Prove that tan(x)sin(x) = sec(x)-cos(x) tan(x)sin(x) = [sin(x) / cos (x)] sin(x) = sin2(x) / cos(x) = [1-cos2(x)] / cos(x) = 1/cos(x) - cos2(x)/ cos(x) = sec(x)-cos(x) Q.E.D
cos2 x + sin2 x = 1 cos2 x = 1 - sin2 x
cos(x)-cos(x)sin2(x)=[cos(x)][1-sin2(x)]cos(x)-cos(x)sin2(x)=[cos(x)][cos2(x)]cos(x)-cos(x)sin2(x)=cos3(x)
If you mean: sin2(x) cos2(x) then it can be simplified by noting that the square of the sine of x is equal to (1 - cos(2x)) ÷ 2 and the square of the cosine of x is equal to (1 + cos(2x)) ÷ 2. We can then simplify further: sin(x)2cos(x)2 = [(1 - cos(2x)) / 2][(1 + cos(2x)) / 2] = (1 - cos(2x))(1 + cos(2x)) / 2 = (1 - cos2(2x)) / 2 Also note that 1 - cos2(x) = sin2(x), so we can then say: = sin2(2x) / 2
The deriviative of sin2 x + cos2 x is 2 cos x - 2 sin x
cos 2x = cos2 x - sin2 x = 2 cos2 x - 1; whence, cos 2x / cos x = 2 cos x - (1 / cos x) = 2 cos x - sec x.
Sin2(x)/Cos2(x) is an expression, not an equation. Because it is an expression, it cannot be solved. It can be transformed to other, equivalent expressions, but that is as far as you can go. So, Sin2(x)/Cos2(x) = [Sin(x)/Cos(x)]2 = Tan2x or [1/Cos2(x) - 1] or [Sec2(x) - 1]
Integral of [1/(sin x cos x) dx] (substitute sin2 x + cos2 x for 1)= Integral of [(sin2 x + cos2 x)/(sin x cos x) dx]= Integral of [sin2 x/(sin x cos x) dx] + Integral of [cos2 x/(sin x cos x) dx]= Integral of (sin x/cos x dx) + Integral of (cos x/sin x dx)= Integral of tan x dx + Integral of cot x dx= ln |sec x| + ln |sin x| + C
Prove that tan(x)sin(x) = sec(x)-cos(x) tan(x)sin(x) = [sin(x) / cos (x)] sin(x) = sin2(x) / cos(x) = [1-cos2(x)] / cos(x) = 1/cos(x) - cos2(x)/ cos(x) = sec(x)-cos(x) Q.E.D
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Cos(90 - x) = sin(x) so cos2(90 - x) = sin2(x)
cos2 x + sin2 x = 1 cos2 x = 1 - sin2 x
sin2 + cos2 = 1 So, (1 - 2*cos2)/(sin*cos) = (sin2 + cos2 - 2*cos2)/(sin*cos) = (sin2 - cos2)/(sin*cos) = sin2/(sin*cos) - cos2/(sin*cos) = sin/cos - cos-sin = tan - cot
cos x / (1-sin x) = cos x (1 + sin x) / (1 - sin x) (1 + sin x) = cos x (1 + sin x) / (1 - sin2x) = cos x (1 + sin x) / cos2 x = (1 + sin x) / cos x = sec x + tan xcos x / (1-sin x) = cos x (1 + sin x) / (1 - sin x) (1 + sin x) = cos x (1 + sin x) / (1 - sin2x) = cos x (1 + sin x) / cos2 x = (1 + sin x) / cos x = sec x + tan xcos x / (1-sin x) = cos x (1 + sin x) / (1 - sin x) (1 + sin x) = cos x (1 + sin x) / (1 - sin2x) = cos x (1 + sin x) / cos2 x = (1 + sin x) / cos x = sec x + tan xcos x / (1-sin x) = cos x (1 + sin x) / (1 - sin x) (1 + sin x) = cos x (1 + sin x) / (1 - sin2x) = cos x (1 + sin x) / cos2 x = (1 + sin x) / cos x = sec x + tan x
cos(x)-cos(x)sin2(x)=[cos(x)][1-sin2(x)]cos(x)-cos(x)sin2(x)=[cos(x)][cos2(x)]cos(x)-cos(x)sin2(x)=cos3(x)
If you mean: sin2(x) cos2(x) then it can be simplified by noting that the square of the sine of x is equal to (1 - cos(2x)) ÷ 2 and the square of the cosine of x is equal to (1 + cos(2x)) ÷ 2. We can then simplify further: sin(x)2cos(x)2 = [(1 - cos(2x)) / 2][(1 + cos(2x)) / 2] = (1 - cos(2x))(1 + cos(2x)) / 2 = (1 - cos2(2x)) / 2 Also note that 1 - cos2(x) = sin2(x), so we can then say: = sin2(2x) / 2
Use these identities: sin2(x) + cos2(x) = 1, and tan(x) = sin(x)/cos(x) For clarity, the functions are written here without their arguments (the "of x" part). (1 - sin2) = cos2 (1 + tan2) = (1 + sin2/cos2) = (cos2+sin2) / cos2 = 1/cos2 Multiply them: (cos2) times (1/cos2) = 1'QED'