× ($$\frac 45$$)3

= $$\frac {12}5$$ - $$\frac {256}{125}$$

= $$\frac {300 - 256}{125}$$

= $$\frac {44}{125}$$ Ans

Here,

cos$$\frac \theta3$$ = $$\frac 35$$

cos $$\theta$$

= 4 cos3$$\frac \theta3$$ - 3 cos$$\frac \theta3$$

= 4× ($$\frac 35$$)3- 3× $$\frac 35$$

= 4× $$\frac {27}{125}$$ - $$\frac 95$$

= $$\frac {108 - 225}{125}$$

= -$$\frac {117}{125}$$ Ans

Here,

sin$$\frac \theta2$$ = $$\frac 35$$

cos$$\frac \theta2$$

= $$\sqrt {1 - sin^2\frac \theta2}$$

= $$\sqrt {1 - (\frac 35)^2}$$

= $$\sqrt {1-\frac 9{25}}$$

= $$\sqrt {\frac {25 - 9}{25}}$$

= $$\sqrt {\frac {16}{25}}$$

= $$\frac 45$$

Now,

sin$$\theta$$ = 2 sin$$\frac \theta2$$⋅ cos$$\frac \theta2$$ = 2× $$\frac 35$$× $$\frac 45$$ = $$\frac {24}{25}$$ Ans

L.H.S.

=$$\frac {tanx + sinx}{2 tanx}$$

= $$\frac {\frac {sinx}{cosx} + sinx}{\frac {2 sinx}{cosx}}$$

= $$\frac {\frac {sinx + sinx cosx}{cosx}}{\frac {2sinx}{cosx}}$$

= $$\frac {sinx (1 + cosx)}{2 sinx}$$

= $$\frac {1 + cosx}{2}$$

= cos2$$\frac x2$$

Hence, L.H.S. = R.H.S. Proved

L.H.S.

=$$\frac {2 sinA - sin 2A}{2 sinA + sin2A}$$

= $$\frac {2 sinA - 2 sinA cosA}{2 sinA + 2 sinA cosA}$$

= $$\frac {2 sinA (1 - cosA)}{2 sinA (1 + cosA)}$$

= $$\frac {1 - cosA}{1 + cosA}$$

= $$\frac {2 sin^2\frac A2}{2 cos^2\frac A2}$$

= tan2$$\frac A2$$

Hence, L.H.S. = R.H.S. Proved

Here,

sin 45° = $$\frac 1{\sqrt2}$$

We know,

cos45°

= $$\sqrt {1 - sin^245°}$$

= $$\sqrt {1 - (\frac 1{\sqrt 2})^2}$$

= $$\sqrt {1 - \frac 12}$$

= $$\frac 1{\sqrt 2}$$

Now,

cosA= 1 - 2 sin2$$\frac A2$$

Let: A = 45°

or, cos 45° = 1 - 2 sin222$$\frac 12$$°

or, 2 sin222$$\frac 12$$° = 1 - cos 45°

or,2 sin222$$\frac 12$$° = 1 - $$\frac 1{\sqrt 2}$$

or,2 sin222$$\frac 12$$° = $$\frac {\sqrt 2 -1}{\sqrt 2}$$

or,2 sin222$$\frac 12$$° = $$\frac {\sqrt 2 - 1}{\sqrt 2}$$× $$\frac {\sqrt 2}{\sqrt 2}$$

or,2 sin222$$\frac 12$$° = $$\frac {2 - \sqrt 2}{2 × 2}$$

or,2 sin222$$\frac 12$$° = $$\frac {2 - \sqrt 2}{4}$$

or,2 sin222$$\frac 12$$° =± $$\sqrt {\frac {2 - \sqrt 2}{4}}$$

∴2 sin222$$\frac 12$$° =± $$\frac 12$$ $$\sqrt {2 - \sqrt 2}$$ Ans

L.H.S.

=cos2($$\frac \pi4$$ - $$\frac \theta4$$) - sin2($$\frac \pi4$$ - $$\frac \theta4$$)

= cos 2($$\frac \pi4$$ - $$\frac \theta4$$) [$$\because$$ cos2A - sin2A = cos 2A]

= cos($$\frac \pi2$$ - $$\frac \theta2$$)

= cos (90 - $$\frac \theta2$$)

= sin$$\frac \theta2$$

Hence, L.H.S. = R.H.S. Proved

Here,

cos$$\frac \theta3$$ = $$\frac 12$$(a + $$\frac 1a$$)

L.H.S.

