Transformation of Trigonometric Formulae
Subject: Optional Mathematics
Overview
| 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 of Trigonometric Formulae
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 |
Things to remember
| 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 |
- 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.
Videos for Transformation of Trigonometric Formulae
Factoring Trinomials Using the Product-and-Sum Method
Sums and scalar multiples of linear transformations | Linear Algebra | Khan Academy
Questions and Answers
Prove that:
sin 50° - sin 70° + sin 10° = 0
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
Express sin 36° sin 24° as difference.
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
Express sin 50° cos 32° as sum.
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
Prove that:
\(\frac {cosB - cosA}{cosA + cosB}\) = tan\(\frac {A + B}{2}\) tan\(\frac {A - B}2\)
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
Prove that:
\(\frac {sinA + sinB}{cosA + cosB}\) = tan(\(\frac {A + B}2\))
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
Prove that:
\(\frac {sin 5A - sin 3A}{cos 5A + cos 3A}\) = tanA
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
Evaluate without using calculator or table.
sin 70° - cos 80° + cos 140°
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
Without using calculator or table, prove that:
cos 70° + cos 40° = 2 cos 55° cos 15°
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
Without using calculator or table, find the value of cos 15° - cos 75°.
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
Express as a product sin 50° + sin 20°.
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
Express as a sum or difference: sin 25° cos 75°.
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
Prove that:
sin 5\(\theta\) + sin 3\(\theta\) = 2 sin 4\(\theta\) cos\(\theta\)
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
Prove that:
\(\frac {sinA + sinB}{sinA - sinB}\) = tan\(\frac {A + B}2\) cot\(\frac {A - B}2\)
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
Prove that:
\(\frac {sin 3A - sinA}{cosA - cos 3A}\) = cot 2A
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
Prove that:
\(\frac {sin 2A + sin 5A - sinA}{cos 2A + cos 5A + cosA}\) = tan 2A
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
Prove that:
sin 36° sin 72° sin 108° sin 144° = \(\frac 5{16}\)
L.H.S.
=sin 36° sin 72° sin 108° sin 144°
= \(\frac 12\)(2 sin 36° sin 144°)× \(\frac 12\)(2 sin 72° sin 108°)
= \(\frac 14\) [cos(36° - 144°) - cos(36° + 144°)] [cos (72° - 108°) - cos (72° + 108°)]
= \(\frac 14\) [cos(-108°) - cos(180°)] [cos(-36°) - cos(108°)]
= \(\frac 14\) [cos (90° + 18°) - (-1)] [cos 36° - (-1)] [\(\because\) cos (-\(\theta\)) = cos\(\theta\)]
= \(\frac 14\) [sin 18° + 1] [cos 36° + 1]
= \(\frac 14\) [-\(\frac {\sqrt 5 - 1}4\) + 1] [\(\frac {\sqrt 5 + 1}4\) + 1]
[\(\because\) sin 18° = \(\frac {-\sqrt 5 - 1}4\), cos 36° = \(\frac {\sqrt 5 + 1}4\)]
= \(\frac 14\) (\(\frac {-\sqrt 5 + 1 + 4}4\)) (\(\frac {\sqrt 5 + 1 + 4}4\))
= \(\frac 14\) (\(\frac {5 - \sqrt 5}4\)) (\(\frac {5 + \sqrt 5}4\))
= \(\frac 14\)× \(\frac {(5)^2 - (\sqrt 5)^2}{16}\)
= \(\frac 14\)× \(\frac {25 - 5}{16}\)
= \(\frac {20}{4 × 16}\)
= \(\frac 56\)
Hence, L.H.S. = R.H.S. Proved
Prove that:
cosA cos(60° - A) cos(60° + A) = \(\frac 14\) cos 3A
L.H.S.
=cosAcos(60° - A) cos(60° + A)
= \(\frac 12\) cosA [2cos(60° - A) cos(60° + A)]
= \(\frac 12\) cosA [cos(60° + A + 60° - A) + cos(60° + A - 60° + A )]
= \(\frac 12\) cosA [cos 120° + co 2A]
= \(\frac 12\) cosA× (-\(\frac 12\)) + \(\frac 12\) cosA cos 2A
= -\(\frac 14\) cosA + \(\frac 12\)× \(\frac 12\) (2 cosA cos 2A)
= -\(\frac 14\) cosA + \(\frac 14\) [cos(2A + A) + cos(2A - A)]
= -\(\frac 14\) cosA + \(\frac 14\) (cos 3A + cosA)
= -\(\frac 14\) cosA + \(\frac 14\) cos 3A + \(\frac 14\) cosA
= \(\frac 14\) cos 3A
Hence, L.H.S. = R.H.S. Proved
Find the value of sin 20° sin 40° sin 80°.
