How do you prove Sinx is less than X?

limit of c/x as x becomes greater than zero is c/∞, which will eventually be 0. If the ratio of c/x becomes zero, then sin(x) is less than or equal to x for values of x greater than 0.

Is sin x greater than 1?

sin(bi) is imaginary, and so there is no well defined notion of a less than or greater than comparison to real numbers. sin(a + bi) is complex, and so there is no well defined notion of a less than or greater than comparison to real numbers. In any case, sin(x) is never greater than 1.

Is Sinx COSX X is linearly independent?

You proved that there were no non-trivial (ie, with at least one non-zero coefficient) linear combination of sin and cos identically zero on [0,2π]. That by definition means that {sin,cos} is linearly independent on [0,2π].

Why is sin xx less than 1?

Sin x and x are both 0 when x = 0. The derivative of x is 1 so the line steadily increases. The derivative of sin x is cos x. That derivative starts with 1 and steadily becomes less.

Is Sinx less than 1?

Keep in mind that sinx is always greater than or equal to zero and less than or equal to 1.

Why is sin x less than or equal to 1?

Simply put, the sine and cosine of an angle are the lengths of the opposite and adjacent sides of one vertex of a right triangle divided by the length of the hypotenuse. Since the hypotenuse of a right triangle is ALWAYS the longest side of the triangle the values of sine and cosine HAVE to be less than one.

How do you show Sinx and COSX are linearly independent?

a1cos(x)+a2sin(x)=θ(x)=0. If this linear combination has only the zero solution a1=a2=0, then the set {cos(x),sin(x)} is linearly independent. The equality (*) should be true for any values of x∈[−π,π].

Is Sinx COSX a vector space?

It is clear that the set {cosx,sinx} spans the vector space V by definition. Thus, it suffices to prove that {cosx,sinx} is a linearly independent set.

Why differentiation of Sinx is COSX?

So the derivative can be viewed as the slope of the tangent line. So for example at this point right over here, it looks like the slope of our tangent line should be zero. So our derivative function should be zero at that x value. And it is indeed the case that the derivative of sine of x is equal to cosine of x.

How do you simplify sin(cos−1 x)?

How do you simplify sin(cos−1 x)? Let’s draw a right triangle with an angle of a = cos−1(x). As we know cos(a) = x = x 1 we can label the adjacent leg as x and the hypotenuse as 1. The Pythagorean theorem then allows us to solve for the second leg as √1 −x2.

What does $$\\sin x$ mean in calculus?

In an introductory calculus course, $(\\sin x,\\cos x)$ is probably defined to be the point we reach after traveling $x$ units counterclockwise along the unit circle from $(1,0)$.$\\endgroup$

How do you find sin(cos^ (-1)(x)) using the Pythagorean theorem?

The Pythagorean theorem then allows us to solve for the second leg as sqrt (1-x^2). With this, we can now find sin (cos^ (-1) (x)) as the quotient of the opposite leg and the hypotenuse. sin (cos^ (-1) (x)) = sin (a) = sqrt (1-x^2)/1 = sqrt (1-x^2)