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sinθ = O/H
cosθ = A/H
tanθ = O/A
cosθ*tanθ = A/H * O/A = AO/AH = O/H
Therefore, sinθ = cosθtanθ = O/H
(Notes: O=opposite, A=adjacent, H=hypotenuse)
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because sin(2x) = 2sin(x)cos(x)
What is the exact value of cos theta if csc theta -4 with theta in quadrant III?
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If a cos theta plus b sin theta equals 8 and a sin theta - b cos theta equals 5 show that a squared plus b squared equals 89?
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How do you simplify tan theta cos theta?
Remember that tan = sin/cos. So your expression is sin/cos times cos. That's sin(theta).
How do you simplify cos theta times csc theta divided by tan 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
De Morgan's law in complex number?
(Sin theta + cos theta)^n= sin n theta + cos n theta
What is the identity for tan theta?
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What is sin theta cos theta?
It's 1/2 of sin(2 theta) .
In which quardant are the terminal arms of the angle lie when sin thita is less then zero and cos thita is greater then zero?
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
If cos and theta 0.65 what is the value of sin and theta?
You can use the Pythagorean identity to solve this:(sin theta) squared + (cos theta) squared = 1.
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because sin(2x) = 2sin(x)cos(x)
Prove that sin theta power 8 - cos theta power 8 equal to sin sq Theta -cos sq x 1-sin sq Theta cos sq theta?
The equation cannot be proved because of the scattered parts.
What is the derivative of sin under root theta?
The derivative of (sin (theta))^.5 is (cos(theta))/(2sin(theta))
What is sec theta - 1 over sec theta?
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
