# mathDetermine the value of cot15 and sin75.

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### 2 Answers

You also may use half angle identity to evaluate `cot 15^o` , such that:

`cot 15^o = cot (30^o/2)`

`cot (30^o/2) = sqrt((1 + cos 30^o)/(1 - cos 30^o))`

`cot (30^o/2) = sqrt((2 + sqrt3)/(2 - sqrt3))`

You may evaluate `sin 15^o` such that:

`1 + cot^2 15^o = 1/(sin^2 15^o)`

`1 + (2 + sqrt3)/(2 - sqrt3) = 1/(sin^2 15^o)`

`(2 - sqrt 3 + 2 + sqrt 3)/(2 - sqrt3) = 1/(sin^2 15^o)`

`4/(2 - sqrt3) = 1/(sin^2 15^o) => sin^2 15^o = (2 - sqrt3)/4`

`sin 15^o = (sqrt(2 - sqrt3))/2`

`cos 15^o = sqrt(1 - (2 - sqrt3)/4) => cos 15^o = (2 + sqrt3)/2`

You may evaluate `sin75^o` such that:

`sin 75^o = sin(90 - 15^o) = cos 15^o`

Since `cos 15^o = (2 + sqrt3)/2` yields `sin 75^o = (2 + sqrt3)/2.`

**Hence, evaluating `cot 15^o` yields `cot 15^o = sqrt((2 + sqrt3)/(2 - sqrt3))` and `sin 75^o = (2 + sqrt3)/2` .**

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