How do you find the exact values of sin 15 degrees using the half angle formula?
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"Suppose that I don't have a formula for #g(x)# but I know that #g(1)
= 3# and #g'(x) = sqrt(x^2+15)# for all x. How do I use a linear approximation to estimate #g(0.9)# and #g(1.1)#?"
I found: #sin(15°)=0.258#
Using the Half Angle Formula:
#color(red)(sin^2(x)=1/2[1-cos(2x)])#
with #x=15°# and #2x=30°#
you get:
#sin^2(15°)=1/2[1-cos(30°)]#
knowing that: #cos(30°)=sqrt(3)/2#:
#sin^2(15°)=1/2[1-sqrt(3)/2]#
#sin^2(15°)=(2-sqrt(3))/4=0.067#
So:
#sin(15°)=+-sqrt(0.067)=+-0.258#
We choose the positive one.
#sin(15^@)=sqrt(2-sqrt(3))/2=(sqrt(6)-sqrt(2))/4#
See below.
Find exact value of #sin(15^@)# with half-angle formula.
Consider the half-angle formula for sine: #sin(theta/2)=sqrt((1-cosx)/2)#
Since we know that 15 is half of 30, we can plug #30^@# in as #theta# and simplify:
#sin(15^@)=sin(30^@/2)=sqrt((1-cos(30^@))/2)#
#=sqrt((1-sqrt(3)/2)/2)#
#=sqrt(((2-sqrt(3))/2)/2)#
#=sqrt((2-sqrt(3))/4)#
or see slightly more advanced method to remove nested root (at the bottom)
#=sqrt(2-sqrt(3))/2#
which is our answer, but has a nested root
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Removing nested root from answer:
Let us multiply both the numerator and denominator inside the square root by #2#:
#=sqrt((2-sqrt(3))/4*2/2)#
#=sqrt((4-2sqrt(3))/8)#
Now we can write #4-2sqrt(3)# as a square in the numerator:
#=sqrt((sqrt(3)-1)^2/8)#
We can take out a #sqrt(3)-1# from the numerator and a #2# from the denominator:
#=((sqrt(3)-1)/2)*1/sqrt(2)#
#=(sqrt(3)-1)/(2sqrt(2))#
We can rationalize the denominator:
#=(sqrt(2)(sqrt(3)-1))/4#
#=(sqrt(6)-sqrt(2))/4#
which is our answer without a nested root.
Use
#sin(A-B) = sinA*cosB - SinB*cosA#
So
#sin 15 = sin(45-30) = sin45*cos30 - sin30*cos45#
We know
#sin45 = cos45 = (sqrt2)/2#
We also know
#cos30 = (sqrt3)/2#
#sin30 = 1/2#
Plugging in the values, we get
#sin15 = (sqrt2)/2*((sqrt3)/2-1/2)#
Simplifying, we get
#(sqrt6-sqrt2)/4#
