Prove if
sinA + 2cosA = 1, then prove that
2sinA - cosA = 2
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sin(A) = 1 - 2*cos(A)
Squaring both sides:
sin²(A) = 1 - 4*cos(A) + 4*cos²(A)
1 - cos²(A) = 1- 4*cos(A) + 4*cos²(A)
5*cos²(A) - 4*cos(A) = 0
cos(A)*(5*cos(A) - 4) = 0
cos(A) = 0 and cos(A) = 4/5
When cos(A) = 0, sin(A) = 1
2*sin(A) - cos(A) = 2(1) - 0 = 2
Squaring both sides:
sin²(A) = 1 - 4*cos(A) + 4*cos²(A)
1 - cos²(A) = 1- 4*cos(A) + 4*cos²(A)
5*cos²(A) - 4*cos(A) = 0
cos(A)*(5*cos(A) - 4) = 0
cos(A) = 0 and cos(A) = 4/5
When cos(A) = 0, sin(A) = 1
2*sin(A) - cos(A) = 2(1) - 0 = 2
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