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y = x^3 sinx / 5cosx
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what about it ?
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Prove the following:
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LS = (1 + sinx)/(1 + 1/sinx) = (1 + sinx)/( ( sinx + 1)/sinx ) = (1 + sinx) ( sinx/(1+sinx)) = sinx
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2(sinx)^2-5cosx-4=0
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change sin^2x to 1-cos^2x, then 2-2cos^2x-5cosx-4=0 let u= cosx 2-2u^2-5u-4=0 2u^2+5u+2=0
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x^3(sinx)/ 5cosx
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To simplify the expression (x^3*sin(x))/(5*cos(x)), we can divide the numerator and denominator
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2(sinx)^2-5cosx-4=0
Solve for all solutions between [0,2pi]
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change sin^2x to 1-cos^2x, then 2-2cos^2x-5cosx-4=0 let u= cosx 2-2u^2-5u-4=0 2u^2+5u+2=0
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Mathematics - Trigonometric Identities - Reiny, Friday, November 9, 2007 at 10:30pm
(sinx - 1 -cos^2x) (sinx + 1 - cos^2x) should
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sin^2x - sin^4x = cos^2x - cos^4x
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Compute inverse functions to four significant digits.
cos^2x=3-5cosx rewrite it as.. cos^2x + 5cosx -3=0 now you have a
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To find the inverse function, let's first solve for cos(x) in the quadratic equation cos^2x + 5cosx
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the problem is
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although, after looking at this problem i'm still not sure why i had to change the original problem?
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Compute inverse functions to four significant digits.
cos^2x=3-5cosx cos^2x + 5cosx -3=0 now you have a quadratic, solve for cos
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You're welcome! I'm glad I could help you understand the solution to the problem. Yes, the answer is
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Hello! Can someone please check and see if I did this right? Thanks! :)
Directions: Find the exact solutions of the equation in
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that is not an answer. It is a restatement of the problem, using only sines. 2sin^2x-sinx-1 == 0
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tanx+secx=2cosx
(sinx/cosx)+ (1/cosx)=2cosx (sinx+1)/cosx =2cosx multiplying both sides by cosx sinx + 1 =2cos^2x sinx+1 =
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The mistake occurred when you simplified the equation (sinx+1)/cosx = 2cosx to sinx + 1 = 2cos^2x.
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