# Find the exact value of tan 15 and cos 105 without using tables or a calculator.I honestly don't know how to work this out. I can't find an example in my books. So if someone could show me the...

Find the exact value of tan 15 and cos 105 without using tables or a calculator.

I honestly don't know how to work this out. I can't find an example in my books. So if someone could show me the steps, it would be super loved! I have a couple of questions like this, so explainations would be awesome!

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

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

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

tan 15 = (sqrt3 - 1)/(sqrt3 + 1)

tan 15 = (sqrt3 - 1)^2/(3 - 1)

tan 15 = (4-2sqrt3)/2

**tan 15 = 2-sqrt3**

We'll calculate cos 105 = cos (15+90)

cos (15+90) = cos15*cos90 - sin15*sin90

cos 90 = 0

sin 90 = 1

sin 15 = sqrt2(sqrt3 - 1)/4

cos (15+90) = - sqrt2(sqrt3 - 1)/4

**cos 105 = sqrt2( 1-sqrt3)/4**

We kw that 15 degree .= (1/2)30 degree

Also we know that tan2a = 2tana/(1-tan^2 a).

tan30 = 2tan15/(1-tan^2(15))

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

(1-t^2) = 2tsqrt3.

t^2+(2sqrt3)t -1 = 0. We solve for t as in usual quadratic equation:

t1 = {-2sqrt3 + sqrt[(2sqr3)^2 +4]}/2

t1 = (-2sqrt3 + sqrt14)/2 = (-sqrt3+2)

So t1 = tan 15 = 2-sqrt3.

e2 = (-sqrt3-2) cannot be elligible for 15 degree.

Also tan(105) = tan (90+15) = -cot (15) = - (1/(2-sqrt3) = -(2+sqrt3).

Therefore cos(105) = -1/sqrt(1+(tan105)^2 = -1/(sqrt{1+(2+sqrt3)^2} = -1/sqrt(8+4sqrt3) = -1/(sqrt6+sqrt2) = (sqrt2-sqrt6)/4

So cos(105) = (sqrt2-sqrt6)/4 = -0.258819045 approx.