SinA ➕ 2cosA=1 prove that 2sinA ➖ cosA=2
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sin ϴ + 2 cos ϴ = 1
Squaring both the sides
(sin ϴ + 2 cos ϴ) ² = (1) ²
sin² ϴ + 4 cos² ϴ + 4 sin ϴ cos ϴ = 1
because sin² ϴ = 1 - cos² ϴ & cos² ϴ=1- sin² ϴ
So replacing sin² ϴ by 1 - cos² ϴ and cos² ϴ by 1- sin² ϴ
we get
1 - cos² ϴ + 4 ( 1 - sin² ϴ ) + 4sin ϴ cos ϴ = 1
1 - cos² ϴ + 4 – 4sin² ϴ + 4 sin ϴ cos ϴ = 1
5 – 1 = cos² ϴ +4sin² ϴ - 4 sin ϴ cos ϴ
or
( cos ϴ – 2 sin ϴ ) ² = 4
cos ϴ -2sin ϴ = ± 2 or simply 2 ignoring -2
I Hope It Will Help!
sin ϴ + 2 cos ϴ = 1
Squaring both the sides
(sin ϴ + 2 cos ϴ) ² = (1) ²
sin² ϴ + 4 cos² ϴ + 4 sin ϴ cos ϴ = 1
because sin² ϴ = 1 - cos² ϴ & cos² ϴ=1- sin² ϴ
So replacing sin² ϴ by 1 - cos² ϴ and cos² ϴ by 1- sin² ϴ
we get
1 - cos² ϴ + 4 ( 1 - sin² ϴ ) + 4sin ϴ cos ϴ = 1
1 - cos² ϴ + 4 – 4sin² ϴ + 4 sin ϴ cos ϴ = 1
5 – 1 = cos² ϴ +4sin² ϴ - 4 sin ϴ cos ϴ
or
( cos ϴ – 2 sin ϴ ) ² = 4
cos ϴ -2sin ϴ = ± 2 or simply 2 ignoring -2
I Hope It Will Help!
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