## Note on Transformation of Trigonometric Formulae

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We can transform the product of the trigonometric ratios of the angles into the sum or the difference of the trigonometric ratios of the compound angles and vice versa.

### Transformation of products into sum or difference

(a) We know that,

sinA. cosB + cosA sinB = sin(A + B) .................(i)

sinA cosB - cosA sinB = sin(A - B) ...................(ii)

Adding (i) and (ii), we get

2sinA cosB = sin(A + B) + sin(A - B)

Subtracting (ii) from (i) we get.

2cosA sinB = sin(A + B) - sin(A - B)

(b) Again we know that,

cosA cosB - sinA sinB = cos(A + B) .............(iii)

cosA cosB + sinA sinB = cos(A - B) ..............(iv)

Adding (iii) and (iv) we get

2cosA cosB = cos(A + B) + cos(A - B)

Subtracting (iii) from (iv) we get

2sinA sinB = cos(A - B) - cos(A + B)

Now, the following formulae transform the product of the trigonometric ratios of the angles into the sum or the difference of the trigonometric ratios of the compound angles:

2 sinA cosB = sin(A + B) - sin(A - B)

2 cosA sinB = sin(A + B) - sin(A - B)

2 cosA cosB = sin(A + B) = cos(A - B)

2 sinA sinB = cos(A - B) - cos(A + B)

It will be convenient to remmember the above formulae in the form

2 sin cos = sin + sin

2 sin cos = sin - sin

2 cos cos = cos + cos

2 sin sin = cos - cos

### Transformation of sum or difference into product

From the above formula, we have

sin(A + B) + sin(A - B) = 2 sinA cosB ...........(i)

sin(A + B) - sin(A - B) = 2 cosA sinB ........... (ii)

cos(A + B) + cos (A - B) = 2 cosA cosB ...............(iii)

cos(A - B) - cos(A + B) = 2 sinA sinB ..............(iv)

Suppose A + B = C and A - B = D

Adding the two we get, 2A = C + D

or, A =$$\frac{C + D}{2}$$

Again subtracting second from first, we get 2B = C - D

or, B =$$\frac{C - D}{2}$$

Now, substituting the values of A, B, A + B and A - B in (i), (ii), (iii) and (iv), we have

sinC + sinD = 2 sin ($$)\frac{C + D}{2}$$ cos ($$\frac{C - D}{2}$$)

sinC - sinD = 2 cos ($$\frac{C + D}{2}$$) sin ($$\frac{C - D}{2}$$)

cosC + cosD = 2 cos ($$\frac{C + D}{2}$$) cos ($$\frac{C - D}{2}$$)

cosD - cosC = 2 sin ($$\frac{C + D}{2}$$) sin ($$\frac{D - C}{2}$$)

cosC - cosD = -2 sin ($$\frac{C + D}{2}$$) sin ($$\frac{C - D}{2}$$)

These formulae transform the sum or difference of trigonometric ratios into the products of trigonometric ratios.

It will be convenient to remember the above formulae in the form

sin + sin = 2 sin. cos

sin - sin = 2 cos. sin

cos + cos = 2 cos. sin

cos - cos = 2 sin. sin

 Transformation formulae Key to remember 2sinA cosB = sin(A + B) + sin(A - B) 2 sin. cos = sin + sin 2 cosA sinB = sin(A + B) - sin(A - B) 2 cos. sin = sin - sin 2 cosA cosB = cos(A + B) + cos(A - B) 2 cos. cos = cos + cos 2 sinnA sinB = cos (A - B) - cos(A + B) 2 sin. sin = cos - cos sinC + sinD = 2sin($$\frac{C + D}{2}$$) cos ($$\frac{C - D}{2}$$) sin + sin = 2sin. cos sinC - sinD = 2 cos ($$\frac{C + D}{2}$$) sin ($$\frac{C - D}{2}$$) sin - sin = 2cos. sin cosC + cosD = 2 cos ($$\frac{C + D}{2}$$) cos ($$\frac{C - D}{2}$$) cos + cos = 2cos. cos cosC - cosD = -2 sin ($$\frac{C + D}{2}$$) sin ($$\frac{C - D}{2}$$) cos - cos = 2sin. sin

