Rewrite sec x as 1/cos x. Then, sec x sin x = (1/cos x)(sin x) = sin x/cos x. By definition, this is equal to tan x.
Cos(x) = Sin(2x) Using angle-addition, we have Sin(a+b) = Sin(a)Cos(b) + Sin(b)Cos(a). From that, we see Sin(2x) = Sin(x)Cos(x)+Sin(x)Cos(x) = 2Sin(x)Cos(x) Cos(x) = 2Sin(x)Cos(x) If Cos(x) = 0, then the two sides are equal. This occurs at x= Pi/2 + nPi, where n is an integer and Pi is approximately 3.14. If Cos(x) doesn't equal 0, then we can divide it out. Then, 1 = 2 Sin(x) , or 1/2 = Sin(x) This occurs when x = Pi/6 or 5Pi/6, plus or minus any multiples of 2 Pi.
The easiest way to approach this problem is by rewriting the left hand side entirely in terms of sin and cos and then simplifying. To do so, use the fact that cot(x)=cos(x)/sin(x) to get that 2*cot(x)*sin(x)*cos(x)=2*cos(x)/sin(x)*sin(x)*cos(x)=2*cos(x)² Next, we will try to simplify the right hand side by factoring and utilizing the formula cos(x)²+sin(x)²=1 which implies that 1-sin(x)²=cos(x)² 2-2sin(x)²=2*(1-sin(x)²)=2*cos(x)² Since both sides can be simplified to equal the same thing, both sides must always be equal, and the equation 2*cot(x)*sin(x)*cos(x)=2-2sin(x)² must be an identity
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
(tan x - sin x)/(tan x sin x) = (tan x sin x)/(tan x + sin x)[sin x/cos x) - sin x]/[(sin x/cos x)sin x] =? [(sin x/cos x)sin x]/[sin x/cos x) + sin x][(sin x - sin x cos x)/cos x]/(sin2 x/cos x) =? (sin2 x/cos x)/[(sin x + sin x cos x)/cos x)(sin x - sin x cos x)/sin2 x =? sin2 x/(sin x + sin x cos x)[sin x(1 - cos x)]/sin2 x =? sin2 x/[sin x(1 + cos x)(1 - cos x)/sin x =? sin x/(1 + cos x)(1 - cos x)/sin x =? [(sin x)(1 - cos x)]/[(1 + cos x)(1 - cos x)](1 - cos x)/sin x =? [(sin x)(1 - cos x)]/[1 - cos2 x)(1 - cos x)/sin x =? [(sin x)(1 - cos x)]/[1 - (1 - sin2 x)](1 - cos x)/sin x =? [(sin x)(1 - cos x)]/sin2 x(1 - cos x)/sin x = (1 - cos x)/sin x True
The derivative of cos(x) equals -sin(x); therefore, the anti-derivative of -sin(x) equals cos(x).
Rewrite sec x as 1/cos x. Then, sec x sin x = (1/cos x)(sin x) = sin x/cos x. By definition, this is equal to tan x.
2 x cosine squared x -1 which also equals cos (2x)
(2 sin^2 x - 1)/(sin x - cos x) = sin x + cos x (sin^2 x + sin^2 x - 1)/(sin x - cos x) =? sin x + cos x [sin^2 x - (1 - sin^2 x)]/(sin x - cos x) =? sin x + cos x (sin^2 x - cos^2 x)/(sin x - cos x) =? sin x + cos x [(sin x - cos x)(sin x + cos x)]/(sin x - cos x) =? sin x + cos x sin x + cos x = sin x + cos x
You can't. tan x = sin x/cos x So sin x tan x = sin x (sin x/cos x) = sin^2 x/cos x.
Cos(x) = Sin(2x) Using angle-addition, we have Sin(a+b) = Sin(a)Cos(b) + Sin(b)Cos(a). From that, we see Sin(2x) = Sin(x)Cos(x)+Sin(x)Cos(x) = 2Sin(x)Cos(x) Cos(x) = 2Sin(x)Cos(x) If Cos(x) = 0, then the two sides are equal. This occurs at x= Pi/2 + nPi, where n is an integer and Pi is approximately 3.14. If Cos(x) doesn't equal 0, then we can divide it out. Then, 1 = 2 Sin(x) , or 1/2 = Sin(x) This occurs when x = Pi/6 or 5Pi/6, plus or minus any multiples of 2 Pi.
Sin[x] = Cos[x] + (1/3)
If x = sin θ and y = cos θ then: sin² θ + cos² θ = 1 → x² + y² = 1 → x² = 1 - y²
No. Tan(x)=Sin(x)/Cos(x) Sin(x)Tan(x)=Sin2(x)/Cos(x) Cos(x)Tan(x)=Sin(x)
It isn't. The derivate of sin x = cos x.It isn't. The derivate of sin x = cos x.It isn't. The derivate of sin x = cos x.It isn't. The derivate of sin x = cos x.
-1
The easiest way to approach this problem is by rewriting the left hand side entirely in terms of sin and cos and then simplifying. To do so, use the fact that cot(x)=cos(x)/sin(x) to get that 2*cot(x)*sin(x)*cos(x)=2*cos(x)/sin(x)*sin(x)*cos(x)=2*cos(x)² Next, we will try to simplify the right hand side by factoring and utilizing the formula cos(x)²+sin(x)²=1 which implies that 1-sin(x)²=cos(x)² 2-2sin(x)²=2*(1-sin(x)²)=2*cos(x)² Since both sides can be simplified to equal the same thing, both sides must always be equal, and the equation 2*cot(x)*sin(x)*cos(x)=2-2sin(x)² must be an identity