Find two solutions to solve sin15 + sin75 .

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hala718 | High School Teacher | (Level 1) Educator Emeritus

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sin15+sin75

Method (1)

We know that sinx +siny = 2sin (x+y)/2 * cos(x-y)/2

sin15+sin75 = 2sin(45) * cos (-30)

               = 2(sqrt(2)/2) (sqrt(3)/2)

               = sqrt(6)/2

Method (2)

sin15+sin75

we will rewrite sin75 as sin(90-15)= cos15

sin15+cos15= = sin(45-30) + cos(45-30)

 = sin45cos30-sin30cos45+ sin45sin30+cos45cos30

= sqrt2/2 * sqrt(3)/2 - 1/2 *sqrt(2)/2 + sqrt(2)/2 *1/2 +sqrt(2)/2 *sqrt(3)/2

= sqrt(6)/4 -sqrt(2)/4 + sqrt(2)/4 + sqrt(6)/4

= 2sqrt(6)/4= sqrt(6)/2

neela's profile pic

neela | High School Teacher | (Level 3) Valedictorian

Posted on

To find 2 solutions to sin15+sin75

Solution:

sin15+sin75  = sin75+sin 15 = 2sin (75+15)/2 * cos (75-15)/2 = 2sin 45*cos30

=2 (1/sqrt2)*(sqrt3/2) =  2(sqrt2*sqrt3)/4 = (sqrt6)/2.

2nd  metod to solve:

sin15 + sin75 = sin15+ sin  (90 -15) = sin15+cos15 = {(1/sqrt2)sin15 + (1/sqrt2) cos 15}sqrt2

 (sin45*sin15 + cos45cos15)sqrt2

= cos (45-15) sqrt2

= (cos 30 )sqrt2

= {(sqrt3)/2} sqrt2

= (sqrt6)/2

Top Answer

giorgiana1976's profile pic

giorgiana1976 | College Teacher | (Level 3) Valedictorian

Posted on

One method would be to consider the fact that being an addition of two alike functions, we'll transform the addition into a product, in this way:

sin a + sin b = 2sin [(a+b)/2]cos [(a-b)/2]

sin 15 + sin 75 = 2sin[(15+75)/2] cos [(15-75)/2]

sin 15 + sin 75 = 2sin45cos30

sin 15 + sin 75 = 2*[(sqrt2)/2]*[(sqrt3)/2]

sin 15 + sin 75 = sqrt(2*3)/2=sqrt(6)/2

Another manner of solving would be to write the angles:

15 = 45 - 30

75 = 45 + 30

sin (45 - 30) = sin45*cos30 - sin30*cos45

                  =(sqrt2/2)(sqrt3/2) - sqrt2/4

                  = (sqrt6 - sqrt2)/4

sin (45 + 30) = sin45*cos30 + sin30*cos45

                  = (sqrt6 + sqrt2)/4

So,

sin 15 + sin 75 = (sqrt6 - sqrt2+sqrt6 + sqrt2)/4

sin 15 + sin 75 = 2sqrt6/4

sin 15 + sin 75 = sqrt6/2

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