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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).

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Q: What is sine squared?

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(1 - cos(2x))/2, where x is the variable. And/Or, 1 - cos(x)^2, where x is the variable.

The anti derivative of negative sine is cosine.

a^2b^2c^2 ^2 is squared

3 squared = 9 4 squared = 16 9 x 16 = 144 3 squared x 4 squared = 144

sine graph will be formed at origine of graph and cosine graph is find on y-axise

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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.

6,561 (i solved it by using this sentence: (9x9) x (9x9)= 81x81=6,561

It is 1.

.5(x-sin(x)cos(x))+c

A sine wave is a mathematical function that describes a smooth repetitive oscillation which looks like a wave going from 1 to -1 and back to 1. A normal sine wave is much like a sine wave but has been normalized for practical uses like in electronics creating a "squared" sine wave A perfect sine wave does not exist in reality, it only exists in the minds of mathematicians.

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)

No.Remember: sin2 + cos2 = 1So, in place of (1 - 2 sin2) we can write (sin2 + cos2 - 2 sin2).Massage that around slightly: (sin2 + cos2 - 2 sin2) = cos2 - sin2That's not equal to (sin2 - cos2), which is the original question we were asked.

if that 144 is the peak voltage if its a sine wave the rms voltage is that voltage divided by sqrt(2) if not a sine wave (modified) you must find the area under the curve by integrating a cycle of that wave shape (root mean squared)

(1 - cos(2x))/2, where x is the variable. And/Or, 1 - cos(x)^2, where x is the variable.

The answer is 1. sin^2 x cos^2/sin^2 x 1/cos^2 cos^2 will be cancelled =1 sin^2 also will be cancelled=1 1/1 = 1

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