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It's difficult for me to be sure what function you are asking about, because of the limitations of answers.com.
I am considering y = -1 + csc (x)
csc (x) = 1/sin(x) and sin is a periodic function of x.
Ignoring what happens for negative values of sin(x), csc(x) is at local minima for maximal values of sin(x), which occur at x = (2k+1/2)pi for i any integer.
Putting the -1 into -1 + csc(x) simply 'lowers' the function without changing the positions of these minima.
PS: Incidentally, using GeoGebra can be a big help in solving problems like these. Free. Easy to use.
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How do you verify the identity sinx cscx 1?
sinx cscx = 1 is the same thing as sinx(1/sinx) = 1 which is the same as sinx/sinx = 1. This evaluates to 1=1, which is true.
How do you take the derivative of a trig function?
Trig functions have their own special derivatives that you will have to memorize. For instance: the derivative of sinx is cosx. The derivative of cosx is -sinx The derivative of tanx is sec2x The derivative of cscx is -cscxcotx The derivative of secx is secxtanx The derivative of cotx is -csc2x
Parenthesis 1 plus tanx end parenthesis divided by sinx equals cscx plus secx?
(1 + tanx)/sinxMultiply by sinx/sinxsinx + tanxsinxDivide by sin2x (1/sin2x) = cscxcscx + tan(x)csc(x)tanx = sinx/cosx and cscx = 1/sinxcscx + (sinx/cosx)(1/sinx)sinx cancels outcscx + 1/cosx1/cosx = secxcscx + secx
Is 1 plus sinX divided by 1 plus cscX equal to sinX?
2
What is the anti-derivative of co secant x?
According to Wolfram Alpha, input:integral csc x it is -log[cot(x) + csc(x)] + constant You can verify this by taking the derivative of the purported integral.
What is the integral of cscx?
- ln (cscx + cotx) + C You use u substitution.
How do you verify the identity sinx cscx 1?
sinx cscx = 1 is the same thing as sinx(1/sinx) = 1 which is the same as sinx/sinx = 1. This evaluates to 1=1, which is true.
What is the derivative of cscx?
d/dx csc(x) = - csc(x) tan(x)
How do you find the derivative of - csc x - sin x?
d/dx (-cscx-sinx)=cscxcotx-cosx
How do you take the derivative of a trig function?
Trig functions have their own special derivatives that you will have to memorize. For instance: the derivative of sinx is cosx. The derivative of cosx is -sinx The derivative of tanx is sec2x The derivative of cscx is -cscxcotx The derivative of secx is secxtanx The derivative of cotx is -csc2x
How do you rewrite y equals cscx to graph it?
You could try y = 1/sin(x) but I do not see how that helps.
What is the derivative of -cscxcotx?
Using the u substitution method of derivation (selecting sinx as u and cosxdx as du), you get f'(x)=cscx.
Parenthesis 1 plus tanx end parenthesis divided by sinx equals cscx plus secx?
(1 + tanx)/sinxMultiply by sinx/sinxsinx + tanxsinxDivide by sin2x (1/sin2x) = cscxcscx + tan(x)csc(x)tanx = sinx/cosx and cscx = 1/sinxcscx + (sinx/cosx)(1/sinx)sinx cancels outcscx + 1/cosx1/cosx = secxcscx + secx
How do you prove that the derivative of csc x is equals to -csc x cot x?
d/dx cscx = d/dx 1/sinx = d/dx (sinx)-1= -(sinx)-2 cosx = -cosx/sin2x = -1/sinx.cosx/sinx = -cscx cotx I suggest you copy this out onto paper so it is more clear. The / signs make it harder to see what is happening compared to horizontal divide lines.
Is 1 plus sinX divided by 1 plus cscX equal to sinX?
2
What is the anti derivative of the square root of 1-x2?
-1
What is the anti-derivative of co secant x?
According to Wolfram Alpha, input:integral csc x it is -log[cot(x) + csc(x)] + constant You can verify this by taking the derivative of the purported integral.
