To figure this out, graph y = sin x and y = cos2 x on the same piece of paper or graphing utility and find the points where the graphs intersect.
To determine this analytically, take the equation sin x = cos2 x and divide both sides by cos x to obtain the equivalent expression tan x = cos x. What this means is that the ratio of the opposite side of the triangle formed by angle x and the triangle's adjacent side is equal to the ratio of the triangle's adjacent side divided by its hypotenuse. This can be expressed by using the variables x and y to represent the opposite side and the adjacent side, respectively, by the expression
x2/y2 = y2/(x2 + y2)
where x2 + y2 equals the hypotenuse squared and x2/y2 equals the tangent squared. Now you can solve for either y or x by plugging in any number into the other (here we will solve for x by plugging 1 into y to obtain the expression
x2 = 1/(x2 + 1)
This can be rearranged to the form
x4 + x2 - 1 = 0
We can now solve for x2 using the quadratic formula.
[-1 + or - (1 + 4)1/2]/2
This equals negative one plus or minus the square root of five, all divided by two. Since we need to find the square root of x2, and negative numbers don't have real square roots, and negative one minus the square root of five divided by two is negative, we shall take the square root of negative one plus the square root of five divided by two, since this yields a real value. This approximately equals 0.6180. This, divided by our chosen y value, is the tangent of all angles for which the sine equals the cosine squared, and since our chosen y-value is 1, we can simply use a calculator's tan-1 function to find the value within the domain 0 x = 31.7175 + 180n where n is any integer, positive or negative.
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.
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.
(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
Because the slope of the curve of sin(x) is cos(x). Or, equivalently, the limit of sin(x) over x tends to cos(x) as x tends to zero.
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
Sin(x) cos(x) = 1/2 of sin(2x)
The derivative of cos(x) equals -sin(x); therefore, the anti-derivative of -sin(x) equals 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.It isn't. The derivate of sin x = cos x.
Yes. Multiplication is commutative for real and complex numbers. (meaning the order can be changed with the same result) so [cos(x)] * [sin(x)] = [sin(x)] * [cos(x)]
2
because sin(2x) = 2sin(x)cos(x)
(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
if tan x = cos x then sin x / cos x = cos x => sin x = cos x cos x => sin x = cos2 x => sin x = 1 - sin2x => sin2x + sin x - 1 = 0 Using the quadratic formula => 1. sin x = 0.61803398874989484820458683436564 => x = sin-1 (0.61803398874989484820458683436564) or => 2. sin x = -1.6180339887498948482045868343656 => x = sin-1 (-1.6180339887498948482045868343656)
Cos^2 x = 1 - sin^2 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.
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
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.