# Angles:Calculate sin105 degrees , sin7deg30'

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### 1 Answer

We'll evaluate sin 105 degrees as being the sine of the sum of 2 angles:

105 = 75 + 30

We'll apply sine function both sides:

sin 105 = sin (75 + 30)

To calculate sin (75+30), we'll use the identity:

sin (a+b) = sin a*cos b + sin b*cos a

We'll put a = 75 and b = 30

sin (75 + 30) = sin 75*cos 30 + sin 30*cos 75

We know the values for sin 30 and cos 30. We have to calculate the values for sin 75 and cos 75.

To calculate sin 75, we'll write 75 as the sum of 2 angles:

75 = 30 + 45

We'll apply sine function both sides:

sin 75 = sin (30+45)

We'll put a = 30 and b = 45

sin (30+45) = sin 30*cos 45 + sin 45*cos 30

We'll substitute sin 30; sin 45; cos 30; cos 45 by their values:

sin 30 = 1/2

cos 30 = sqrt3/2

sin 45 = cos 45 = sqrt2/2

sin (30+45) = (1/2)*(sqrt2/2) + (sqrt2/2)*(sqrt3/2)

We'll factorize by (sqrt2/2):

sin (30+45) = (sqrt2/2)[(1+sqrt3)/2]

sin (30+45) = sqrt2*(1+sqrt3)/4

sin 75 = sqrt2*(1+sqrt3)/4

cos 75 = sqrt[1 - 2*(1+sqrt3)^2/16]

cos 75 = sqrt(16 - 2 - 4sqrt3 - 6)/4

cos 75 = sqrt4(2-sqrt3)/4

cos 75 = 2sqrt(2-sqrt3)/4

cos 75 = sqrt(2-sqrt3)/2

sin 105 = sin 75*cos 30 + sin 30*cos 75

sin 105 = sqrt6*(1+sqrt3)/8 + sqrt(2-sqrt3)/4

To calculate cos 7deg30min, we'll write the formula for the half-angle:

cos (a/2) = sqrt [(1+cos a)/2]

Let a = 7deg30min,

2a = 2*7deg30min = 7deg30min + 7deg30min = 14deg + 1deg = 15 degrees

cos 7deg30min = sqrt [(1+cos 15)/2]

cos 15 = cos (30/2) = sqrt [(1+cos 30)/2]

cos (30/2) = sqrt [(2+sqrt3)]/2

cos7deg30min = sqrt [(1 + sqrt [(2+sqrt3)/2]/2]