So, we want the limit of (sin2(x))/x as x approaches 0. We can use L'Hopital's Rule: If you haven't learned derivatives yet, please send me a message and I will both provide you with a different way to solve this problem and teach you derivatives!
Using L'Hopital's Rule yields: the limit of (sin2(x))/x as x approaches 0=the limit of (2sinxcosx)/1 as x approaches zero. Plugging in, we, get that the limit is 2sin(0)cos(0)/1=2(0)(1)=0. So the original limit in question is zero.
No, they do not.
Because infinity is not a umber, it is usually not treated as a number when computing functions. Instead, you can look for a limit of a function as it approaches infinity. For example, the limit as x approaches infinity of 1/x is 0. Because sine oscillates, it's value constantly moves up and down, and it's value as it approaches infinity is not defined because it does not converge on any one number, as some other functions (like 1/x) do.
It is 1.
Yes, sine is a trig function, it is opposite over hypotenuse.
The answer depends on what information you do have. For instance, if you have the sine, the cosecant is simply 1 over the sine. Formally, the cosecant is hypotenuse over opposite.
Cosine squared theta = 1 + Sine squared theta
No, they do not.
Your question is insufficiently precise, but I'll try to answer anyway. "Sine squared theta" usually means "the value of the sine of theta, quantity squared". "Sine theta squared" usually means "the value of the sine of the quantity theta*theta". The two are not at all the same.
Because infinity is not a umber, it is usually not treated as a number when computing functions. Instead, you can look for a limit of a function as it approaches infinity. For example, the limit as x approaches infinity of 1/x is 0. Because sine oscillates, it's value constantly moves up and down, and it's value as it approaches infinity is not defined because it does not converge on any one number, as some other functions (like 1/x) do.
6,561 (i solved it by using this sentence: (9x9) x (9x9)= 81x81=6,561
Sine does not converge but oscillates. As a result sine does not tend to a limit as its argument tends to infinity. So sine(infinity) is not defined.
It is 1.
Answer 1 Put simply, sine squared is sinX x sinX. However, sine is a function, so the real question must be 'what is sinx squared' or 'what is sin squared x': 'Sin(x) squared' would be sin(x^2), i.e. the 'x' is squared before performing the function sin. 'Sin squared x' would be sin^2(x) i.e. sin squared times sin squared: sin(x) x sin(x). This can also be written as (sinx)^2 but means exactly the same. Answer 2 Sine squared is sin^2(x). If the power was placed like this sin(x)^2, then the X is what is being squared. If it's sin^2(x) it's telling you they want sin(x) times sin(x).
.5(x-sin(x)cos(x))+c
Yes, sine is a trig function, it is opposite over hypotenuse.
its short for sine. theres sine, cosine, and tangent. sine is opposite over adjacent for the sides of a triangle (or angles)
To determine what negative sine squared plus cosine squared is equal to, start with the primary trigonometric identity, which is based on the pythagorean theorem...sin2(theta) + cos2(theta) = 1... and then solve for the question...cos2(theta) = 1 - sin2(theta)2 cos2(theta) = 1 - sin2(theta) + cos2(theta)2 cos2(theta) - 1 = - sin2(theta) + cos2(theta)