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

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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