Here,
sin 20° sin 40° sin 80°
= sin 20° . \(\frac 12\) [2 sin 40° sin 80°]
= \(\frac 12\) sin 20° [cos(40° - 80°) - cos(40° + 80°)]
= \(\frac 12\) sin 20° [cos (-40°) - cos 120°]
= \(\frac 12\) sin 20° cos 40° + \(\frac 14\) sin 20° [\(\because\) cos(-\(\theta\)) = cos\(\theta\)]
= \(\frac 12\)× \(\frac 12\) [2 sin 20° cos 40°] + \(\frac 14\)sin 20°
= \(\frac 14\) [sin (20° + 40°) + sin (20° - 40°)] + \(\frac 14\)sin 20°
= \(\frac 14\) [sin 60° + sin (-20°)] + \(\frac 14\)sin 20°
= \(\frac 14\) (\(\frac {\sqrt 3}2\) - sin 20°) + \(\frac 14\) sin 20° [\(\because\) sin(-\(\theta\)) = - sin\(\theta\)]
= \(\frac {\sqrt 3}8\) - \(\frac 14\)sin 20° + \(\frac 14\)sin 20°
= \(\frac {\sqrt 3}8\)
Hence, L.H.S. = R.H.S. Proved
Without using calculator or table, find the numerical value of:
8 sin 20° . sin 40° . sin 80°
Here,
8 sin 20° . sin 40° . sin 80°
= 4 sin 20°× (2 sin 40° sin 80°)
= 4 sin 20° [cos (40° - 80°) - cos (40° + 80°)]
= 4 sin 20° [cos (-40°) - cos 120°]
= 4 sin 20° [cos 40° - (-\(\frac 12\))] [\(\because\) cos (-\(\theta\)) = cos\(\theta\)]
= 4 sin 20° cos 40° + 2 sin 20°
= 2× (2 sin 20° cos 40°) + 2 sin 20°
= 2 [sin (20° + 40°) + sin (20° - 40°)] + 2 sin 20°
= 2 [sin 60° + sin (-20°)] + 2 sin 20°
= 2 [\(\frac {\sqrt 3}2\) - sin 20°] + 2 sin 20° [\(\because\) sin (-\(\theta\)) = - sin\(\theta\)]
= \(\sqrt 3\) - 2 sin 20° + 2 sin 20°
= \(\sqrt 3\) Ans
Prove that:
cos 20° cos 40° cos 60° cos 80° = \(\frac 1{16}\)
L.H.S.
= cos 20° cos 40° cos 60° cos 80°
= cos 20° cos 40° \(\frac 12\) cos 80°
= \(\frac 14\) cos 20° (cos 40° cos 80°)
= \(\frac 14\) cos 20° [cos (40° + 80°) + cos (40° - 80°)]
= \(\frac 14\) cos 20° [cos 120° + cos (-40°)]
= \(\frac 14\) cos 20° [cos 120° + cos 40°]
= \(\frac 14\) cos 20° [-\(\frac 12\) + cos 40°]
= -\(\frac 18\) cos 20° + \(\frac 14\) cos 20° cos 40°
= -\(\frac 18\) cos 20° + \(\frac 18\) [2 cos 20° cos 40°]
= -\(\frac 18\) cos 20° + \(\frac 18\) [cos (20° + 40°) + cos (20° - 40°)]
= -\(\frac 18\) cos 20° + \(\frac 18\) [cos 60° + cos (-20°)]
= -\(\frac 18\) cos 20° + \(\frac 18\) [\(\frac 12\) + cos 20°]
= -\(\frac 18\) cos 20° + \(\frac 1{16}\) + \(\frac 18\) cos 20°
= \(\frac 1{16}\)
Hence, L.H.S. = R.H.S. Proved
Prove that:
cos 20° cos 30° cos 40° cos 80° = \(\frac {\sqrt 3}{16}\)
L.H.S.