=cos$$\theta$$

= 4 cos3$$\frac \theta3$$ - 3 cos$$\frac \theta3$$

= 4 [$$\frac 12$$(a + $$\frac 1a$$)]3 - 3 [$$\frac 12$$(a + $$\frac 1a$$)]

= 4× $$\frac 18$$ (a + $$\frac 1a$$)3 - 3×$$\frac 12$$(a + $$\frac 1a$$)

= $$\frac 12$$[(a + $$\frac 1a$$)3 - 3× (a + $$\frac 1a$$)]

= $$\frac 12$$ [a3 + $$\frac 1{a^3}$$ + 3.a.$$\frac 1a$$ - 3(a + $$\frac 1a$$)]

= $$\frac 12$$(a3 + $$\frac 1{a^3}$$)

Hence, L.H.S. = R.H.S. Proved

R.H.S.

= $$\frac 12$$(cot$$\frac A2$$ - tan$$\frac A2$$)

= $$\frac 12$$($$\frac {cos\frac A2}{sin \frac A2}$$ - $$\frac {sin\frac A2}{cos\frac A2}$$)

= $$\frac 12$$($$\frac {cos^2\frac A2 - sin^2\frac A2}{sin\frac A2 cos\frac A2}$$)

= $$\frac {cosA}{sinA}$$

= cot A Proved

Here,

cos$$\frac A3$$ = $$\frac 12$$(p + $$\frac 1p$$)

L.H.S.

=cosA

= 4 cos3$$\frac A3$$ - 3 cos$$\frac A3$$

= 4 [$$\frac 12$$(p + $$\frac 1p$$)]3 - 3 [$$\frac 12$$(p + $$\frac 1p$$)]

= 4× $$\frac 18$$ (p + $$\frac 1p$$)3 - 3×$$\frac 12$$(p + $$\frac 1p$$)

= $$\frac 12$$[(p + $$\frac 1p$$)3 - 3× (p + $$\frac 1p$$)]

= $$\frac 12$$ [p3 + $$\frac 1{p^3}$$ + 3.p.$$\frac 1p$$ - 3(p + $$\frac 1p$$)]

= $$\frac 12$$(p3 + $$\frac 1{p^3}$$)

Hence, L.H.S. = R.H.S. Proved

Here,

sin$$\frac \alpha3$$ = $$\frac 35$$

sin $$\alpha$$

= 3 sin$$\frac \alpha3$$ - 4 sin3$$\frac \alpha3$$

= 3 × $$\frac 35$$ - 4 × ($$\frac 35$$)3

= $$\frac 95$$ - 4 × $$\frac {27}{125}$$

= $$\frac {225 - 108}{125}$$

= $$\frac {117}{125}$$ Ans

Here,

sin$$\alpha$$ = $$\frac 35$$

cos$$\alpha$$ = $$\sqrt {1 - sin^2\alpha}$$ = $$\sqrt {1 - (\frac 35)^2}$$ = $$\sqrt {1 - \frac 9{25}}$$ = $$\sqrt {\frac {25 - 9}{25}}$$ = $$\sqrt {\frac {16}{25}}$$ = $$\frac 45$$

Now,

cos 2$$\alpha$$

= cos2$$\alpha$$ - sin2$$\alpha$$

= ($$\frac 45$$)2- ($$\frac 35$$)2

= $$\frac {16}{25}$$ - $$\frac {9}{25}$$

= $$\frac {16 - 9}{25}$$

= $$\frac 7{25}$$

Again,

sin 3$$\alpha$$

= 3 sin$$\alpha$$ - 4 sin3$$\alpha$$

= 3× $$\frac 35$$ - 4× ($$\frac 35$$)3

= $$\frac 95$$ - 4 × $$\frac {27}{125}$$

= $$\frac 95$$ - $$\frac {108}{125}$$

= $$\frac {225 - 108}{125}$$

= $$\frac {117}{125}$$

∴ cos 2$$\alpha$$ = $$\frac 7{25}$$ and sin 3$$\alpha$$ = $$\frac {117}{125}$$ Ans

L.H.S.

