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.
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Questions and Answers
If tan\(\frac {\theta}{2}\) = \(\frac 34\), prove that:
cos\(\theta\) = \(\frac 7{25}\)
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
If cos\(\frac {\theta}{2}\) = \(\frac 23\), prove that:
cos\(\theta\) = \(\frac {-22}{27}\)
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
If tan\(\frac {\alpha}3\) = \(\frac 15\), prove that:
tan\(\alpha\) = \(\frac {37}{55}\)
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
Prove that:
\(\frac {1 + cos\theta}{1 - cos\theta}\) = cot2\(\frac {\theta}2\)
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
Prove that:
\(\frac {1 - tan^2({\frac \pi4} - {\frac \theta4})}{1 + tan^2({\frac \pi4} - {\frac \theta4})}\) = sin\(\frac \theta2\)
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
Prove that:
\(\frac {1 + sin\theta - cos\theta}{1 + sin\theta + cos\theta}\) = tan\(\frac \theta2\)
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
Prove that:
\(\frac {sin^3 \frac \theta2 + cos^3 \frac \theta2}{sin \frac \theta2 + cos \frac \theta2}\) = 1 - \(\frac 12\)sin\(\theta\)
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
Prove that:
\(\frac {sin\frac A2 + sinA}{1 + cos\frac A2 + cosA}\) = tan \(\frac A2\)
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
If sin \(\frac \theta3\) = \(\frac 12\), find the value of sin\(\theta\).
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
Prove that:
\(\frac {cos\alpha}{1 - sin\alpha}\) = \(\frac {1 + tan\frac \alpha2}{1 - tan\frac \alpha2}\)
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
Prove that:
\(\frac {sin 2A}{1 + cos 2A}\) × \(\frac {cosA}{1 + cosA}\) = tan\(\frac A2\)
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
If cos 30° = \(\frac {\sqrt 3}2\), find the value of sin 15°.
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
Prove that:
\(\frac {2 sin\beta + sin 2\beta}{2 sin\beta - sin 2\beta}\) = cot2\(\frac \beta2\)
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
If sin \(\frac \theta3\) = \(\frac 45\), find the value of sin\(\theta\).
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
If cos\(\frac \theta3\) = \(\frac 35\), find the value of cos\(\theta\).
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
Find the value of sin\(\theta\) if sin\(\frac \theta2\) = \(\frac 35\).
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
Prove that:
\(\frac {tanx + sinx}{2 tanx}\) = cos2\(\frac x2\)
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
Prove that:
\(\frac {2 sinA - sin 2A}{2 sinA + sin2A}\) = tan2\(\frac A2\)
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
If sin 45° = \(\frac 1{\sqrt2}\), find the value of sin22\(\frac 12\)°.
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
Prove that:
cos2(\(\frac \pi4\) - \(\frac \theta4\)) - sin2(\(\frac \pi4\) - \(\frac \theta4\)) = sin\(\frac \theta2\)
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
If cos\(\frac \theta3\) = \(\frac 12\)(a + \(\frac 1a\)), prove that:
cos\(\theta\) = \(\frac 12\)(a3 + \(\frac 1{a^3}\))
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
Prove that:
cotA = \(\frac 12\)(cot\(\frac A2\) - tan\(\frac A2\))
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
If cos\(\frac A3\) = \(\frac 12\)(p + \(\frac 1p\)), prove that:
cosA = \(\frac 12\)(p3 + \(\frac 1{p^3}\))
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
If sin\(\frac \alpha3\) = \(\frac 35\), find the value of sin\(\alpha\).
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
If sin\(\alpha\) = \(\frac 35\), find the values of cos 2\(\alpha\) and sin 3\(\alpha\).
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
Prove that:
tan(\(\frac {\pi^c}{4}\) - \(\frac A2\)) = \(\sqrt {\frac {1 - sinA}{1 + sinA}}\)
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
Prove that:
cot22\(\frac 12\)° - tan22\(\frac 12\)° = 2
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
Find the value of sin 18°.
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
Prove that:
tan(\(\frac {\pi}4 + \frac A2\)) = secA + tanA
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
Prove that:
\(\frac {sin 2A}{1 + cos 2A} × \frac {cosA}{1 + cosA}\) = tan\(\frac A2\)
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
Prove that:
sec(\(\frac \pi4\) + \(\frac \theta2\)) × sec(\(\frac \pi4\) - \(\frac \theta2\)) = 2 sec\(\theta\)
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
Prove that:
cot(\(\frac A2\) + 45°) - tan(\(\frac A2\) + 45°) = \(\frac {2 cosA}{1 + sinA}\)
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
Prove that:
tan (\(\frac \pi4\) + \(\frac A2\)) = \(\sqrt {\frac {1 + sinA}{1 - sinA}}\) = \(\frac {1 + sinA}{cosA}\)
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
Prove that:
cos4\(\frac \pi8\) + cos4\(\frac {3\pi}8\) + cos4\(\frac {5\pi}8\) + cos4\(\frac {7\pi}8\) = \(\frac 32\)
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
Without using table or calculator. Find the value of sin 18° and cos 36°.
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
Quiz
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