Videos Related with Transformation of Trigonometric Formulae

Note on Transformation of Trigonometric Formulae

  • Note
  • Things to remember
  • Videos
  • Exercise
  • Quiz

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{-2}{-1})


    (sqrtfrac{3}{2})


    (sqrtfrac{-3}{-2})


    (sqrtfrac{2}{3})


  • sin 75(^o) - sin 105(^o)

    -2


    -1


    11.12


    22


  • cos15(^o) - cos 75(^o)

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


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


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


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


  • sin 70(^o) - cos80(^o) + cos140(^o)

    1.996


    2


    -2


    -101


  • 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{4}{1})


    (frac{1}{4})


    (frac{1}{-2})


    (frac{-1}{-4})


  • cos105(^o) cos15(^o)

    (frac{-1}{-4})


    (frac{1}{4})


    - (frac{1}{4})


    (frac{4}{1})


  • sin105(^o) . sin 15(^o)

    (frac{1}{4})


     - (frac{1}{4})


    (frac{-1}{-4})


    (frac{1}{-4})


  • 4sin 105(^0) sin 15(^o)

    (frac{1}{2})


    -(frac{1}{4})


    -1


    1


  • 4 cos 75(^o) sin 105(^o)

    (frac{1}{4})


    1


    -1


     - (frac{1}{4})


  • sin 75(^o) - sin 105(^o)

    -1


    1.222


    1.220


    -2


  • sin 75(^o) + sin 105(^o)

    -1


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


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


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


  • sin 105(^o) - cos 15(^o) 

    1.222


    -1


    1


    -2


  • cos 105(^o) + cos15(^o) 

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


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


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


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


  • sin 15(^o) . sin 105(^o)

    (frac{1}{4})


    (frac{-1}{4})


    (frac{1}{-4})


    - (frac{1}{4})


  • cos 15(^0) cos 75(^o)

    (frac{1}{4})


    (frac{1}{-4})


    (frac{-1}{4})


    1


  • You scored /16


    Take test again

DISCUSSIONS ABOUT THIS NOTE

You must login to reply

Forum Time Replies Report


You must login to reply

5sinA-4cosA


You must login to reply

Hrignath

Tan70 - tan20 = 2tan40 4tan10Prove this identity.


You must login to reply

Siddhant shrestha

tan3A tan2A tanA=0 ....... Find the value of A....


You must login to reply