 Transformation formulae Key to remember 2sinA cosB = sin(A + B) + sin(A - B) 2 sin. cos = sin + sin 2 cosA sinB = sin(A + B) - sin(A - B) 2 cos. sin = sin - sin 2 cosA cosB = cos(A + B) + cos(A - B) 2 cos. cos = cos + cos 2 sinnA sinB = cos (A - B) - cos(A + B) 2 sin. sin = cos - cos sinC + sinD = 2sin($$\frac{C + D}{2}$$) cos ($$\frac{C - D}{2}$$) sin + sin = 2sin. cos sinC - sinD = 2 cos ($$\frac{C + D}{2}$$) sin ($$\frac{C - D}{2}$$) sin - sin = 2cos. sin cosC + cosD = 2 cos ($$\frac{C + D}{2}$$) cos ($$\frac{C - D}{2}$$) cos + cos = 2cos. cos cosC - cosD = -2 sin ($$\frac{C + D}{2}$$) sin ($$\frac{C - D}{2}$$) cos - cos = 2sin. sin
.

### Very Short Questions

L.H.S.

=sin 50° - sin 70° + sin 10°

= 2 cos($$\frac {50° + 70°}2$$) sin ($$\frac {50° - 70°}2$$) + sin 10°

= 2 cos($$\frac {120°}2$$) sin($$\frac {-20°}2$$) + sin 10°

= - 2 cos 60° sin 10° + sin 10°

= - 2× $$\frac 12$$ sin 10° + sin 10°

= - sin 10° + sin 10°

= 0

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

sin 36° sin 24°

= $$\frac 12$$ [2 sin 36° sin 24°]

= $$\frac 12$$ [cos (36° - 24°) - cos(36° + 24°)]

= $$\frac 12$$ [cos 12° - cos 60°] Ans

sin 50° cos 32°

= $$\frac 12$$ (2 sin 50° cos 32°)

=$$\frac 12$$ [sin (50° + 32°) + sin (50° - 32°)]

= $$\frac 12$$[sin 82° + sin 18°] Ans

L.H.S.

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

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

= tan$$\frac {A + B}{2}$$ tan$$\frac {A - B}2$$

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

L.H.S.

=$$\frac {sinA + sinB}{cosA + cosB}$$

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

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

= tan($$\frac {A + B}2$$)

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

L.H.S.

=$$\frac {sin 5A - sin 3A}{cos 5A + cos 3A}$$

=$$\frac {2 cos(\frac {5A + 3A}2) sin(\frac {5A - 3A}2)}{2 cos(\frac {5A + 3A}2) cos(\frac {5A - 3A}2)}$$

= $$\frac {cos4A sinA}{cos 4A cosA}$$

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

= tanA

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

sin 70° - cos 80° + cos 140°

= sin 70° + cos 140° - cos 80°

= sin 70° - 2 sin$$\frac {140° + 80°}2$$ sin$$\frac {140° - 80°}2$$

= sin 70° - 2 sin$$\frac {220°}2$$ sin$$\frac {60°}2$$

= sin 70° - 2 sin 110° sin 30°

= sin 70° - 2 sin(180° - 70°)× $$\frac 12$$

= sin 70° - sin 70°

= 0 Ans

L.H.S.

= cos 70° + cos 40°

= 2 cos$$\frac {70° + 40°}2$$ cos$$\frac {70° - 40°}2$$

= 2 cos$$\frac {110°}2$$ cos$$\frac {30°}2$$

= 2 cos 55° cos 15°

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

Here,

cos 15° - cos 75°

= - 2 sin$$\frac {15 + 75}2$$ sin$$\frac {15 - 75}2$$

= - 2 sin$$\frac {90°}2$$ sin$$\frac {(-60°)}2$$

= - 2 sin 45°× - sin 30°

= 2× $$\frac 1{\sqrt 2}$$× $$\frac 12$$

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

Here,

sin 50° + sin 20°

= 2 sin$$\frac {50° + 20°}2$$ cos$$\frac {50° - 20°}2$$

= 2 sin$$\frac {70°}2$$ cos$$\frac {30°}2$$

= 2 sin 35° cos 15° Ans

Here,

sin 25° cos 75°

= $$\frac 22$$ sin 25° cos 75°

= $$\frac 12$$ (2 sin 25° cos 75°)

= $$\frac 12$$ [sin (25° + 75°) + sin (25° - 75°)]

= $$\frac 12$$ [sin 100° + sin (-50°)]

= $$\frac 12$$ [sin 100° - sin 50°] Ans

L.H.S.

