How do you use the angle sum or difference identity to find the exact value of #sin((23pi)/12)#?

Question

5229 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2016-09-02T02:09:56

#sin((23pi)/12) =(sqrt(2) - sqrt(6))/4#

Let's start by converting to degrees, which are commonly easier to work with. Use the conversion factor #180/pi#.

#(23pi)/12 xx 180/pi = (23 xx 180)/12 = 345˚#

Now, #345˚# can be written as #300˚ + 45˚#, or for the sake of the problem, #sin(300˚ + 45˚)#.

We must next expand this using the sum formula #sin(A + B) = sinAcosB + cosAsinB#

Write #sin(300˚ + 45˚)# in this form and evaluate.

#sin(345˚)#

#=> sin(300˚ + 45˚) #

#=> sin300˚cos45˚ + cos300˚sin45˚#

#=>-sqrt(3)/2 xx 1/sqrt(2) + 1/2 xx 1/sqrt(2)#

#=>-sqrt(3)/(2sqrt(2)) + 1/(2sqrt(2))#

#=> (1 - sqrt(3))/(2sqrt(2))#

Make sure to rationalize the denominator:

#=> (1 - sqrt(3))/(2sqrt(2)) xx (2sqrt(2))/(2sqrt(2))#

#=>(2sqrt(2) - 2sqrt(6))/(4(2))#

#=>(2sqrt(2) - 2sqrt(6))/8#

#=>(sqrt(2) - sqrt(6))/4#

Hopefully this helps!