This is for my trig class. Thanks for the help!

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A) sin 105 degrees using the addition identity

sin 105=sin(45+60))

=sin45cos60+cos45sin60

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

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

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

multiply numerator and denominator by sqrt[2]

=(sqrt[2]+sqrt[6])/4

if you want another answer we have (1.414213+2.449489)/4=3.863702/4

=0.965925

B)tan 15 degrees using the subtraction identity

tan15=tan(60-45)

=(tan60-tan45)/(1+tan60tan45)

try to finish up the problem; if no other tutor does the problem I will finish this later

ok, let's finish the problem

tan15=tan(60-45)

=(tan60-tan45)/(1+tan60tan45)

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

multiply numerator and denominator by (sqrt[3]-1)

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

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

=(4-2sqrt[3])/2

=2-sqrt[3]

or 0.267949

You can use the angle difference formula for sin 15°

sin 15° = sin(60° - 45°) = sin 60° cos 45° - cos 60° sin 45° = (√6 - √2) / 4

cos 15° = cos(60° - 45°) = cos 60° cos 45° + sin 60° sin 45° = (√6 + √2) / 4

tan 15° = sin 15°/cos 15°) = = (√6 - √2)/(√6 + √2) = 2 - √3

sin 105° = sin 75° = cos 15° = (√6 + √2) / 4

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