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

Question

This is for my trig class, thanks for the help!

Answer 1

sin105 = sin(150-45) = sin150cos45 - sin45cos150

= (1/2)sqr2/2- sqr2/2(-sqr3/2) = (sqr2+sqr6)/4

= about .966

check the answer with a calculator: sin105=0.9659

tan15= about 0.268

tan15 = tan(30/2) = sin30/(1+cos30) =(1/2)/(1+sqr3/2) = (1/2)/(2+sqr3)/2)= 1/(2+sqr3) = 0.268

tan15 = tan(45-30) = (tan45-tan30)/(1+tan45tan30) = (1-1/sqr3)/(1+(1)(1/sqr3) = (sqr3-1)/(sqr3 +1) = 0.732/2.732 = 0.268

tan15 = tan(60-45) = (tan60-tan45)/(1+tan60tan45) = (sqr3-1)/(1+sqr3) = 0.268

Answer 2

In general the easy angles are 30, 45, 60 and 90. So sin 105 can be broken into 60 and 45.

 

Using the sum formula for sin:

sin (a+b) = sin a cos b + sin b cos a where a=60 and b=45

In this case cos 45 = sin 45= 1/√2.

cos 60 = 1/2

sin 60 = √3/2

 

so sin (60+45) = √3/2√2 + 1/2√2 = .6124 + .3535 = .965

 

tan 15 is (tan 45-30)  where a = 45 and b=30 in the following formula:

tan (a-b) = (tan a - tan b ) / (1+ tan a tan b)

Now tan 45 = 1 and tan 30 = 1/√3

so tan (45-30) = (1- 1/√3) / (1+1/√3) = .4226/1.577 = .2679