# Determine sin105 degrees , sin7deg30'.

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### 2 Answers

To determine sin 105 deg .

Let 105 deg = x

We know that sin2x = 2sinx*cosx

Therefore sin210 = 2sinx*sqrt(1-sin^2x)

-0.5 - 2sinx sqrt(1-sin^2x)

We square both sides:

0.25 = 4sin^2x - 4sin^4x

4sin^4x -4sin^2x+0.25

sin^2x = {4+or-sqrt(16-4)}/8 = (2+or-sqrt(3))/4

sin x = (1/2)sqrt{2+sqrt3}.

Therefore sin 105 = (1/2){sqrt(2+sqrt3)}.

But sin105 = sin (90+15) = cos 15.

Therefore cos15 = (1/2)sqrt(2+sqrt3) = 0.965925826

To find sin7deg 30 min = 7.5 deg.

We know cos2a = 1-2sin^2a. Or sin^2 a = (1-cos2a)/2.

Therefore sin^2 (7.5deg) = (1- cos15)/2 = {1- (1/2)sqrt(2+sqrt3)}/2 =

sin(7.5) = sqrt {1/2-(1/4)sqrt(2+sqrt3)} = 0.130526192

To calculate sin 105, we'll write 105 as 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 apply the formula:

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]

We'll put 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]**