clintmyster
Prophet 9 FTW
- Joined
- Nov 12, 2007
- Messages
- 1,067
- Gender
- Male
- HSC
- 2009
- Uni Grad
- 2015
you mean replace t now with t=7.5?lyounamu said:Now you put that back in and factorise. You should get that.
you mean replace t now with t=7.5?lyounamu said:Now you put that back in and factorise. You should get that.
the top should be like 4(tan 7.5)(1-tan^2(7.5)) right? can you guide me through it please? do you change the tan^2(7.5) to like sec^2(7.5) - 1 and proceed from there?lyounamu said:Yeah. On top, you should get cos^2(7.5) because 1-sin^2(x) = cos^2(x)
And on the bottom you should have sec^(x) because 1+tan^2(x) = sec^2(x).
the question is 2tan15/(1+tan^2 15) , and sin@ = 2t/(1+t^2), where t = tan @/2clintmyster said:i know that but how do you use that in this scenario..i dont see many ways of making this in terms of sin unless you convert tan...is this question supposed to be hard? its supposedly the "easier" questions in yr12 3u!
I will do it again. I am a bit confused now. Because I am doing two questions simultaneously (from another thread).clintmyster said:the top should be like 4(tan 7.5)(1-tan^2(7.5)) right? can you guide me through it please? do you change the tan^2(7.5) to like sec^2(7.5) - 1 and proceed from there?
goshh do each question need this kind of working out? its huge!
Rather than making t = tan (7.5) let t = tan 15.clintmyster said:i know that but how do you use that in this scenario..i dont see many ways of making this in terms of sin unless you convert tan...is this question supposed to be hard? its supposedly the "easier" questions in yr12 3u!
I thought this question was supposed to be "prove". Damn! Why did I even go further when I didn't need to?lolokay said:the question is 2tan15/(1+tan^2 15) , and sin@ = 2t/(1+t^2), where t = tan @/2
so let tan @/2 = tan15, @ = 30
therefore 2tan15/(1+tan^2 15) = sin30 = 1/2
this is definitely an easy question since its just sub'ing into the t formula for sin@
what im still not getting is why did you choose to sub the angle in sin @...why not cos? how did that come about?lolokay said:the question is 2tan15/(1+tan^2 15) , and sin@ = 2t/(1+t^2), where t = tan @/2
so let tan @/2 = tan15, @ = 30
therefore 2tan15/(1+tan^2 15) = sin30 = 1/2
this is definitely an easy question since its just sub'ing into the t formula for sin@
If you look at the question closely, it resemblesclintmyster said:what im still not getting is why did you choose to sub the angle in sin @...why not cos? how did that come about?
oh that makes a lot of sense now! does that mean for like 4f) it resembles tan and 3b,d) sine again?lyounamu said:If you look at the question closely, it resembles
2t/(1+t^2)
where t = tan15 (which is tan 30/2)
Therefore, the answer is basically sin 30.
Yeah. I thought they were "prove" questions. Otherwise, they are just too simple that you only write like 2 lines and finish.clintmyster said:oh that makes a lot of sense now! does that mean for like 4f) it resembles tan and 3b,d) sine again?
well, even if it was a prove question you could have made things easier by proving that sin@ = 2t/(1 + t^2). though wouldn't you be able to use the t identities in a proof?lyounamu said:I thought this question was supposed to be "prove". Damn! Why did I even go further when I didn't need to?
i got proving ones coming up but like how hard could they be? this is my very last question i gotta do. just steps to guide me through it will be enough i think?lyounamu said:Yeah. I thought they were "prove" questions. Otherwise, they are just too simple that you only write like 2 lines and finish.
on the lhs sub (1+t^2)/(2t) for cosec@ and 1-t^2/2t for cot@ and then get rid of the fractions by multiplying top and bottom by 2t. It should rearrange to (1+t)/(1-t)clintmyster said:
No. t=tan (1/2@), so if you have (1+t)/(1-t), you can sub in t=tan(1/2@) which gives you 1+tan(1/2@)/1-tan(1/2@)clintmyster said:i got the LHS alright but with the rhs do you sub in 2t/1-t^2 for tan1/2@?
Lol, ok.lolokay said:well, even if it was a prove question you could have made things easier by proving that sin@ = 2t/(1 + t^2). though wouldn't you be able to use the t identities in a proof?
clintmyster said:edit: nevermind i am realy acting stupid today not seeing simple things like that tan1/2@ becomes t omgg...this is like the only trig i dont like LOLL..its in my way of a perfect score grrr!
yeah i see all that now..s many mediocre mistakes in this...and its not even remotely challenging or hard to get my head around..maybe i dont feel like doing this or somethingmidifile said:No. t=tan (1/2@), so if you have (1+t)/(1-t), you can sub in t=tan(1/2@) which gives you 1+tan(1/2@)/1-tan(1/2@)
What you were doing was right you just did it for the wrong question Hope your brain feels better tomorrowlyounamu said:Lol, ok.
Well, my brain is not functioning well today. I had an exam today, that's probably why.
was it a math exam? howd you go?lyounamu said:Lol, ok.
Well, my brain is not functioning well today. I had an exam today, that's probably why.