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.
Yes. Except where sin x = 0, because then you would be dividing by zero so the quotient is undefined.
(cos x sin x) / (cos x sin x) = 1. The derivative of a constant, such as 1, is zero.
There will be 4 possible solutions if all you are looking for is the angles, so you will need to find out which quadrant your angle is in.2sin²x = sin x **Subtract sin x from both sides.2sin²x - sin x = 0 **Then factor out sin x.sin x(2sin x - 1) = 0 **Set each equal to zero. (AB=0 is the same as A=0 OR B=0).sin x = 0 or 2sin x - 1 = 0 to 2sin x = 1 to sin x = 1/2At this point all that is left to do is find out where sin x = 0 or 1/2, which is 0, 180 for sin x = 0 or 30, 150 for sin x = 1/2.
Zero
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.
Yes. Except where sin x = 0, because then you would be dividing by zero so the quotient is undefined.
(cos x sin x) / (cos x sin x) = 1. The derivative of a constant, such as 1, is zero.
Solve sin² x = sin x. The answer is x = 0 or π/2 or π. Here's how you can find this result: Let s = sin x. Then, we have, s² - s = 0; whence, s(s - 1) = 0. Now, when the product of two numbers is zero, then one or other of them must equal zero. Hence, either, s = 0 OR s - 1 = 0; that is, sin x = 1 or 0. For 0 ≤ x < 360°, x = 0° or 90° or 180°; those being the only angles whose sine equals either 1 or 0. In term of radian measure, x = 0 or π/2 or π.
It is equal to zero.
sin(x) = tan(x) when x equal 0
The derivative of cos(x) equals -sin(x); therefore, the anti-derivative of -sin(x) equals cos(x).
The limit as x approaches zero of sin(x) over x can be determined using the squeeze theorem.For 0 < x < pi/2, sin(x) < x < tan(x)Divide by sin(x), and you get 1 < x/sin(x) < tan(x)/sin(x)That is the same as 1 < x/sin(x) < 1/cos(x)But the limit as x approaches zero of 1/cos(x) is 1,so 1 < x/sin(x) < 1which means that the limit as x approaches zero of x over sin(x) is 1, and that also means the inverse; the limit as x approaches zero of sin(x) over x is 1.You can also solve this using deriviatives...The deriviative d/dxx is 1, at all points. The deriviative d/dxsin(x) at x=0 is also 1.This means you have the division of two functions, sin(x) and x, at a point where their slope is the same, so the limit reduces to 1 over 1, which is 1.
There will be 4 possible solutions if all you are looking for is the angles, so you will need to find out which quadrant your angle is in.2sin²x = sin x **Subtract sin x from both sides.2sin²x - sin x = 0 **Then factor out sin x.sin x(2sin x - 1) = 0 **Set each equal to zero. (AB=0 is the same as A=0 OR B=0).sin x = 0 or 2sin x - 1 = 0 to 2sin x = 1 to sin x = 1/2At this point all that is left to do is find out where sin x = 0 or 1/2, which is 0, 180 for sin x = 0 or 30, 150 for sin x = 1/2.
Note that an angle should always be specified - for example, 1 - cos square x. Due to the Pythagorean formula, this can be simplified as sin square x. Note that sin square x is a shortcut of (sin x) squared.
Any number to the power zero is equal to 1 - except zero to the power zero, which is undefined. So, if x is not equal to zero, the answer is 1.Any number to the power zero is equal to 1 - except zero to the power zero, which is undefined. So, if x is not equal to zero, the answer is 1.Any number to the power zero is equal to 1 - except zero to the power zero, which is undefined. So, if x is not equal to zero, the answer is 1.Any number to the power zero is equal to 1 - except zero to the power zero, which is undefined. So, if x is not equal to zero, the answer is 1.
sin x times sin x. or 1/cosec2(x) or 1 - cos2(x) or tan2(x)*cos2(x) etc, etc.