## Trigonometric Ratios of Sub Multiple Angles

Subject: Optional Mathematics

#### Overview

If A is an angle, then $\frac{A}{2}$, $\frac{A}{3}$, $\frac{A}{4}$ etc. are called sub - multiple angles of A.

##### Trigonometric Ratios of Sub Multiple Angles

If A is an angle, 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 13sin3A = 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

##### Things to remember

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 $\frac{A}{2}$

• It includes every relationship which established among the people.
• There can be more than one community in a society. Community smaller than society.
• It is a network of social relationships which cannot see or touched.
• common interests and common objectives are not necessary for society.
##### Trigonometric Ratios Of Multiple and Sub Multiple Angles Example #11 / Maths Trigonometry

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