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Q: What is sin theta cos theta?

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Remember that tan = sin/cos. So your expression is sin/cos times cos. That's sin(theta).

'csc' = 1/sin'tan' = sin/cosSo it must follow that(cos) (csc) / (tan) = (cos) (1/sin)/(sin/cos) = (cos) (1/sin) (cos/sin) = (cos/sin)2

(Sin theta + cos theta)^n= sin n theta + cos n theta

You can use the Pythagorean identity to solve this:(sin theta) squared + (cos theta) squared = 1.

because sin(2x) = 2sin(x)cos(x)

The fourth Across the quadrants sin theta and cos theta vary: sin theta: + + - - cos theta: + - - + So for sin theta < 0, it's the third or fourth quadrant And for cos theta > 0 , it's the first or fourth quadrant. So for sin theta < 0 and cos theta > 0 it's the fourth quadrant

The derivative of (sin (theta))^.5 is (cos(theta))/(2sin(theta))

The equation cannot be proved because of the scattered parts.

sin/cos

Let 'theta' = A [as 'A' is easier to type] sec A - 1/(sec A) = 1/(cos A) - cos A = (1 - cos^2 A)/(cos A) = (sin^2 A)/(cos A) = (tan A)*(sin A) Then you can swap back the 'A' with theta

(/) = theta sin 2(/) = 2sin(/)cos(/)

For such simplifications, it is usually convenient to convert any trigonometric function that is not sine or cosine, into sine or cosine. In this case, you have: sin theta / sec theta = sin theta / (1/cos theta) = sin theta cos theta.

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