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## Use the sum or difference formulas to determine: A. sin105 degrees B. tan15 degrees

This is for my trig class. Thanks for the help!
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# 2 Answers

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