SUM, DIFFERENCE & PRODUCT FORMULAS
PRODUCT OF SINES & COSINES
Sometimes we need to transform the product of sin or cos of two angles into sum or difference of two trigonometric ratios (very common in calculus). Web havec earlier discussed about the baddition – subtraction formulas which are b c b c c sin α + β , sin α @ β , cos α + β , cos α @ β and their expanded forms. As the
expansion of sin α + β
b
minus sign), we shall try to add and subtract them and see what happens.
Since b c b c sin α + β + sin α @ β = 2 sinα cosβ Thus
c b c 1f b f sinα cosβ = f sin α + β + sin α @ β 2 Since b c b c sin α + β @ sin α @ β D E b
c
and sin α @ β are somewhat similar (they only differ by a b c
= sinα cosβ + cosα sinβ + sinα cosβ @ cosα sinβ
c b
c
= sinα cosβ + cosα sinβ @ sinα cosβ @ cosα sinβ = 2 cosα sinβ
b c b
c
Thus
c b c 1f b f cosα sinβ = f sin α + β @ sin α @ β 2 D E
So, whenever we have a product of sin or cos of two angles, and we want to convert the product into sum or difference of two ctrigonometric ratios, we can use these. Similarly b b c combining the formulas for cos α + β and cos α @ β we can derive the following :
Since b c b c cos α + β + cos α @ β = 2 cosαcosβ Thus
D b
= cosαcosβ @ sinαsinβ + cosαcosβ + sinαsinβ
c b
c
c b c 1f b f cosαcosβ = f cos α + β + cos α @ β 2
E
Since b c b c cos α + β @ cos α @ β = @ 2 sinαsinβ Thus
D b
= cosαcosβ @ sinαsinβ @ cosαcosβ + sinαsinβ
c b
c
c b c 1f b f sinαsinβ = @ f cos α + β @ cos α @ β 2
E
These are summarized below .
Product-to-Sum Identities :
sinαcosβ = cosαsinβ = cosαcosβ = sinαsinβ =
c b c 1f b f f sin α + β + sin α @ β 2 D c b cE 1f b f f sin α + β @ sin α @ β 2 D c b cE 1f b f f cos α + β + cos α @ β 2 D c b cE 1f b f f @ cos α + β @ cos α @ β 2 D E
Example : Simplify 8 sin 15 ° sin 75 ° Solution :
8 sin 15 ° sin 75 °
B a ` aC 1f ` f = 8 B@ f cos 15 ° + 75 ° @ cos 15 ° @ 75 ° 2 B ` aC = @ 4 cos 90 ° @ cos @ 60 °
c b c 1f b f using sinαsinβ = @ f cos α + β @ cos α @ β 2 D
E
= @ 4 cos 90 ° @ cos 60 °
F G 1f f =@4 0 @ f 2 =2 @
A
using cos @ α = cosα
` a
Example : Simplify cos 75 ° sin 15 ° Solution :
c b c 1f b f using cosαsinβ = f sin α + β @ sin α @ β 2 D E
cos 75 ° sin 15 °
B a ` aC 1f ` f = f sin 75 ° + 15 ° @ sin 75 ° @ 15 ° 2 @ A 1f f = f sin 90 ° @ sin 60 ° 2 w w w w w w pff F 1f f f f3 f fff ffG ff = 1@ 2 2
2ffffff @ p3 ffffff ffffff = ffffff 4
w w w w w w
Example : Show that d Solution :
πf πf f f f f 2 cos ff α cos ff α + @ 4 4
e d
e
= cos 2α
πf πf ff f f ff f f 2 cos + α cos @α 4 4
d e d
e
c b c 1f b f f using cosαcosβ = cos α + β + cos α @ β 2 D
E
d e d eG F 1f πf πf πf πf f f ff f f ff f f ff f f ff f f cos + α + @ α + cos +α@ +α 2 4 4 4 4 πf ff f f = cos + cos 2α 2 = 0 + cos 2α = cos 2α
= 2B
SUM & DIFFERENCE OF SINES & COSINES
Sometimes we need to do the opposite of what we have done above, that is, we may need to change the sum or difference of sin or cos of two angles into product of two sin or cos ratios.
αffff +f αfffff @β fffβf fffff ffff fffff fffff ffff and B = 2 2 αffff αfffff +fff fffff βf fffβf @ff fffff ff f ff f ff then A + B = + =α 2 2 @β αffff αfffff +f fffβf fffff fffff fffff ffff ffff @ =β and A @ B = 2 2 If we take two angles A = sinα + sinβ = sin A + B + sin A @ B ` a ` a = sin A cos B + cos A sin B + sin A cos B @ cos A sin B = 2 sin A cos B αffff +β αfffff @β fffff ffff fffff ffff = 2 sin fffffcos fffff 2 2
` a ` a
sinα @ sinβ = sin A + B @ sin A @ B ` a ` a = sin A cos B + cos A sin B @ sin A cos B @ cos A sin B = 2 cos A sin B αffff αfffff +β @β fffff ffff fffff ffff = 2 cos fffffsin fffff 2 2
` a ` a
cosα + cosβ = cos A + B + cos A @ B ` a ` a = cos A cos B @ sin A sin B + cos A cos B + sin A sin B = 2 cos A cos B αffff +β αfffff @β fffff ffff fffff ffff = 2 cos fffffcos fffff 2 2
` a ` a
cosα @ cosβ = cos A + B @ cos A @ B ` a ` a = cos A cos B @ sin A sin B @ cos A cos B + sin A sin B = @ 2 sin A sin B αffff αfffff +β @β fffff ffff fffff ffff = @ 2 sin fffffsin fffff 2 2
` a ` a
These are summarized below .
