sin2x / (1-cos x) = (1-cos2x) / (1-cos x) = (1-cos x)(1+cos x) / (1-cos x) = (1+cos x)
sin2x=1-cos2x as sin2x+cos2x=1
1-cos2x = (1-cos x)(1+cos x) as a2-b2=(a-b)(a+b)
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Cos(360 - X) = Trig. Identity Cos(360)Cos(x) + Sin(360)Sin(x) => 1CosX + 0Sinx => CosX + o => CosX
Remember SecX = 1/CosX Substitute SinX X 1 /CosX = SinX / CosX = TanX
Given: = cosx - sinxJust isolate sx= sx(co-in)
I suggest you convert everything to sines and cosines, and then try to simplify. For example, sec x = 1 / cos x, tan x = sin x / cos x, etc. Then - depending on the problem requirements - you either verify whether they are always equal or not, or determine for what values of x they are equal.
sin 2x + cos x = 0 (substitute 2sin x cos x for sin 2x)2sin x cos x + cos x = 0 (divide by cos x each term to both sides)2sin x + 1 = 0 (subtract 1 to both sides)2sin x = -1 (divide by 2 to both sides)sin x = -1/2Because the period of the sine function is 360⁰, first find all solutions in [0, 360⁰].Because sin 30⁰ = 1/2 , the solutions of sin x = -1/2 in [0, 360] arex = 180⁰ + 30⁰ = 210⁰ (the sine is negative in the third quadrant)x = 360⁰ - 30⁰ = 330⁰ (the sine is negative in the fourth quadrant)Thus, the solutions of the equation are given byx = 210⁰ + 360⁰n and x = 330⁰ + 360⁰n, where n is any integer.