=sin 5$$\theta$$ + sin 3$$\theta$$

= 2 sin$$\frac {5\theta + 3\theta}2$$ cos$$\frac {5\theta - 3\theta}2$$

= 2 sin$$\frac {8\theta}2$$ cos$$\frac {2\theta}2$$

= 2 sin 4$$\theta$$ cos$$\theta$$

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

L.H.S.

=$$\frac {sinA + sinB}{sinA - sinB}$$

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

= tan$$\frac {A + B}2$$ cot$$\frac {A - B}2$$

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

L.H.S.

=$$\frac {sin 3A - sinA}{cosA - cos 3A}$$

= $$\frac {2 cos\frac {3A + A}2 . sin\frac {3A - A}2}{2 sin\frac {3A + A}2 . sin\frac {3A - A}2}$$

= $$\frac {cos\frac {4A}2 . sin\frac {2A}2}{sin\frac {4A}2 . sin\frac {2A}2}$$

= $$\frac {cos 2A}{sin 2A}$$

= cot 2A

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

L.H.S.

=$$\frac {sin 2A + sin 5A - sinA}{cos 2A + cos 5A + cosA}$$

= $$\frac {sin 2A + 2 cos\frac {5A + A}2 . sin\frac {5A - A}2}{cos 2A + 2 cos\frac {5A + A}2 . cos\frac {5A - A}2}$$

= $$\frac {sin 2A + 2 cos 3A . sin 2A}{cos 2A + 2 cos3A . cos 2A}$$

= $$\frac {sin 2A (1 + 2 cos 3A)}{cos 2A (1 + 2 cos 3A)}$$

= $$\frac {sin 2A}{cos 2A}$$

= tan 2A

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

0%
• ### Find The value of :sin 75(^0) + sin 15(^0)

(sqrtfrac{3}{2})

(sqrtfrac{-2}{-1})

(sqrtfrac{-3}{-2})

(sqrtfrac{2}{3})

-2

22

11.12

-1

• ### cos15(^o) - cos 75(^o)

(frac{2}{sqrt{3}})

(frac{1}{sqrt{2}})

(frac{1}{sqrt{-2}})

(frac{-1}{sqrt{3}})

1.996

-101

2

-2

• ### Find the value of sin 75(^0) . sin 15(^0)

(frac{1}{4})

(frac{-1}{2})

(frac{4}{1})

(frac{-1}{-4})

• ### sin 75(^o) . sin 15 (^o)

(frac{1}{4})

(frac{1}{-2})

(frac{4}{1})

(frac{-1}{-4})

• ### cos105(^o) cos15(^o)

(frac{1}{4})

(frac{4}{1})

- (frac{1}{4})

(frac{-1}{-4})

• ### sin105(^o) . sin 15(^o)

- (frac{1}{4})

(frac{-1}{-4})

(frac{1}{-4})

(frac{1}{4})

• ### 4sin 105(^0) sin 15(^o)

-1

1

(frac{1}{2})

-(frac{1}{4})

• ### 4 cos 75(^o) sin 105(^o)

- (frac{1}{4})

1

(frac{1}{4})

-1

-1

1.220

1.222

-2

• ### sin 75(^o) + sin 105(^o)

(frac{1+sqrt{3}}{sqrt{2}})

(frac{1+sqrt{3}}{sqrt{-2}})

(frac{-1+sqrt{-3}}{sqrt{-2}})

-1

-2

1

1.222

-1

• ### cos 105(^o) + cos15(^o)

(frac{-1}{sqrt{2}})

(frac{-1}{sqrt{-2}})

(frac{1}{sqrt{2}})

(frac{2}{sqrt{1}})

• ### sin 15(^o) . sin 105(^o)

(frac{1}{-4})

(frac{1}{4})

- (frac{1}{4})

(frac{-1}{4})

• ### cos 15(^0) cos 75(^o)

1

(frac{1}{4})

(frac{-1}{4})

(frac{1}{-4})

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##### Hrignath

Tan70 - tan20 = 2tan40 4tan10Prove this identity.