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 2 years ago
Question on Advanced function grade 12
Determine the value of "a" in the following equation :
2 tanx  tan2x + 2a = 1  tan2x ･ tan^2x
Hint: Double Angle Formula for Tangent is :
tan2x = 2tanx/1tan^2x
 2 years ago
Question on Advanced function grade 12 Determine the value of "a" in the following equation : 2 tanx  tan2x + 2a = 1  tan2x ･ tan^2x Hint: Double Angle Formula for Tangent is : tan2x = 2tanx/1tan^2x

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bahrom7893
 2 years ago
Best ResponseYou've already chosen the best response.1Hint: apply the double angle formula

alexeis_nicole
 2 years ago
Best ResponseYou've already chosen the best response.0would i matter which side i start first ? :S

satellite73
 2 years ago
Best ResponseYou've already chosen the best response.2i am confused why don't you just add \(\tan(2x)\) to both sides and get rid of it

satellite73
 2 years ago
Best ResponseYou've already chosen the best response.2oh because it is times doh

alexeis_nicole
 2 years ago
Best ResponseYou've already chosen the best response.0so wait... 2 tanx + 2a  1 = tan2x  tan2x tan^2x does that work?

alexeis_nicole
 2 years ago
Best ResponseYou've already chosen the best response.0if i bring tan2x to the right then bring 1 to the left side?

satellite73
 2 years ago
Best ResponseYou've already chosen the best response.2i think it is a bunch of algebra i can try it with pencil and paper if you like

alexeis_nicole
 2 years ago
Best ResponseYou've already chosen the best response.0yes please? i want to understand how to do it.. i have a test coming up

alexeis_nicole
 2 years ago
Best ResponseYou've already chosen the best response.0im trying to do it myself as well

satellite73
 2 years ago
Best ResponseYou've already chosen the best response.2ok i screwed up already, let me try it again

RadEn
 2 years ago
Best ResponseYou've already chosen the best response.0from : tan2x = 2tanx/(1tan^2 x) 2tanx = tan2x(1tan^2 x) 2tanx = tan2x  tan2x tan^2 x now, apply it to left side

satellite73
 2 years ago
Best ResponseYou've already chosen the best response.2this has nothing whatsoever to do with trig so fora as i can tell

satellite73
 2 years ago
Best ResponseYou've already chosen the best response.2if we replace \(\tan(x)\) by \(x\) you get \[2x\frac{2x}{1x^2}+2a=1\frac{2x^3}{1x^2}\] now we add up on the left and right get \[\frac{2x(1x^2)2x+2a(1x^2)}{1x^2}=\frac{1x^22x^3}{1x^2}\]

satellite73
 2 years ago
Best ResponseYou've already chosen the best response.2ignoring the denominators, you ge t \[2x2x^32x+2a2ax^2=1x^22x^3\] \[2a2ax^2=1x^2\]

satellite73
 2 years ago
Best ResponseYou've already chosen the best response.2and therefore \(a=\frac{1}{2}\)

alexeis_nicole
 2 years ago
Best ResponseYou've already chosen the best response.0wth! LOL that's the right answer... omg i was so off.... replacing tan(x) with "x" helped. thank you!

satellite73
 2 years ago
Best ResponseYou've already chosen the best response.2yeah that is a good trick to remember when you have a ton of annoying algebra to do it is good for all kinds of trig identities because almost all of the work is algebra, and you get too bogged down writing \(\sin(x)\) and \(\tan(x)\) when \(a\) or \(b\) would do
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