Trigonometric Ratios And Identities

MathsTrigonometryFor JEE aspirants

Trigonometric Ratios of Compound Angles

The algebraic sum of two or more angles are generally called compound angles and the angles are known as the constituent angle . some standard formulas of compound angles have been given below.

$sin (A + B) = sin A cos B + cos A sin B$
$sin (A - B) = sin A cos B - cos A sin B$
$cos (A + B) = cos A cos B - sin A sin B$
$cos (A - B) = cos A cos B + sin A s in B$
$t an (A + B) = \frac{tan A + tan B}{1 - tan A tan B}$
$tan (A - B) = \frac{tan A - tan B}{1 + tan A tan B}$
$cot (A + B) = \frac{cot A cot B - 1}{cot B + cot A}$
$cot (A - B) = \frac{cot A co t B + 1}{cot B - cot A}$
$sin (A + B) sin (A - B) = sin^{2} A - sin^{2} B = cos^{2} B - cos^{2} A$
$cos (A + B) cos (A - B) = cos^{2} A - sin^{2} B = cos^{2} B - sin^{2} A$
$sin (A + B + C) = cosA cos B s in C + cos A sin B cos C + sinA cos B cos C - sin A sin B sin C$ $or$

sin (A + B + C) = cos A cos B cos C (tan A + tan B + tan C - tan A tan B tan C)

12.

cos (A + B + C) = cos A cos B cos C - sin A sin B cos C - sin A cos B sin C - cos A sin B sin C

or

cos (A + B + C) = cos A cos B cos C (1 - tan A tan B - tan B tan C - tan C tan A)

13.

tan (A + B + C) = \frac{tan A + tan B + tan C - tan A tan B tan C}{1 - tan A tan B - tan B tan C - tan C tan A}

A + B + C = 0,

then

tan A + tan B + tan C = tan A tan B tan C

14. (a)

sin (A_{1} + A_{2} + ........ + A_{n}) = (cos A_{1} cos A_{2} cos A_{3} ..... cos A_{n}) \times (S_{1} - S_{3} + S_{5} + S_{7} ....)

(b)

cos (A_{1} + A_{2} + ...... + A_{n}) = (cos A_{1} cos A_{2} cos A_{3} .... cos A_{n}) \times (1 - S_{2} + S_{4} - S_{6} + ....)

(c)

\tan \left( {{A}_{1}}+{{A}_{2}}+.....+{{A}_{n}} \right)=\dfrac{{{S}_{1}}-{{S}_{3}}+{{S}_{5}}-{{S}_{7}}+#x2026..}{1-{{S}_{2}}+{{S}_{4}}-{{S}_{6}}+.....}

Where,

S_{1} = tan A_{1} + tan A_{2} + ....... + tan A_{n}

[sum of the tangents of the separate angles]

S_{2} = tan A_{1} tan A_{2} + tan A_{2} tan A_{3} + \dots\dots

[sum of the tangents taken two at a time]

S_{3} = tan A_{1} tan A_{2} tan A_{3} + tan A_{2} tan A_{3} tan A_{4} + \dots\dots

[sum of the tangents taken three at a time]Transformation Formulae

$2 sin A cos B = sin (A + B) + sin (A - B)$
$2 cos A sin B = sin (A + B) - sin (A - B)$
$2 cos A cos B = cos (A + B) + cos (A - B)$
$2 sin A sin B = cos (A - B) - cos (A + B)$
$sin C + sin D = 2 sin (\frac{C + D}{2}) cos (\frac{C - D}{2})$
$sin C - sin D = 2 cos (\frac{C + D}{2}) sin (\frac{C - D}{2})$
$cos C + cos D = 2 cos (\frac{C + D}{2}) cos (\frac{C - D}{2})$
$cos C - cos D = - 2 sin (\frac{C + D}{2}) sin (\frac{C - D}{2})$ $= 2 sin (\frac{C + D}{2}) sin (\frac{D - C}{2})$

