A. sin22.5 degrees

and

B. tanpi/12 degrees

Tutors, sign in to answer this question.

A) sin(22.5)degrees=sin(45/2)degrees

sin(45/2)=+sqrt[(1-cos(45))/2]

=+sqrt[(1-1/sqrt(2))/2]

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

=+sqrt[{2-sqrt(2)/2}/2]

=+sqrt[{2-sqrt(2)}/4]

=+sqrt[{2-1.414213}/2

=+sqrt{0.585786}/2

=+0.765366/2

=0.382683

B) tan pi/12 degrees

1 degree=pi/180

15pi/(15)(12)=15*(pi/180)=15 degrees

tan15 degrees=0.26794919

getting the answer using half-angle formulas gives us the following:

tan(30/2)=sqrt[(1-cos30)/(1+cos30)]

=sqrt[(1-sqrt(3)/2)/(1+sqrt(3)/2)]

simplify 1-sqrt(3)/2 and 1+sqrt(3)/2

1-sqrt(3)/2=[2-sqrt(3)]/2

1+sqrt(3)/2=[2+sqrt(3)]/2

divide[2-sqrt(3)]/2 by [2+sqrt(3)]/2

we get [2-sqrt(3)]/[2+sqrt(3)]

rationalize the denominator

we get [2-sqrt(3)][2-sqrt(3)]/[2-sqrt(3)][2+sqrt(3)]

this gives us [4-4sqrt(3)+3]/[4-3] which simplifies to

[7-4sqrt(3)]

[7-4sqrt(3)]=7-4(1.7320508)=7-6.9282032=0.0717968

now we take the square root of 0.0717968

sqrt(0.0717968)=0.26794919 which we got before

tan(pi/12)=0.26794919

cos45=1/√2

1-2sin^{2}22.5=cos45=1/√2

sin^{2}22.5=(2-√2)/2

2tan(pi/12)/(1-tan^{2}(pi/12))=tan(pi/6)=1/√3

1-tan^{2}(pi/12)=2√3tan(pi/12)

tan^{2}(pi/12)+2√3tan(pi/12)-1=0

Already have an account? Log in

By signing up, I agree to Wyzant’s terms of use and privacy policy.

Or

To present the tutors that are the best fit for you, we’ll need your ZIP code.

Your Facebook email address is associated with a Wyzant tutor account. Please use a different email address to create a new student account.

Good news! It looks like you already have an account registered with the email address **you provided**.

It looks like this is your first time here. Welcome!

To present the tutors that are the best fit for you, we’ll need your ZIP code.

Please try again, our system had a problem processing your request.