Sum-to-Product Identities:
αffff +β αfffff @β fffff ffff fffff ffff = 2 sin fffffcos fffff 2 2 αffff αfffff +fff fffff βf @ff ff f ff ff β sinα @ sinβ = 2 cos fffffsin fffff 2 2 αffff +fff αfffff βf @β ff f ff fffff ffff cosα + cosβ = 2 cos fffffcos fffff 2 2 αffff αfffff +fff fffff βf @β ff f ff ffff cosα @ cosβ = @ 2 sin fffffsin fffff 2 2 sinα + sinβ
w w w cos15ff+fffffff w fffffffffffff w fffffffffffff w fff ° f sin15 ° Example : Show that fffffffffffff= p3 cos15 ° @ sin15 °
Solution :
cos15ff+fffffff fffffffffffff fffffffffffff ffff° ffffffff fff ff sin15 ° cos15 ° @ sin15 ° ` a sinffffffffffffffffff ffffffffffffffffffff ffffffffffffffffffff fffffffffffffffffff f 90 ° @ 15 ° + sin15 ° a = ` sin 90 ° @ 15 ° @ sin15 ° sin75ff+fffff°f ffffffffffff fffffffffff f fff° f sin15 f = fffffffffffff sin75 ° @ sin15 °
d
e d e 75ffffffff 75ffffffff f fff fff fff fff f ff+ ff f f° ff ff f ff15 ° f fff15f f f ff ff ff ff fff f f°@ f f° f ff fff 2 sin cos 2 2 ffffffffffffffffffffffffff ffffffffffffffffffffffffff fffffffffffffffffffffffff fffffffffffffffffffffffff d e d e = 75ffffffff 75ffffffff fffffff f ffffff f f+ fff f° f fff fff15 ° f fff15f f f ff ff ff ff fff f f°@ f f° f ff fff 2 cos sin 2 2
αffff +f αfffff @β fffβf fffff ffff fffff ffff Lusing sinα + sinβ = 2 sin M cos fffff L 2 2 M L M L M αffff αfffff +f @β J K fffβf fffff fffff ffff fffff ffff sinα @ sinβ = 2 cos sin 2 2
H I
sinfff°ffffffff ffffffffffff ffff fffffff f45 fcos 30 ° = ffffffffffff cos 45 ° sin 30 ° B fffffff fffffff ff 2 2w fffff w w w w = ffffffff pff 1f 2 fff f fff f fff f ff B
2 w w w w w w p3 = 2
w w w w w w w w w w pff pff 2 3 fff fff fff fff fff fff ff ff
Example : Show that cos 20 ° + cos 100 ° + cos 140 ° = 0 Solution : cos 20 ° + cos 100 ° + cos 140 ° = cos 140 ° + cos 100 ° + cos 20 °
` a f g f F
140ffffffff 140ffffffff ffffffffff ffffffffff ff° + 100 ° ffffffffff ffffffffff ff° @ 100 ° = 2 cos fffffffffff cos fffffffffff + cos 20 ° 2 2 = 2 cos ` ° cos 20 a + cos 20 ° 120 ° = 2 cos 180 ° @ 60 ° cos 20 ° + cos 20 ° ` a = 2 @ cos 60 ° cos 20 ° + cos 20 °
g
αffff +β αfffff @βG fffff ffff fffff ffff using cosα + cosβ = 2 cos fffffcos fffff 2 2
1f f f cos 20 ° + cos 20 ° 2 = @ cos 20 ° + cos 20 ° =0 =2 @
f g
Example : For any angle α , show that sinα + 2 sin 3α + sin 5α = 4 cos 2 αsin 3α
Solution :
sinα + 2 sin 3α + sin 5α ` a = sin 5α + sinα + 2 sin 3α
f g f
5αfffff 5αf+fff ffffff ffffff fffff f fα ffffff ffffff ffffff f@α cos + 2 sin 3α = 2 sin 2 2
g
= 2 sin 3αcos 2α + 2 sin 3α ` a = 2 sin 3α cos 2α + 1 = 2 sin 3α 2 cos 2 α = 4 cos 2 αsin 3α
b b c
F
using sinα + sinβ = 2 sin
αffff +f αfffff @βG fffβf fffff ffff fffff fffff ffff cos 2 2
= 2 sin 3α 2 cos 2 α @ 1 + 1
c
B
using cos 2α = 2 cos 2 α @ 1
C