=tan($$\frac {\pi^c}{4}$$ - $$\frac A2$$)

= $$\frac {tan\frac \pi4 - tan\frac A2}{1 + tan\frac \pi4 . tan\frac A2}$$

= $$\frac {1 - tan\frac A2}{1 + tan\frac A2}$$

= $$\frac {1 - \frac {sin\frac A2}{cos\frac A2}}{1 + \frac {sin\frac A2}{cos\frac A2}}$$

= $$\frac {\frac {cos\frac A2 - sin\frac A2}{cos\frac A2}}{\frac {cos\frac A2 + sin\frac A2}{cos\frac A2}}$$

= $$\frac {cos\frac A2 - sin\frac A2}{cos\frac A2 + sin\frac A2}$$×$$\frac {cos\frac A2 - sin\frac A2}{cos\frac A2 -sin\frac A2}$$

= $$\frac {(cos\frac A2 - sin\frac A2)^2}{cos^2\frac A2 - sin^2\frac A2}$$

= $$\frac {cos^2\frac A2 + sin^2\frac A2 - 2 sin\frac A2.cos\frac A2}{cosA}$$

= $$\frac {1 - sinA}{\sqrt {1 - sin^2A}}$$

= $$\sqrt {\frac {(1 - sinA) (1 - sinA)}{(1 + sinA) (1 - sinA)}}$$

= $$\sqrt {\frac {1 - sinA}{1 + sinA}}$$

Hence, L.H.S. = R.H.S. Proved

L.H.S.

=cot22$$\frac 12$$° - tan22$$\frac 12$$°

= $$\frac {cos 22\frac 12°}{sin 22\frac 12°}$$ -$$\frac {sin 22\frac 12°}{cos 22\frac 12°}$$

= $$\frac {cos^2 22\frac 12° - sin^2 22\frac 12°}{sin 22\frac 12° cos 22\frac 12°}$$

= $$\frac {cos 2.22\frac 12°}{\frac 12 × 2 sin 22\frac 12° cos 22\frac 12°}$$

= $$\frac {cos 45°}{\frac 12 sin 2.22\frac 12°}$$

= $$\frac {2 × \frac 1{\sqrt 2}}{sin 45°}$$

= $$\frac {2 × \frac 1{\sqrt 2}}{\frac 1{\sqrt 2}}$$

= 2

Hence, L.H.S = R.H.S. Proved

Let: A = 18°

Then,

5A = 90°

or, 2A + 3A = 90°

or, 3A = 90° - 2A

Puting cos on both sides:

cos 3A = cos (90° - 2A)

or, 4 cos3A - 3 cosA = sin 2A

or, cosA(4 cos2A - 3) = 2 sinA cosA

or, 4 (1 - sin2A) - 3 = 2 sinA

or, 4 - 4 sin2A - 3 - 2 sinA = 0

or, - 4 sin2A - 2 sinA + 1 = 0

or, -(4 sin2A + 2 sinA - 1) = 0

or,4 sin2A + 2 sinA - 1 = 0

Comparing the above condition with ax2 + bx + c = 0

a = 4

b = 2

c = -1

x = $$\frac {-b ± \sqrt {b^2 - 4ac}}{2a}$$

Now,

sin 18°

=$$\frac {-2 ± \sqrt {2^2 - 4 × 4 × (-1)}}{2 × 4}$$

=$$\frac {-2 ± \sqrt {4 + 16}}{8}$$

=$$\frac {-2 ± \sqrt {20}}{8}$$

= $$\frac {2 (-1 + \sqrt 5)}{8}$$

= $$\frac {-1 + \sqrt 5}{4}$$ Ans

L.H.S.

=tan($$\frac {\pi}4 + \frac A2$$)

= $$\frac {tan \frac \pi4 + tan\frac A2}{1 - tan \frac \pi4 tan \frac A2}$$

= $$\frac {tan 45° + tan\frac A2}{1 - tan 45° tan\frac A2}$$

= $$\frac {1 + \frac {sin\frac A2}{cos\frac A2}}{1 - 1 × \frac {sin \frac A2}{cos\frac A2}}$$

= $$\frac {\frac {cos\frac A2 + sin\frac A2}{cos\frac A2}}{\frac {cos\frac A2 - sin\frac A2}{cos\frac A2}}$$

= $$\frac {cos\frac A2 + sin\frac A2}{cos\frac A2 - sin\frac A2}$$× $$\frac {cos\frac A2 + sin\frac A2}{cos\frac A2 + sin\frac A2}$$

= $$\frac {(cos\frac A2 + sin\frac A2)^2}{cos^2\frac A2 - sin^2\frac A2}$$

= $$\frac {cos^2\frac A2 + sin^2\frac A2 + 2 sin\frac A2 cos\frac A2}{cosA}$$

= $$\frac {1 + sinA}{cosA}$$

= $$\frac 1{cosA}$$ + $$\frac {sinA}{cosA}$$

= secA + tanA

Hence, L.H.S. = R.H.S. Proved

L.H.S.