= cos 20° cos 30° cos 40° cos 80°
= cos 30°× \(\frac 12\) cos 20° (2 cos 40° cos 80°)
= \(\frac {\sqrt 3}{2}\) cos 20° [cos 120° + cos (-40°)]
= \(\frac {\sqrt 3}4\) cos 20° [(-\(\frac 12\)) + cos 40°]
= -\(\frac {\sqrt 3}8\) cos 20° + \(\frac {\sqrt 3}4\) cos 20° cos 40°
= -\(\frac {\sqrt 3}8\) cos 20° + \(\frac {\sqrt 3}4\)× \(\frac 12\) (2 cos 20° cos 40°)
= -\(\frac {\sqrt 3}8\) cos 20° + \(\frac {\sqrt 3}8\) cos (20° + 40°) + cos (20° - 40°)
= -\(\frac {\sqrt 3}8\) cos 20° + \(\frac {\sqrt 3}8\) (cos 60° + cos (-20°))
= -\(\frac {\sqrt 3}8\) cos 20° + \(\frac {\sqrt 3}{16}\) + \(\frac {\sqrt 3}8\) cos 20°
= \(\frac {\sqrt 3}{16}\)
Hence, L.H.S. = R.H.S. Proved
Prove that:
cos2\(\theta\) + cos2(\(\theta\) - 120°) + cos2(\(\theta\) + 120°) = \(\frac 32\)
L.H.S.
=cos2\(\theta\) + cos2(\(\theta\) - 120°) + cos2(\(\theta\) + 120°)
= \(\frac 12\) [2 cos2\(\theta\) + 2 cos2(\(\theta\) + 120°) + 2 cos2(\(\theta\) + 120°)]
= \(\frac 12\) [1 + cos 2\(\theta\) + 1 + cos 2(\(\theta\) - 120°) + 1 + cos 2(\(\theta\) + 120°)]
= \(\frac 12\) [3 + cos 2\(\theta\) + cos (2\(\theta\) + 240°) + cos (2\(\theta\) + 240°)]
= \(\frac 12\) [3 + cos 2\(\theta\) + 2 cos (\(\frac {2\theta - 240° + 2\theta + 240°}2\)) cos (\(\frac {2\theta - 240° - 2\theta - 240°}2\))]
= \(\frac 12\) [3 + cos 2\(\theta\) + 2 cos\(\frac {4\theta}2\) cos\(\frac {-480°}2\)]
= \(\frac 12\) [3 + cos 2\(\theta\) + 2 cos 2\(\theta\) cos 240°]
= \(\frac 12\) [3 + cos 2\(\theta\) + 2 cos 2\(\theta\) × -\(\frac 12\)]
= \(\frac 12\) [3 + cos 2\(\theta\) - cos 2\(\theta\)]
= \(\frac 32\)
Hence, L.H.S. = R.H.S. Proved
Prove that:
sin\(\theta\) sin (60° - \(\theta\)) . sin (60° + \(\theta\)) = \(\frac 14\) sin 3\(\theta\)
L.H.S.
=sin\(\theta\) sin (60° - \(\theta\)) . sin (60° + \(\theta\))
= sin\(\theta\) \(\frac 12\) [2 sin (60° - \(\theta\)) sin (60° + \(\theta\))]
= sin\(\theta\) \(\frac 12\) [cos (60° - \(\theta\) - 60° - \(\theta\)) - cos (60° - \(\theta\) + 60° + \(\theta\))]
= sin\(\theta\) \(\frac 12\) [cos (-2\(\theta\)) - cos 120°]
= \(\frac 12\) sin\(\theta\) [cos 2\(\theta\) - (-\(\frac 12\))]
= \(\frac 12\) sin\(\theta\) cos 2\(\theta\) + \(\frac 14\) sin\(\theta\)
= \(\frac 12\)× \(\frac 12\) [2 sin\(\theta\) cos 2\(\theta\)] + \(\frac 14\) sin\(\theta\)
= \(\frac 14\) [sin (\(\theta\) + 20°) - sin (\(\theta\) - 2\(\theta\))] + \(\frac 14\) sin\(\theta\)
= \(\frac 14\) [sin 3\(\theta\) + sin(-\(\theta\))] + \(\frac 14\) sin\(\theta\)
= \(\frac 14\) sin 3\(\theta\) - \(\frac 14\) sin\(\theta\) + \(\frac 14\)sin\(\theta\)
= \(\frac 14\) sin 3\(\theta\)
Hence, L.H.S. = R.H.S. Proved
Prove that:
(cosA + cosB)2 + (sinA + sinB)2 = 4 cos2\(\frac {A - B}2\)
L.H.S.
