Calculate tan75 in two methods.

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hala718's profile pic

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

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tan75:

method (1):

tan75 = sin75/cos75

            = sin(45+30)/cos(45+30)

             = (sin45cos30+sin30cos45)/(cos45cos30 - sin45sin30)

             = [sqrt2/2*sqrt3/2 + (1/2)*sqrt2/2]/(sqrt2/2*sqrt3/2 - sqrt2/2*(1/2)

             = (sqrt(6)/6 + sqrt2/4)/(sqrt6/4)- sqrt2/4)

             = (2sqrt6 + 3sqrt2)/12 / (sqrt6/4 - sqrt2/4)

             = (2sqrt6 + 3sqrt2)/12 / (sqrt6-sqrt2)/4

             = (2sqrt6 + 3sqrt2)/3(sqrt6-sqrt2)

Multiply by sqrt6 + sqrt2

==> (sqrt6+sqrt2)(2sqrt6+3sqrt2) / 3(6-2)

==> (2*6 + 3*2sqrt3 + 2*2sqrt3 + 3*2)/12

==> (18 + 10sqrt3)/12

==> (9+ 5sqrt3)6

neela's profile pic

neela | High School Teacher | (Level 3) Valedictorian

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tan75 = tan(45+30) = (tan45+tan30)/(1-tan45*tan30)

= (1+ 1/sqrt3)/(1+1*1/sqrt3)

= (sqrt3+1)/( sqrt3-1) = (srt3+1)^2/(3-1) = (4+2sqrt3)/2 = 2+sqrt3.

 

tan2*75 = 2t/(1-t^2) = tan 150 deg = tan (180-30) = -tan30 = -1/2, where t = tan75

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

2sqr3*t = -(1-t^2)

t^2-2sqrt3*t-1 = 0

t = {(-  -2sqr3  +or- sqrt( 14+4)}/2 = sqrt3 +2 Or sqrt3-2.

Therefore tan75 = sqrt3+2  positive is valid as 75 deg is in 1st quadrant.

Top Answer

giorgiana1976's profile pic

giorgiana1976 | College Teacher | (Level 3) Valedictorian

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An indicated method of calculating the value of tan 75:

We could consider 75 degrees as the result of the sum between 2 angles: 45 and 30 degrees.

tan 75 = tg(45+30)= (tan 45 + tan 30)/[1 - (tan 45*tan30)]

tan 75 =(1+sqrt 3/3)/[1 - (1*sqrt3/3)]

tan 75 = [3+sqrt(3)]/[3-sqrt(3)]

We'll multiply the ratio with the adjoint of the denominator, which is [3+sqrt(3)] and the result will be:

tan 75= [3+sqrt(3)]^2/(9-3)

We'll develop the binomial at the numerator:

tan 75= [9+3+6*sqrt(3)]/6

tan 75 = [12 + 6*sqrt(3)]/6

tan 75= 2 + sqrt 3

thewriter's profile pic

thewriter | College Teacher | (Level 1) Valedictorian

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One way you can find tan 75 is by using the relation                       tan(a+b) = {tan(a) +tan(b)}/{1-tan(a)tan(b)}.

Take a as 45 and b as 30.

tan 75= tan(45+30) = {tan(30) +tan(45)}/{1-tan(45)(tan(30)}

Now tan 45 is 1 and tan 30 is 1/3^(1/2)

Substituting these we get {1/3^(1/2) +1}/{1-1/3^(1/2)}

Or tan 75= {1+1/3^(1/2)}/{1-1/3^(1/2)}

 

Another method would be using

tan 75=sin 75/cos 75=sin(45+30)/cos(45+30)= (sin45cos30+cos45sin30)/(cos45cos30-sin45sin30)

Now, cos45=sin45, so we can eliminate these terms and we get (cos30+sin30)/(cos30-sin30)

Now cos 30=3^(1/2)/2 and sin 30=1/2

So substituting:

{3^(1/2)/2+1/2}/{3^(1/2)/2-1/2}={3^(1/2)+1}/{3^(1/2)-1}

tan 75= {3^(1/2)+1}/{3^(1/2)-1}

 

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