=$$\frac {sin 2A}{1 + cos 2A} × \frac {cosA}{1 + cosA}$$

= $$\frac {2 sinA cosA}{1 + 2 cos^2A - 1}× \frac {cosA}{1 + cosA}$$

= $$\frac {2 sinA cosA}{2 cos^2A}× \frac {cosA}{1 + cosA}$$

= $$\frac {sinA}{1 + cosA}$$

= $$\frac {2 sin\frac A2 cos\frac A2}{1 + 2 cos^2\frac A2 - 1}$$

= $$\frac {2 sin\frac A2 cos\frac A2}{2 cos^2\frac A2}$$

= $$\frac {sin\frac A2}{cos\frac A2}$$

= tan$$\frac A2$$

Hence, L.H.S. = R.H.S. Proved

L.H.S.

=sec($$\frac \pi4$$ + $$\frac \theta2$$)×sec($$\frac \pi4$$ - $$\frac \theta2$$)

= $$\frac 1{cos (\frac \pi4 + \frac \theta2)}$$×$$\frac 1{cos (\frac \pi4 - \frac \theta2)}$$

= $$\frac 1{(cos\frac \pi4 cos\frac \theta2 - sin\frac \pi4 sin\frac \theta2)(cos\frac \pi4 cos\frac \theta2 + sin\frac \pi4 sin\frac \theta2)}$$

= $$\frac 1{(\frac 1{\sqrt 2} cos\frac \theta2 - \frac 1{\sqrt 2} sin\frac \theta2)(\frac 1{\sqrt 2} cos\frac \theta2 + \frac 1{\sqrt 2} sin\frac \theta2)}$$

= $$\frac 1{\frac 1{\sqrt 2} (cos\frac \theta2 - sin\frac \theta2) × \frac 1{\sqrt 2} (cos\frac \theta2 + sin\frac \theta2)}$$

=$$\frac 1{\frac 12 (cos^2\frac \theta2 - sin^2\frac \theta2)}$$

= $$\frac 2{cos\theta}$$

= 2 sec$$\theta$$

Hence, L.H.S. = R.H.S. Proved

L.H.S.

=cot($$\frac A2$$ + 45°) - tan($$\frac A2$$ + 45°)

= $$\frac {cot\frac A2 cot 45° - 1}{cot 45° + cot\frac A2}$$ -$$\frac {tan\frac A2 tan 45°}{1 + tan\frac A2 tan 45°}$$

= $$\frac {\frac {cos\frac A2}{sin\frac A2} ×1 - 1}{1 + \frac {cos\frac A2}{sin\frac A2}}$$ -$$\frac {\frac {sin\frac A2}{cos\frac A2} - 1}{1 + \frac {sin\frac A2}{cos\frac A2}×1}$$

= $$\frac {\frac {cos\frac A2 - sin\frac A2}{sin\frac A2}}{\frac {sin\frac A2 + cos\frac A2}{sin\frac A2}}$$ -$$\frac {\frac {sin\frac A2 - cos\frac A2}{cos\frac A2}}{\frac {cos\frac A2 + sin\frac A2}{cos\frac A2}}$$

=$$\frac {cos\frac A2 - sin\frac A2}{sin\frac A2 + cos\frac A2}$$- $$\frac {sin\frac A2 - cos\frac A2}{cos\frac A2 + sin\frac A2}$$

= $$\frac {cos\frac A2 - sin\frac A2 - sin\frac A2 + cos\frac A2}{cos\frac A2 + sin\frac A2}$$

= $$\frac {2(cos\frac A2 - sin\frac A2)}{cos\frac A2 + sin\frac A2}$$× $$\frac {cos\frac A2 - sin\frac A2}{cos\frac A2 - sin\frac A2}$$

= $$\frac {2(cos^2\frac A2 + sin^2\frac A2 - 2sin\frac A2 cos\frac A2)}{cos^2\frac A2 - sin^2\frac A2}$$

= $$\frac {2(1 - sinA)}{cosA}$$× $$\frac {cosA}{cosA}$$

= $$\frac {2(1 - sinA) cosA}{1 - sin^2A}$$

= $$\frac {2 cosA (1 - sinA)}{(1 + sinA) (1 - sinA)}$$

= $$\frac {2 cosA}{1 + sinA}$$

Hence, L.H.S. = R.H.S. proved

L.H.S.

