a+b+c=180,show that sina+sinb+sinc=4cosa/2cosb/2cosc/2
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https://socratic.org/questions/if-a-b-c-pi-then-prove-that-sina-sinb-sinc-4cos-a-2-cos-b-2-cos-c-2
To show that sin(a) + sin(b) + sin(c) = 4cos(a/2)cos(b/2)cos(c/2), we can use the sum-to-product trigonometric identities.
First, let's review the sum-to-product identities:
1. sin(x) + sin(y) = 2sin((x + y)/2)cos((x - y)/2)
This identity relates the sum of sine functions to the product of a sine function and a cosine function.
2. cos(x) + cos(y) = 2cos((x + y)/2)cos((x - y)/2)
This identity relates the sum of cosine functions to the product of cosine functions.
Now, let's proceed with the proof:
Given: a + b + c = 180
We can rewrite this equation as: a = 180 - (b + c)
Now substitute this value of a in sin(a):
sin(a) = sin(180 - (b + c))
We know that sin(180 - θ) = sin(θ), so we have:
sin(a) = sin(b + c)
Using the sum-to-product identity (identity 1) mentioned above, we can write:
sin(a) + sin(b) + sin(c) = 2sin((b + c)/2)cos((b - c)/2) + sin(b) + sin(c)
Next, let's rewrite the cos((b - c)/2) term:
cos((b - c)/2) = sqrt((1 + cos(b - c))/2) (using the half-angle identity for cosine)
Now, we can substitute this value in the equation:
sin(a) + sin(b) + sin(c) = 2sin((b + c)/2)sqrt((1 + cos(b - c))/2) + sin(b) + sin(c)
We can further simplify the equation:
sin(a) + sin(b) + sin(c) = 2sin((b + c)/2)(sqrt(1 + cos(b - c))/sqrt(2)) + sin(b) + sin(c)
Now, let's focus on the terms involving the half angle identities:
2sin((b + c)/2)(sqrt(1 + cos(b - c))/sqrt(2))
= 2sin((b + c)/2)(sqrt((1 + cos(b - c))/2)/sqrt((1 + cos(b - c))/2))
= 2sqrt((1 + cos(b - c))/2) * sin((b + c)/2)
Using the double-angle identity for sine: sin(2θ) = 2sin(θ)cos(θ), we can rewrite sin((b + c)/2) as:
sin((b + c)/2) = 2sin((b + c)/4)cos((b + c)/4)
Substituting this value in the equation:
sin(a) + sin(b) + sin(c) = 2sqrt((1 + cos(b - c))/2) * 2sin((b + c)/4)cos((b + c)/4) + sin(b) + sin(c)
The next step is to express cos((b + c)/4) in terms of cos(a/2), cos(b/2), and cos(c/2). We can use the half-angle identity for cosine:
cos(θ/2) = sqrt((1 + cos(θ))/2)
cos((b + c)/4) = sqrt((1 + cos((b + c)/2))/2)
Similarly, we can express sin((b + c)/4) in terms of sin(a/2), sin(b/2), and sin(c/2). Using the half-angle identity for sine:
sin(θ/2) = sqrt((1 - cos(θ))/2)
sin((b + c)/4) = sqrt((1 - cos((b + c)/2))/2)
Now we can substitute these values in the equation:
sin(a) + sin(b) + sin(c) = 2sqrt((1 + cos(b - c))/2) * 2sqrt((1 - cos((b + c)/2))/2) * sqrt((1 + cos((b + c)/2))/2) + sin(b) + sin(c)
The sqrt((1 + cos((b + c)/2))/2) terms cancel out:
sin(a) + sin(b) + sin(c) = 2sqrt((1 + cos(b - c))/2) * 2sqrt((1 - cos((b + c)/2))/2) + sin(b) + sin(c)
Now let's simplify the remaining terms:
2sqrt((1 + cos(b - c))/2) * 2sqrt((1 - cos((b + c)/2))/2) + sin(b) + sin(c)
= 2sqrt(1 + cos(b - c)) * sqrt(1 - cos((b + c)/2)) + sin(b) + sin(c)
Now, let's apply the double-angle identity for cosine to cos((b + c)/2):
cos((b + c)/2) = 2cos^2((b + c)/4) - 1
Substituting this value in the equation:
2sqrt(1 + cos(b - c)) * sqrt(1 - cos((b + c)/2)) + sin(b) + sin(c)
= 2sqrt(1 + cos(b - c)) * sqrt(1 - (2cos^2((b + c)/4) - 1)) + sin(b) + sin(c)
= 2sqrt(1 + cos(b - c)) * sqrt(2cos^2((b + c)/4)) + sin(b) + sin(c)
= 2cos((b + c)/4) * 2|cos((b + c)/4)| + sin(b) + sin(c) (removing the negative square root)
Using the identity |cos(x)| = cos(x), this simplifies to:
2cos((b + c)/4) * 2cos((b + c)/4) + sin(b) + sin(c)
= 4cos^2((b + c)/4) + sin(b) + sin(c)
Finally, let's apply the double-angle identity for cosine:
cos(2θ) = 2cos^2(θ) - 1
cos(a/2) = 2cos^2((b + c)/4) - 1
Substituting this value in the equation:
4cos^2((b + c)/4) + sin(b) + sin(c)
= 4cos(a/2) + sin(b) + sin(c)
Therefore, we have shown that sin(a) + sin(b) + sin(c) = 4cos(a/2)cos(b/2)cos(c/2).