# CalculationCalculate the following functions: sin (7 degrees30 minutes) tg (37 degrees30minutes) cos 165 degrees ctg 195 degrees

sciencesolve | Teacher | (Level 3) Educator Emeritus

Posted on

You need to notice that if you double the angle of `7^o 30'` yields:

`2*7^o 30' = 14^o 60' = 14^o + 1^o = 15^o`

Hence, you need to use the half angle identity, such that:

`2sin^2 7^o 30' = 1 - cos (2*7^o 30' ) => 2sin^2 7^o 30' = 1 - cos 15^o`

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

`cos 15^o = cos (45^o - 30^o) => cos 15^o = cos 45^o*cos 30^o + sin 45^o*sin 30^o`

`cos 15^o = (sqrt2)/2*(sqrt3/2) + (sqrt2)/2*(1/2)`

`cos 15^o = (sqrt2)/2*(sqrt3/2 + 1/2)`

Substituting ` (sqrt2)/2*(sqrt3/2 + 1/2)` for `cos 15^o` yields:

`2sin^2 7^o 30' = 1 -(sqrt2)/2*(sqrt3/2 + 1/2) => sin^2 7^o 30' = 1/2 - sqrt2(sqrt3 + 1)/8`

`sin 7^o 30' = sqrt(1/2 - sqrt2(sqrt3 + 1)/8)`

Hence, evaluating `sin 7^o 30'` yields `sin 7^o 30' = sqrt(1/2 - sqrt2(sqrt3 + 1)/8).`

You need to evaluate `tan 36^o 30'` , hence you need to use the identity `tan alpha = (sin alpha)/(cos alpha)`

`sin^2 36^o 30' = (1 - cos 75^o)/2`

`cos 75^o = cos (30^o + 45^o) => cos 75^o = cos 30^o*cos45^o - sin 30^o*sin 45^o`

`cos 75^o = sqrt2/2((sqrt3 - 1)/2)`

`sin^2 36^o 30' = 1/2 - sqrt2(sqrt3 - 1)/8`

`sin 36^o 30' = sqrt(1/2 - sqrt2(sqrt3 - 1)/8)`

`cos 36^o 30' = sqrt(1/2 + sqrt2(sqrt3 - 1)/8)`

`tan 36^o 30' = sqrt((1/2 - sqrt2(sqrt3 - 1)/8)/(1/2 + sqrt2(sqrt3 - 1)/8))`

Hence, evaluating `tan 36^o 30'` yields `tan 36^o 30' = sqrt((1/2 - sqrt2(sqrt3 - 1)/8)/(1/2 + sqrt2(sqrt3 - 1)/8)).`

neela | High School Teacher | (Level 3) Valedictorian

Posted on

1)To find sin(7.5 deg).

sin30=1/2= 2x(1-x^2)^(1/2)., where x= sin15.Squaring and rearranging,

16(x^2)^2-16x^2+1. Therefore,

x= sin15= sqrt{[2-3^(1/2)]}/2 =0.258819045

sin(15 deg) = sqrt[2-3^(1/2)]/4=y(1-y^2), where y=sin7.5 deg. Squaring and rearranging,

(16y^2)^2-16y^2 +{2-3^(1/2)}/4=0

y=sin(7.5 deg)= sqrt {2-sqrt(2+sqrt3)}/2=0.130526192

2)

To find tg37.5 degree:

tan (7.5) = sqrt{2sqrt(2+sqrt3)}/sqrt[2+sqrt(2+sqrt3)]

=0.131652497.

tan30=1/3^(1/2)=0.577350269

Tan(37.5)= (tan30+tan7.5]/1-tan30)(tan7.5)

=(0.577350269+0.131652497)/[1-(0.577350269)(0.131652497)]

=0.767326986

3)

To find cos 165:

cos 165= -cos(180-15) -cos15 =-sqrt(1- sin^2 (15))

=-sqrt(1-(2-sqr3)/2}=-sqrt(2+sqrt3)/2 =0.965925826

4)

To find ctg 195:

ctg(195) = ctg(180+15)=ctg 15= sqrt(2+sqrt3)/sqrt(2-sqr3) = 2+sqrt3 =3.732050808

giorgiana1976 | College Teacher | (Level 3) Valedictorian

Posted on

For sin (7 degrees30 minutes) and tg (37 degrees30minutes),

you have to apply the following formula:

sin (a/2) = [(1-cos a)/2]^1/2

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

tg 2a = [2tg a]/[1  + tg^2(a)]

sin (7 degrees30 minutes)=sin(15 degrees/2)=

=[(1-cos 15)/2]^1/2

But cos15= cos(30/2)=[(1+cos 30)/2]^1/2=

=[(1+(3^1/2)/2)/2]^1/2={[2 +(3^1/2) ]^1/2}/2

So, [(1-cos 15)/2]^1/2={2-[2 +(3^1/2) ]^1/2}^1/2/2

cos 165 degrees = cos (180-15) = cos(180)cos(15) - sin(180)sin(15)=(-1)*{[2 +(3^1/2) ]^1/2}/2 - 0*sin15 =

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

ctg 195 degrees = ctg(180+15)=[ctg(180)ctg(15)-1]/[ctg^2(180)+ctg^2(15)]

We know the fact that function ctg is a ratio between cos ans sin functions:

ctg(15)=cos(15)/sin(15)