Prelim 2016 Maths Help Thread (3 Viewers)

[InteGrand]

so just switch the tan one around?

Yeah, cot(x+y) = 1/tan(x+y) = [1 - tan(x)tan(y)]/[tan(x) + tan(y)]. Now convert things to cot.

 

[Green Yoda]

Yeah, cot(x+y) = 1/tan(x+y) = [1 - tan(x)tan(y)]/[tan(x) + tan(y)]. Now convert things to cot.

Why is it that you switch the numerator and denominator around?
What I did was 1/(tan(x)+tan(y))/(1-tan(x)tan(y))

 

[InteGrand]

Why is it that you switch the numerator and denominator around?
What I did was 1/(tan(x)+tan(y))/(1-tan(x)tan(y))

This is because switching them around is the same thing. For fractions, 1/(a/b) = b/a.

 

[Green Yoda]

This is because switching them around is the same thing. For fractions, 1/(a/b) = b/a.

oh yes ofc..lol..what a brain fart
After than what do you mean by convert things to cot?

 

[InteGrand]

oh yes ofc..lol..what a brain fart
After than what do you mean by convert things to cot?

We have things like tan(x) and tan(y) now, but want the answer in terms of cot(x) and cot(y) instead. So we replace each tan with a 1/cot, and simplify the whole thing.

 

[Green Yoda]

We have things like tan(x) and tan(y) now, but want the answer in terms of cot(x) and cot(y) instead. So we replace each tan with a 1/cot, and simplify the whole thing.

ah ok thanks man ur a legend

 

[Green Yoda]

Prove:
(tan(2θ)-tan(θ))/tan(2θ)+cot(θ)=tan^2(θ)

I keep getting tan(θ) by simplifying LHS and not tan^2(θ)

 

[Paradoxica]

Prove:
(tan(2θ)-tan(θ))/tan(2θ)+cot(θ)=tan^2(θ)

I keep getting tan(θ) by simplifying LHS and not tan^2(θ)

Brackets. BRACKETS

 

[Green Yoda]

How did you get (tan^2(θ)(1+tan^2(θ))/(1+tan^2(θ)) at the end?
what I did was:
(2tan(θ)-tan(θ)+tan^3(θ))/(1+tan^2(θ)) to (tan(θ)+tan^3(θ))/(1+tan^2(θ))
and the simplified it to (tan(θ)(1+tan^2(θ))/(1-tan^2(θ)) which = tan(θ)

 

[parad0xica]

How did you get (tan^2(θ)(1+tan^2(θ))/(1+tan^2(θ)) at the end?

 

[eyeseeyou]

In maths induction there are 4 steps

1. Show true for n=1 (or whenever it starts)
2. Assume true for n=k
3. Show true for n=k+1. ALWAYS involve assumption
4. Conclusion

WHy do we need these 4 steps?

 

[InteGrand]

In maths induction there are 4 steps

1. Show true for n=1 (or whenever it starts)
2. Assume true for n=k
3. Show true for n=k+1. ALWAYS involve assumption
4. Conclusion

WHy do we need these 4 steps?

The fourth one is only to avoid losing cheap marks from the markers (in other words, "because the HSC says so"). The first three are the ones needed for a mathematically valid induction proof of that kind.

 

[Green Yoda]

simplify:
(sin(θ)+sin(θ/2))/(1+cos(θ)+cos(θ/2))

 

[eyeseeyou]

HELP ME PLZ

Find the acute angle between 2x+y-3=0 and x+1=0

 

[InteGrand]

HELP ME PLZ

Find the acute angle between 2x+y-3=0 and x+1=0

Use the (acute) angle between two lines formula. Have you learnt this yet? It is:



provided that m₁m₂ is not equal to -1. (m₁ and m₂ refer to the slopes of the two lines. Obviously if m₁m₂ were equal to -1, the lines would be perpendicular, and the angle between them would be 90 deg.)

 

[eyeseeyou]

Use the (acute) angle between two lines formula. Have you learnt this yet? It is:



provided that m₁m₂ is not equal to -1. (m₁ and m₂ refer to the slopes of the two lines. Obviously if m₁m₂ were equal to -1, the lines would be perpendicular, and the angle between them would be 90 deg.)

yeah but there's no y for the second equation. That confuses me

 

[InteGrand]

yeah but there's no y for the second equation. That confuses me

 

[InteGrand]

 

[InteGrand]

simplify:
(sin(θ)+sin(θ/2))/(1+cos(θ)+cos(θ/2))

 

[Green Yoda]

for the question;
let t=tan(θ/2), solve for the equation 12tan(θ)=5 for 180<θ<270
I got the answers as t=1/5 or t=-5 but the answer given by the textbook is only -5..why?