Find two ways to calculate tg15 .

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

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We need to calculate: tg15

Method (1):

We know that tg=sin/cos

==> tg15= sin15/cos15

             = sin(45-30)/cos(45-30)

We know that sin(x-y)=sinxcosy-cosxsiny

 and we know cos(x-y)=cosxcosy+sinxsiny

             = sin45cos30-cos45sin30/ cos45cos30-sin45sin30

=[sqrt(2)/2* sqrt(3)/2 -sqrt(2)/2*1/2]/sqrt(2)/2*sqrt(3)/2 +sqrt(2)/2*1/2

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

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

Now multiply and divide by (sqrt(6)-sqrt(2)

= (sqrt6-sqrt2)^2/ 4

= 6-2sqrt12+2/4

= 8-4sqrt(3)/4

= 2-sqrt(3)

 

Method 2:

tg15 = tg(45-30)

We koknwo that tg(x-y)= tgx-tgy/1+tgx*tgy

Now substitute:

tg(45-30)= tg45-tg30 /1+tg45*tg30

            = (1-sqrt(3)/3)/1+1(sqrt(3)/3)

            = (3-3sqrt(3)*3/3+3sqrt(3)*3

           = 3(1-sqrt(3)/3(1+sqrt(3)

           = 1-sqrt3/1+sqrt3

         Multiply and divide by 1-sqrt(3)

==> (1-sqrt3)^2/ (1+sqrt3)(1-sqrt(3)

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

= 4-2sqrt3/-2

= 2-sqrt3

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neela | High School Teacher | (Level 3) Valedictorian

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We know that  tan30 = 1/sqrt3=2t/(1-t^2),where t = tan15.

So (1-t^2) = (2sqrt3)t Or

t^2+(2sqr32)t -1 = 0. Solving, we get:

t = {-2 sqrt3 +sqrt(12+4)}/2 =  -sqrt3 + 2

2nd methd:

tan15 = tan(60-45) = (tan60-tan45)/(1+tan60*tan45)

= (sqrt3 -1)/{1+sqrt3 *1}.

  Rationilise the denominator:

=(sqrt3 - 1)(sqrt1+ sqrt3)

= (sqrt3-1)^2/ (3-1)

= (4-2sqr3)/2

=2-sqrt3

 

 

giorgiana1976's profile pic

giorgiana1976 | College Teacher | (Level 3) Valedictorian

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We'll consider 15 degrees as the result of the difference between 2 well known angles: 45 and 30 degrees.

tg 15 = tg(45 - 30) = (tg 45 - tg 30)/[1 + (tg 45*tg30)]

tg 15 = {1 - [(3)^1/2]/3}/[1 + (1*(3)^1/2/3)]

tg 15 = [3 - (3)^1/2]/[3 + (3)^1/2]

We'll multiply the ratio with the adjoint of the denominator, which is [3 - (3)^1/2] and the result will be:

tg 15 = [3 - (3)^1/2]^2/9 - 3

We'll develop the binomial at the numerator:

tg 15 = [9 + 3 - 6*(3)^1/2]/6

tg 15 = [12 - 6*(3)^1/2]/6

tg 15 = 2 - sq root 3

Second method:

We could calculate tg 15 as being the half of 30 degrees.

tg 15 = tg (30/2) = sin (30/2)/cos (30/2)

sin (30/2) = sq root [(1 - cos 30)/2] = sq root{[2 - sq root(3)]/4}

cos (30/2) = sq root [(1 + cos 30)/2] = sq root{[2 + sq root(3)]/4}

tg 15 = {sq root[2 - sq root(3)]}/{sq root[2 + sq root(3)]}

tg 15 = 1/2 + sq root(3)

We'll multiply the denominator with it's adjoint 2 - sq root(3)

tg 15 = 2 - sq root(3)/(4 - 3)

tg 15 = 2 - sq root 3

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