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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

 

 

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 \(\frac{A}{2}\)

 

.

Very Short Questions

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

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    (frac{23}{25})


    (frac{24}{25})


    (frac{22}{25})


  • Find the value of cosθ:

    sin(frac{θ}{2}) =(frac{4}{5})

    (-frac{7}{25})


    (frac{7}{25})


    (frac{11}{25})


    (frac{17}{25})


  • Find the value of cosθ:

    cos(frac{θ}{2}) = (frac{3}{5})

    (frac{8}{25})


    (frac{7}{25})


    (-frac{6}{25})


    (-frac{7}{25})


  • Find the value of tanθ:

    cos(frac{θ}{2}) =(frac{3}{5})

    (frac{22}{25})


    (frac{24}{25})


    (frac{27}{25})


    (frac{37}{25})


  • Find the value of sin3θ:

    sinθ =(frac{3}{5})

    (frac{137}{125})


    (frac{127}{125})


    (frac{117}{125})


    (frac{147}{125})


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

    (-frac{67}{125})


    (-frac{47}{125})


    (-frac{49}{125})


    (-frac{46}{125})


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

    (-frac{46}{9})


    (frac{47}{9})


    (frac{46}{9})


    (-frac{47}{9})


  • You scored /14


    Take test again

DISCUSSIONS ABOUT THIS NOTE

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Abinav

Also Submit ....Tan A , sin A in form of sub multiple....


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BIRENDA

Tan70-tan20=2tan40 4tan20


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