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i think the examiners would prefer it if you just played with one side instead of both LHS & RHSacmilan said:(1+cosA-sinA) / (1+cosA+sinA) = cosA/(1+sinA)
If that is true, then:
(1+cosA-sinA)(1+sinA) = cosA(1+cosA+sinA)
1+cosA-sinA + sinA+sinAcosA-sin2A = cosA+cos2+sinAcosA
1-sin2 = cos2
sin2+cos2 = 1
Which we know to be true, hence the original equation is true.
Note: Theres probably a more formal way to do it, i just felt like an algebra bash![]()
ok ok, i said 'think', wasn't implying you were wrongacmilan said:Why? It doesnt say anything like that, and you often get questions where you have to play with both sides to get to some sort of equality. They cant drop marks for using a correct method unless they specify which method to use.
that will probably get u 0 in an examacmilan said:(1+cosA-sinA) / (1+cosA+sinA) = cosA/(1+sinA)
If that is true, then:
(1+cosA-sinA)(1+sinA) = cosA(1+cosA+sinA)
1+cosA-sinA + sinA+sinAcosA-sin2A = cosA+cos2+sinAcosA
1-sin2 = cos2
sin2+cos2 = 1
Which we know to be true, hence the original equation is true.
Note: Theres probably a more formal way to do it, i just felt like an algebra bash![]()
because he said "If that is true, then:"Slide Rule said:Why would it get 0, ngai? He did not use the question to 'prove' the question, after all.
I've never fully understood the issues with the direction of a proof. It seems to be an issue of semantics 'cause both are logically saying the same thing (aren't they?):ngai said:because he said "If that is true, then:"
^ yes the idea is similar, but not the same... your examples play on the fact that the parameters of the equations are not set, so it's not really what i was talking about before - in fact, your statements themselves are not even complete. you have not defined what the symbols "theta", "x", or "y" mean.Originally Posted by KFunk
i think I might get it. Is it the difference between say:
'a' represents sin2θ+cos2θ = 1
'b' represents statement x2 + y2 = 1
where if a is true then not(a) is untrue but if b is true then that does not logically imply that not(b) is untrue?
the logical slip is between "p implies q" and "q is true, ∴ p is true."; you needed to have also stated that "q conversely implies p". but because you didn't, the case is not complete.Case 1: Assume p. p implies q. q is true, ∴ p is true. (as acmillan did)