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July 30

Prove trigonometric identity

Prove that- ( cos²(x) + 4sin(x)-1 ) / ( cos²x+5sin(x)-5 ) = ( sin²(x) + sin(x) ) / ( -cos²x ) . Exclusive Editor _{Notify Me!} 14:34, 30 July 2022 (UTC)[reply]

This looks like a homework problem, but to get you started: Clear the fractions by multiplying both sides by the denominators. Multiply everything out, then it's a matter of repeatedly applying cos²x+sin²x=1. It is, of course, false, and therefore difficult to prove, if either of the denominators is 0. --RDBury (talk) 14:51, 30 July 2022 (UTC)[reply]

You should try to factorize the nominators and denominators. Ruslik_Zero 20:03, 31 July 2022 (UTC)[reply]

Only after getting rid of the cosines. --Lambiam 21:29, 31 July 2022 (UTC)[reply]

Plotting the output of two functions against each other in a 2D graph

This question is more technical than mathematical. I have two formulas, say (y z)/(x y) and ((y z)/(x + y - y z))/(x y) (for x, y, z in [0,1]). How do I represent the relation between the two as a line graph, such that the x axis represents the first formula and the y axis the second? (Say with both axes ranging from 0 to 10.) Can this perhaps be done with WolframAlpha? --Cubefox (talk) 23:13, 30 July 2022 (UTC)[reply]

Unless one of your formulas is injective, you generally don't expect there to be a clean relationship which can be graphed like this. You can make a scatter plot, though. I plotted the example you gave on Desmos using the command

\left[\left({\frac {\left(yz\right)}{\left(xy\right)}},{\frac {\left(yz\right)}{\left(xy\right)\left(x+y-yz\right)}}\right)\ \operatorname {for} \ x=\left[0,.1,...,1\right],\ y\ =\ \left[0,.1,...,1\right],\ z=\left[0,.1,...,1\right]\right]

.--2406:E003:812:5001:9D4:C84F:FDB4:AA8B (talk) 00:42, 31 July 2022 (UTC)[reply]

Where they are defined, these formulas define continuous mappings from

\mathbb {R} ^{3}

to

\mathbb {R} ,

from a higher to a lower dimension. By Brouwer's theorem of the invariance of dimension, such mappings cannot be injective. --Lambiam 06:49, 31 July 2022 (UTC)[reply]

Thank you both! --Cubefox (talk) 09:50, 1 August 2022 (UTC)[reply]