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1/sin(x) is also known as cosec(x).
It looks a bit like a U, starting at "infinity" when x = 0, bottoming out at 1 when x = pi/2 radians and then returning to "infinity" at x = pi. Next, it is an upside down U, below the axis and peaking at -1: between x = pi and 2*pi. These U shapes alternate.
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Sec x times sin x divided by tan x?
1 (sec x)(sin x /tan x = (1/cos x)(sin x)/tan x = (sin x/cos x)/tan x) = tan x/tan x = 1
Integral of sin squared x?
First we look at the double-angle identity of cos2x. We know that: cos2x = cos^2x - sin^2x cos2x = [1-sin^2x] - sin^2x.............. (From sin^2x + cos^2x = 1, cos^2x = 1 - sin^2x) Therefore: cos2x = 1 - 2sin^2x 2sin^2x = 1 - cos2x sin^2x = 1/2(1-cos2x) sin^2x = 1/2 - cos2x/2 And intergrating, we get: x/2 - sin2x/4 + c...................(Integral of cos2x = 1/2sin2x; and c is a constant)
Solution for tan x plus cot x divided by sec x csc x?
(tan x + cot x)/sec x . csc x The key to solve this question is to turn tan x, cot x, sec x, csc x into the simpler form. Remember that tan x = sin x / cos x, cot x = 1/tan x, sec x = 1/cos x, csc x = 1/sin x The solution is: [(sin x / cos x)+(cos x / sin x)] / (1/cos x . 1/sin x) [(sin x . sin x + cos x . cos x) / (sin x . cos x)] (1/sin x cos x) [(sin x . sin x + cos x . cos x) / (sin x . cos x)] (sin x . cos x) then sin x. sin x + cos x . cos x sin2x+cos2x =1 The answer is 1.
2 sin squared x plus sin x minus 1 is equal to 0 solve for x?
2 sin2(x) + sin(x) - 1 = 0(2 sin + 1) (sin - 1) = 0Either 2 sin(x) + 1 = 02sin(x) = -1sin(x) = -0.5x = 210°, 330°or sin(x) - 1 = 0sin(x) = 1x = 90°
What is the simplest form for tanx divided by secx?
Need the fundamental identities here. tan(X) = sin(X)/cos(X) sec(X) = 1/cos(X) so tan(X)/sec(X) same as, sin(X)/cos(X) * cos(X)/1 cancel the cos(X) = sin(X) ---------------simplest form
What is the sin of pi divided by 2?
sin (pi/2) = 1
Verify that sin minus cos plus 1 divided by sin plus cos subtract 1 equals sin plus 1 divided by cos?
[sin - cos + 1]/[sin + cos - 1] = [sin + 1]/cosiff [sin - cos + 1]*cos = [sin + 1]*[sin + cos - 1]iff sin*cos - cos^2 + cos = sin^2 + sin*cos - sin + sin + cos - 1iff -cos^2 = sin^2 - 11 = sin^2 + cos^2, which is true,
What does the graph of sin squared x look like?
cos(2x) = 1 - 2(sin(x)^2), so sin(x)^2 = 1/2 - 1/2*cos(2x).
What is the sin of pi divided by 6?
1/2
How do you solve sinx divided by 1 plus cosx plus cosx divided by sinx?
sin x/(1+cos x) + cos x / sin x Multiply by sin x (1+cos x) =[(sin^2 x + cos x(1+cos x) ] / sin x (1+cos x) = [(sin^2 x + cos x + cos^2 x) ] / sin x (1+cos x) sin^2 x + cos^2 x = 1 = (1+cos x) / sin x (1+cos x) = 1/sin x
What is the derivative of cos x sin x divided by cos x sin x?
(cos x sin x) / (cos x sin x) = 1. The derivative of a constant, such as 1, is zero.
How do you calculate the limit of e5x -1 divided by sin x as x approaches 0?
You can use the L'hopital's rule to calculate the limit of e5x -1 divided by sin x as x approaches 0.
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
What does the product of a number divided by 5 and 2 is 1 look like?
in mathematics
What is sin pie divided by six equals one half?
Sin(pi/6) = 1/2 is a true statement [not pie].
What is the amplitude for y equals sin x divided by 2?
Assuming the question refers to [sin(x)]/2 rather than sin(x/2) the answer is 1.
How do you simplify sin theta times csc theta divided by tan theta?
Since sin(theta) = 1/cosec(theta) the first two terms simply camcel out and you are left with 1 divided by tan(theta), which is cot(theta).
