Prove that
sin3A+sin2A-sinA = 4sinAcosA/2cos3A/2
To prove the equation sin3A + sin2A - sinA = 4sinAcos(A/2)cos(3A/2), we will start by manipulating the left side of the equation:
sin3A + sin2A - sinA
We'll use the double angle formula for sine to rewrite sin2A as 2sinAcosA, and the triple angle formula for sine to rewrite sin3A as 3sinA - 4sin^3(A):
3sinA - 4sin^3(A) + 2sinAcosA - sinA
Next, we can factor out sinA from the first two terms:
sinA(3 - 4sin^2(A)) + 2sinAcosA - sinA
Using the Pythagorean identity sin^2(A) + cos^2(A) = 1, we can rewrite the expression (3 - 4sin^2(A)) as:
3 - 4(1 - cos^2(A))
Simplifying further:
3 - 4 + 4cos^2(A) = 4cos^2(A) - 1
Now we have:
sinA(4cos^2(A) - 1) + 2sinAcosA - sinA
Factoring out sinA from the expression:
sinA(4cos^2(A) - 1 + 2cosA - 1)
Combining like terms:
sinA(4cos^2(A) + 2cosA - 2)
Factoring out a 2 from the second two terms:
2sinA(2cos^2(A) + cosA - 1)
Now, using the double angle formula for cosine, we can rewrite 2cos^2(A) + cosA - 1 as cos(2A):
2sinAcos(2A)
Finally, the left side of the equation is equal to:
2sinAcos(2A)
Therefore, we have proven the equation:
sin3A + sin2A - sinA = 4sinAcos(A/2)cos(3A/2)