Trigonometric Ratios of Multiple Angles

$sin 2 A = 2 sin A cos A = \frac{2 tan A}{1 + tan ^{2} A}$
$cos 2 A = cos^{2} A - sin^{2} A = 2 cos^{2} A - 1$
$1 - 2 sin^{2} A = \frac{1 - tan ^{2} A}{1 + tan ^{2} A}$
$tan 2 A = \frac{2 tan A}{1 - tan ^{2} A}$
$sin 3 A = 3 sin A - 4 sin^{3} A$
$cos 3 A = 4 cos^{3} A - 3 cos A$
$tan 3 A = \frac{3 tan A - tan ^{3} A}{1 - 3 tan ^{2} A}$
$sin A = 2 sin \frac{A}{2} cos \frac{A}{2} = \frac{2 tan \frac{A}{2}}{1 + tan ^{2} \frac{A}{2}}$
$cos A = \frac{1 - tan ^{2} \frac{A}{2}}{1 + tan ^{2} \frac{A}{2}}$
$tan A = \frac{2 tan \frac{A}{2}}{1 - tan ^{2} \frac{A}{2}}$
$1 - cos A = 2 sin^{2} \frac{A}{2}$
$1 + cos A = 2 cos^{2} \frac{A}{2}$
$\frac{1 - cos A}{1 + cos A} = tan^{2} \frac{A}{2}$
$sin (\frac{A}{2}) + cos (\frac{A}{2}) = \pm 1 + sin A$
$sin (\frac{A}{2}) - cos (\frac{A}{2}) = \pm 1 - sin A$

Trigonometric Ratios of some Useful Angles

$sin 75^{o} = \frac{3 + 1}{2 2} = cos 15^{o}$
$cos 75^{o} = \frac{3 - 1}{2 2} = sin 15^{o}$
$tan 75^{o} = 2 + 3 = cot 15^{o}$
$co t 75^{o} = 2 - 3 = tan 15^{o}$
$sin 9^{o} = \frac{3 + 5 - 5 - 5}{4} = cos 81^{o}$
$cos 9^{o} = \frac{3 + 5 + 5 - 5}{4} = sin 81^{o}$

Key Points

A. Product of Trigonometric Ratio

$sin θ sin (60^{o} - θ) sin (60^{o} + θ) = \frac{1}{4} sin 3 θ$
$cos θ cos (60^{o} - θ) cos (60^{o} + θ) = \frac{1}{4} cos 3 θ$
$tan θ tan (60^{o} - θ) tan (60^{o} + θ) = tan 3 θ$
$cos 36^{o} cos 72^{o} = \frac{1}{4}$
$cos A cos 2 A cos 4 A ..... cos 2^{n - 1} A = \frac{1}{2 ^{n} sin A} sin (2^{n} A)$

B. Sum of Trigonometric Ratios

$sin A + sin (A + B) + sin (A + 2 B) + .... + sin (A + (n - 1) B)$

$= \frac{sin { A + ( n - 1 ) \frac{B}{2} } sin \frac{n B}{2}}{sin \frac{B}{2}}$

2. $cos A + cos (A + B) + cos (A + 2 B) + ........ + cos (A + (n - 1) B)$

$= \frac{sin \frac{n B}{2}}{sin \frac{B}{2}} cos {A + \frac{( n - 1 ) B}{2}}$

C. Identities for Angles of a Triangle

If $A, B$ and $C$ are angles of a triangle $(or A + B + C = π)$
$sin (A + B) = sin A$
$cos (B + C) = - cos A$
$sin (\frac{B + C}{2}) = cos \frac{A}{2}$
$cos (\frac{B + C}{2}) = sin \frac{A}{2}$
$sin 2 A + sin 2 B + sin 2 C = 4 sin A sin B sin C$
$cos 2 A + cos 2 B + cos 2 C = - 1 - 4 cos A cos B cos C$
$sin A + sin B + sin C = 4 cos \frac{A}{2} cos \frac{B}{2} cos \frac{C}{2}$
$cos A + cos B + cos C = 1 + 4 sin \frac{A}{2} sin \frac{B}{2} sin \frac{C}{2}$
$tan A + tan B + cos C = tan A tan B tan C$
$cot B cot C + cot C cot A + cot A cot B = 1$
$cot \frac{A}{2} + cot \frac{B}{2} cot \frac{C}{2} = cot \frac{A}{2} cot \frac{B}{2} cot \frac{C}{2}$
$tan \frac{A}{2} tan \frac{B}{2} + tan \frac{B}{2} tan \frac{C}{2} + tan \frac{C}{2} tan \frac{A}{2} = 1$

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