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Determine the value of cot15 and sin75.

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ginkomind | Student, College Freshman | (Level 1) Honors

Posted November 15, 2010 at 3:08 AM via web

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Determine the value of cot15 and sin75.

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giorgiana1976 | College Teacher | (Level 3) Valedictorian

Posted November 15, 2010 at 3:09 AM (Answer #1)

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We could calculate cot 15 = 1/tan 15 = cos 15/ sin 15.

cos 15/ sin 15 = cos (45-30) / sin (45-30)

We can calculate sin 15 = sin (45-30)

 = sin 45*cos 30 - sin 30*cos 45

sin 45 = sqrt2/2 = cos 45

sin 30 = 1/2

cos 30 = sqrt3/2

sin (45-30) = sqrt6/4 - sqrt2/4 = sqrt2(sqrt3 - 1)/4

cos (45-30) = cos 45cos30 + sin45sin30

cos (45-30) = sqrt6/4 + sqrt2/4

cos (45-30) = sqrt2(sqrt3+1)/4

We'll substitute cos 15 and sin 15 by the found values:

cot 15 = [sqrt2(sqrt3 +1)/4]*[4/sqrt2(sqrt3-1)]

cot 15 = (sqrt3 +1)/(sqrt3 - 1)

cot 15 = (sqrt3 + 1)^2/(3 - 1)

cot 15 = (4+2sqrt3)/2

We'll factorize by 2 and we'll simplify:

cot 15 = 2+sqrt3

We'll calculate sin 75 = sin (15+60)

sin (15+60) = sin15*cos60 + cos15*sin60

cos 60 = 1/2

sin 60 = sqrt 3/2

sin 15 = sqrt2(sqrt3 - 1)/4

sin 75 = [sqrt2(sqrt3 - 1)+ sqrt6(sqrt3+1)]/8

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

Posted November 15, 2010 at 3:11 AM (Answer #2)

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We know that cotx = cosx/sinx

==> cot 15 = cos 15 / sin15

Let us rewrite:

==> cot15 = (cos(45-30)/sin(45-3)

We know that:

cos(a-b) = cosa8cosb + sina*sinb

==> cos(45-30) = cos45*cos30 + sin45*sin30

                           = sqrt(2)/2 * sqrt3/2 + sqrt2/2 * 1/2

                             = sqrt6/4 + sqrt2/4

                               = (sqrt6+sqrt2)/4

Also we know that;

sin(a-b) = sina*cosb - sinb*cosa

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

                   = sqrt2/2 * sqrt3/2 - 1/2*sqrt2/2

                   = sqrt6/4 - sqrt2/4

                    = (sqrt6-sqrt2)/4

Now we will subsitute :

cot15 = (sqrt6+sqrt2)/4 / (sqrt6-sqrt2)/4

           = (sqrt6+sqrt2)/(sqrt6-sqrt2)

            = (sqrt6 +sqrt2)^2 / (6-2)

               = 6 + 2sqrt12 + 2 )/ 4

              = (8+ 4sqrt3)/4

                = 2+ sqrt3

==> cot 15 = 2 + sqrt3

 

sin75 = sin(90 - 15)

sin(a-b) =  sina*cosb - sinb*cosa

sin(90-15) = sin90*cos15 - sin15*cos90

                 = 1*cos15 - sin15*0

                    = cos15

==> sin75 = cos15 = (sqrt6+sqrt2)/4

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neela | High School Teacher | (Level 3) Valedictorian

Posted November 15, 2010 at 3:27 AM (Answer #3)

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We know that sin75 = sin(30+45)

sin75 = sin30cos45 +cos30sin45

sin75 = (1/2) (1/sqrt2) + (sqrt3)/2 * 1/sqrt2

sin75 = {1/2){1/sqrt2 + sqrt3/sqrt2}

sin75 = (1/2){sqrt2+sqr6}.

2)

We know that tan 30 = 2tan15/(1+tan^215).

Therefore 1/sqrt3 = 2t/(1-t^2), where t = tan15.

(1-t^2) = 2tsqrt3.

t^2 +2t(sqrt3) -1 = 0

 t = {-2sqrt3 +or-  sqrt{12 +4)}/2

t =  {-sqrt3 +or -  2 }.

 Therefore  t = 2-sqrt3 as  tan 15  is positive .

Therefore cot 15 = 1/(2-sqrt3) = (2+sqrt3)/(2-sqrt3)(2+sqrt3)

cot 15 = 2+sqrt3.

 

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