=(cosA + cosB)2 + (sinA + sinB)2
= (2 cos\(\frac {A + B}2\) . cos\(\frac {A - B}2\))2 +(2 sin\(\frac {A + B}2\) . cos\(\frac {A - B}2\))2
= 4 cos2\(\frac {A + B}2\) . cos2\(\frac {A - B}2\) +4 sin2\(\frac {A + B}2\) . cos2\(\frac {A - B}2\)
= 4 cos2\(\frac {A - B}2\) (cos2\(\frac {A + B}2\) + sin2\(\frac {A + B}2\))
=4 cos2\(\frac {A - B}2\)
Hence, L.H.S. = R.H.S. Proved
If sin\(\alpha\) = k sin\(\beta\), prove that:
\(\frac {\alpha - \beta}2\) = \(\frac {k - 1}{k + 1}\) tan\(\frac {\alpha + \beta}2\)
Given:
\(\frac {sin\alpha}{sin\beta}\) = \(\frac k1\)
By componendo and dividend, we get:
\(\frac {sin\alpha + sin\beta}{sin\alpha - sin\beta}\) = \(\frac {k + 1}{k - 1}\)
or, \(\frac {2 sin\frac {\alpha + \beta}2 . cos\frac {\alpha - \beta}2}{2 cos\frac {\alpha + \beta}2 . sin\frac {\alpha - \beta}2}\) =\(\frac {k + 1}{k - 1}\)
or, tan\(\frac {\alpha + \beta}2\) . cot\(\frac {\alpha - \beta}2\) =\(\frac {k + 1}{k - 1}\)
or, tan\(\frac {\alpha + \beta}2\)=\(\frac {k + 1}{k - 1}\) . \(\frac 1{cot\frac {\alpha - \beta}2}\)
or,tan\(\frac {\alpha + \beta}2\) .\(\frac {k + 1}{k - 1}\) = tan\(\frac {\alpha - \beta}2\)
∴tan\(\frac {\alpha - \beta}2\) =\(\frac {k + 1}{k - 1}\) .tan\(\frac {\alpha + \beta}2\)
Hence, L.H.S. = R.H.S. Proved
Prove that:
\(\frac {cos 7\theta + cos 3\theta - cos 5\theta - cos\theta}{sin 7\theta - sin 3\theta - sin 5\theta +sin\theta}\) = cot 2\(\theta\)
L.H.S.
=\(\frac {cos 7\theta + cos 3\theta - cos 5\theta - cos\theta}{sin 7\theta - sin 3\theta - sin 5\theta +sin\theta}\)
= \(\frac {(cos 7\theta + cos 3\theta) - (cos 5\theta + cos\theta)}{(sin 7\theta - sin 3\theta) - (sin 5\theta - sin\theta)}\)
= \(\frac {2 cos\frac {7\theta + 3\theta}2 . cos\frac {7\theta - 3\theta}2 - 2 cos\frac {5\theta + \theta}2 . cos\frac {5\theta - \theta}2}{2 cos\frac {7\theta + 3\theta}2 . sin\frac {7\theta - 3\theta}2 - 2 cos\frac {5\theta + \theta}2 . sin\frac {5\theta - \theta}2}\)
= \(\frac {2 cos 5\theta. cos 2\theta - 2 cos 3\theta. cos 2\theta}{2 cos 5\theta. sin 2\theta - 2 cos 3\theta. sin 2\theta}\)
= \(\frac {2 cos 2\theta (cos 5\theta - cos 3\theta)}{2 sin 2\theta (cos 5\theta - cos 3\theta)}\)
= cot 2\(\theta\)
Hence, L.H.S. = R.H.S. Proved
Prove that:
\(\frac {cos 3A + 2 cos 5A + cos 7A}{cosA + 2 cos 3A + cos 5A}\) = cos 2A - sin 2A . tan 3A
L.H.S.