=tan ($$\frac \pi4$$ + $$\frac A2$$)

= $$\frac {tan\frac \pi4 + tan\frac A2}{1 - tan\frac \pi4 tan\frac A2}$$

= $$\frac {1 + \frac {sin\frac A2}{cos\frac A2}}{1 -\frac {sin\frac A2}{cos\frac A2}}$$

= $$\frac {\frac {cos\frac A2 + sin\frac A2}{cos\frac A2}}{\frac {cos\frac A2 - sin\frac A2}{cos\frac A2}}$$

= $$\frac {cos\frac A2 + sin\frac A2}{cos\frac A2 - sin\frac A2}$$

= $$\sqrt {\frac {(cos\frac A2 + sin\frac A2)^2}{(cos\frac A2 - sin\frac A2)^2}}$$

= $$\sqrt {\frac {cos^2\frac A2 + 2 cos\frac A2 sin\frac A2 + sin^2\frac A2}{cos^2\frac A2 - 2 cos\frac A2 sin\frac A2 + sin^2\frac A2}}$$

= $$\sqrt {\frac {1 + sinA}{1 - sinA}}$$ M.H.S

Again,

$$\sqrt {\frac {1 + sinA}{1 - sinA}}$$

=$$\sqrt {\frac {1 + sinA}{1 - sinA} × \frac {1 + sinA}{1 + sinA}}$$

= $$\sqrt {\frac {(1 + sinA)^2}{cos^2A}}$$

= $$\frac {1 + sinA}{cosA}$$

Hence, L.H.S = M.H.S = R.H.S. Proved

L.H.S.

=cos4$$\frac \pi8$$ + cos4$$\frac {3\pi}8$$ + cos4$$\frac {5\pi}8$$ + cos4$$\frac {7\pi}8$$

=cos4$$\frac \pi8$$ + cos4$$\frac {3\pi}8$$ + cos4($$\pi$$ - $$\frac {3\pi}8$$) + cos4($$\pi$$ - $$\frac {\pi}8$$)

=cos4$$\frac \pi8$$ + cos4$$\frac {3\pi}8$$ + cos4$$\frac {3\pi}8$$ + cos4$$\frac {\pi}8$$

= 2 cos4$$\frac \pi8$$ + 2 cos4$$\frac {3\pi}8$$

= 2 (cos4$$\frac \pi8$$ + cos4$$\frac {3\pi}8$$)

= 2 [cos4$$\frac \pi8$$ + cos4($$\frac{\pi}2 - \frac {\pi}8$$)]

= 2 [cos4$$\frac \pi8$$ + sin4$$\frac \pi8$$]

= 2 [(cos2$$\frac \pi8$$)2 + (sin2$$\frac \pi8$$)2]

= 2 [(cos2$$\frac \pi8$$ + sin2$$\frac \pi8$$)2 - 2cos2$$\frac \pi8$$sin2$$\frac \pi8$$]

= [2 - 4 cos2$$\frac \pi8$$sin2$$\frac \pi8$$]

= 2 - (2 sin$$\frac \pi8$$cos$$\frac \pi8$$ )2

= 2 - (sin 2 . $$\frac \pi8$$)2

= 2 - (sin $$\frac \pi4$$)2

= 2 - ($$\frac 1{\sqrt 2}$$)2

= 2 - $$\frac 12$$

= $$\frac {4 - 1}2$$

= $$\frac 32$$

Hence, L.H.S. = R.H.S Proved

Let:

$$\theta$$ = 18°

Then:

5$$\theta$$ = 90°

i.e. 2$$\theta$$ + 3$$\theta$$ = 5$$\theta$$

or, 2$$\theta$$ + 3$$\theta$$ = 90°

or, 2$$\theta$$ = 90° - 3$$\theta$$

Now,

sin 2$$\theta$$ = sin (90° - 3$$\theta$$)

or, 2 sin$$\theta$$ cos$$\theta$$ = 4 cos3$$\theta$$ - 3 cos$$\theta$$

or,2 sin$$\theta$$ cos$$\theta$$ = cos$$\theta$$ (4 cos2$$\theta$$ - 3)

or, 2 sin$$\theta$$ = 4 - 4 sin2$$\theta$$ - 3

or, 2 sin$$\theta$$ = 1 - 4sin2$$\theta$$

or, 4sin2$$\theta$$ +2 sin$$\theta$$ - 1 = 0 .................................(i)