=\(\frac {cos 3A + 2 cos 5A + cos 7A}{cosA + 2 cos 3A + cos 5A}\)
= \(\frac {2 cos 5A + (cos 3A + cos 7A)}{2 cos 3A + (cosA + cos 5A)}\)
= \(\frac {2 cos 5A + 2 cos\frac {3A + 7A}2 . cos\frac {3A - 7A}2}{2 cos 3A + 2 cos\frac {A + 5A}2 . cos\frac {A - 5A}2}\)
= \(\frac {2 cos 5A + 2 cos 5A . cos (-2A)}{cos 3A + 2 cos 3A . cos (-2A)}\)
= \(\frac {2 cos 5A + 2 cos 5A . cos 2A}{2 cos 3A + 2 cos 3A . cos 2A}\)
= \(\frac {2 cos 5A (1 + cos 2A)}{2 cos 3A (1 + cos 2A)}\)
= \(\frac {cos 5A}{cos 3A}\)
= \(\frac {cos (2A + 3A)}{cos 3A}\)
= \(\frac {cos 2A . cos 3A - sin 2A . sin 3A}{cos 3A}\)
= \(\frac {cos 2A . cos 3A}{cos 3A}\) - \(\frac {sin 2A . sin 3A}{cos 3A}\)
= cos 2A - sin 2A . tan 3A
Hence, L.H.S. = R.H.S. Proved
Prove that:
cos2x . sin4x = \(\frac 1{32}\) (cos 6x - 2 cos 4x - cos 2x + 2)
L.H.S.
=cos2x . sin4x
= cos2x . sin2x . sin2x
= sin2x \(\frac 14\) [4 sin2x . cos2x]
= sin2x \(\frac 14\) [(2 sinx . cosx)2]
= \(\frac 14\) sin2x . sin22x
= \(\frac 14\) sin2x . \(\frac 12\) . 2 sin22x
= \(\frac 18\) sin2x (2 sin22x)
= \(\frac 18\) sin2x (1 - cos 4x)
= \(\frac 18\) sin2x - \(\frac 18\) sin2x . cos 4x
= \(\frac 18\) sin2x - \(\frac 18\) (1 - cos2x) . cos 4x
= \(\frac 18\) sin2x - \(\frac 18\) cos 4x + \(\frac 18\) cos2x . cos 4x
= \(\frac 18\) sin2x - \(\frac 18\) cos 4x + \(\frac 1{8}\) × \(\frac 12\) [2 cos2x .cos 4x]
= \(\frac 18\) sin2x - \(\frac 18\) cos 4x + \(\frac 1{16}\) cos 4x (1 + cos 2x)
= \(\frac 18\) sin2x - \(\frac 18\) cos 4x + \(\frac 1{16}\) cos 4x + \(\frac 1{16}\) cos 4x . cos 2x
= \(\frac 18\) sin2x + \(\frac 1{16}\) cos 4x + \(\frac 1{32}\) 2 cos 4x . cos 2x - \(\frac 18\) cos 4x
= \(\frac 18\) sin2x - \(\frac 1{16}\) cos 4x + \(\frac 1{32}\)[cos (4x + 2x) + cos (4x - 2x)]
= \(\frac 18\) sin2x - \(\frac 1{16}\) cos 4x + \(\frac 1{32}\)(cos 6x + cos 2x)
= \(\frac 1{16}\) 2 sin2x - \(\frac 1{16}\) cos 4x + \(\frac 1{32}\) cos 6x + \(\frac 1{32}\) cos 2x
= \(\frac 1{16}\) (1 - cos 2x) - \(\frac 1{16}\) cos 4x + \(\frac 1{32}\) cos 6x + \(\frac 1{32}\) cos 2x
= \(\frac 1{16}\) - \(\frac 1{16}\) cos 2x -\(\frac 1{16}\) cos 4x + \(\frac 1{32}\) cos 6x + \(\frac 1{32}\) cos 2x
= \(\frac 1{32}\) (2 - 2 cos 2x - 2 cos 4x + cos 6x + cos 2x)
= \(\frac 1{32}\) (cos 6x - 2 cos 4x - cos 2x + 2)
Hence, L.H.S. = R.H.S. Proved
Quiz
© 2019-20 Kullabs. All Rights Reserved.