Comparingequation (i) with quadratic equation ax2+ bx + c = 0, we get:

sin$$\theta$$

= $$\frac {-b ± \sqrt {b^2 - 4ac}}{2a}$$

= $$\frac {-2 ± \sqrt {2^2 - 4 . 4 . (-1)}}{2 . 4}$$

= $$\frac {-2 ± \sqrt {4 + 16}}{8}$$

= $$\frac {-2 ± \sqrt {20}}{8}$$

= $$\frac {-2 ± 2\sqrt 5}{8}$$

= $$\frac {-1 ± \sqrt 5}{4}$$

∴ sin 18° =$$\frac {-1 ± \sqrt 5}{4}$$

But the value of sin 18° is positive.

Put (+) sign

∴ sin 18° =$$\frac {-1 + \sqrt 5}{4}$$

And

cos 36°

= 1 - 2 sin2$$\frac {36°}2$$ [$$\because$$ cos$$\theta$$ = 1 - sin2$$\frac \theta2$$]

= 1 - 2 sin218°

= 1 - 2 ($$\frac {-1 + \sqrt 5}{4}$$)2

= 1 - 2 ($$\frac {1 - 2\sqrt 5 + 5}{16}$$)

= 1 - ($$\frac {6 - 2\sqrt 5}8$$)

= $$\frac {8 - 6 + 2\sqrt 5}{8}$$

= $$\frac {2 + 2\sqrt 5}{8}$$

= $$\frac {2 (1 + \sqrt 5)}8$$

= $$\frac {1 + \sqrt 5}4$$

∴ sin 18° =$$\frac {-1 + \sqrt 5}{4}$$

and cos 36° =$$\frac {1 + \sqrt 5}{4}$$ Ans

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• ### Express sin A interms of cot (frac{A}{2}).

(frac {2cotfrac A4}{3 + cot^2frac A1})

(frac {1cotfrac A1}{3 + cot^-3frac- A2})

(frac {-2cotfrac A-2}{1 + cot^3frac A3})

(frac {2cotfrac A2}{1 + cot^2frac A2})

22

12

21

24

• ### Find the value of cos2θ:sinθ =(frac{3}{5})

(frac{7}{25})

(frac{17}{25})

(frac{7}{2})

(frac{14}{25})

• ### Find the value of cos2θ:tanθ =(frac{5}{12})

(frac{119}{169})

(frac{71}{25})

(frac{117}{125})

(frac{7}{25})

• ### Find the value of tan2θ:sinθ =(frac{3}{5})

(frac{24}{7})

(frac{7}{25})

(frac{17}{25})

(frac{71}{25})

• ### Find the value of tan2θ:tanθ =(frac{3}{5})

(frac{77}{25})

(frac{67}{25})

(frac{63}{40})

(frac{71}{29})

• ### Find the value of sinθ:sin(frac{θ}{2}) =(frac{3}{5})

(frac{24}{25})

(frac{22}{25})

(frac{27}{25})

(frac{23}{25})

• ### Find the value of cosθ:sin(frac{θ}{2}) =(frac{4}{5})

(-frac{7}{25})

(frac{17}{25})

(frac{7}{25})

(frac{11}{25})

• ### Find the value of cosθ:cos(frac{θ}{2}) = (frac{3}{5})

(frac{8}{25})

(frac{7}{25})

(-frac{7}{25})

(-frac{6}{25})

• ### Find the value of tanθ:cos(frac{θ}{2}) =(frac{3}{5})

(frac{24}{25})

(frac{22}{25})

(frac{27}{25})

(frac{37}{25})

• ### Find the value of sin3θ:sinθ =(frac{3}{5})

(frac{127}{125})

(frac{147}{125})

(frac{137}{125})

(frac{117}{125})

• ### Find the value of cos3θ, when cosθ =(frac{4}{5})

(-frac{67}{125})

(-frac{49}{125})

(-frac{47}{125})

(-frac{46}{125})

• ### Find the value of tan3θ, when tanθ =(frac{2}{3})

(frac{47}{9})

(-frac{47}{9})

(frac{46}{9})

(-frac{46}{9})