Trig and geometry question (1 Viewer)

[Legend-X]

Hey guys, I've been stuck in these 2 questions for like weeks now lol, helps will be very appreciated.

1st question
i) Show that 1-cos2x/1+cos2x = tan^2x
this one is really easy:
LHS= 1-(1-2sin^2x) / 1+(2cos^2x-1)
= 2sin^2x / 2cos^2x
= tan^2x = RHS

but the second part says
ii) Hence find the value of tan 22.5 in simplest exact form
Um, so I got the correct answer which is -1 + √2 but that was by using the t formula. (let x = tan22.5, x^2 + 2x -1 = 0)
So how do I solve this question by using the first part?



2nd question (I know I don't have the picture, sorry)
ABC is a triangle and N is a point on AC.
angle ABN = angle CBN = angle BCN. BC=2a, CA=b, AB=c, BN=CN=d

i. Given that Triangle ABN ||| Triangle ACB, show that c^2 = b^2 - 2ac
ii. Hence show that (a+c)^2 = a^2 + b^2

Funny thing is I didn't even need to solve i) to solve ii).
ii) is really easy, it's just expanding out and rearranging the equation to match the one at i)
a^2 + 2ac + c^2 = a^2 + b^2
2ac + c^2 = b^2
thus c^2 = b^2 - 2ac


but I have no clue about i). How do you get squares involved in this?
The picture looks something like this:

-----------------------------------------A

---------------------------------------------N

------------------------------------B------------C

With ABC being the triangle and N connected to B. (ignore the lines, I had to put them so they actually stay where they are)

Thanks

 

[blaze.bluetane]

The 't' formulae is probably best used for part (i)

hope this helps

 

[scardizzle]

Hey guys, I've been stuck in these 2 questions for like weeks now lol, helps will be very appreciated.

1st question
i) Show that 1-cos2x/1+cos2x = tan^2x
this one is really easy:
LHS= 1-(1-2sin^2x) / 1+(2cos^2x-1)
= 2sin^2x / 2cos^2x
= tan^2x = RHS

but the second part says
ii) Hence find the value of tan 22.5 in simplest exact form
Um, so I got the correct answer which is -1 + √2 but that was by using the t formula. (let x = tan22.5, x^2 + 2x -1 = 0)
So how do I solve this question by using the first part?



2nd question (I know I don't have the picture, sorry)
ABC is a triangle and N is a point on AC. (< = angle)
<abn bc="2a," ca="b," ab="c," bn="CN=d<br">
i. Given that Triangle ABN ||| Triangle ACB, show that c^2 = b^2 - 2ac
ii. Hence show that (a+c)^2 = a^2 + b^2

Funny thing is I didn't even need to solve i) to solve ii).
ii) is really easy, it's just expanding out and rearranging the equation to match the one at i)
a^2 + 2ac + c^2 = a^2 + b^2
2ac + c^2 = b^2
thus c^2 = b^2 - 2ac


but I have no clue about i). How do you get squares involved in this?
The picture looks something like this:

-----------------------------------------A

---------------------------------------------N

------------------------------------B------------C

With ABC being the triangle and N connected to B. (ignore the lines, I had to put them so they actually stay where they are)

Thanks

1st q just sub in x as 22.5

therefore tan^2 (22.5) = </abn>1-cos45/1+cos45
= (1 - 1/root2)/(1 + 1/root 2)

= (root2 -1)/(root2 +1)

= (root2 -1)^2 (rationalizing the denominator

therefore tan22.5 = root 2 -1
<abn bc="2a," ca="b," ab="c," bn="CN=d<br">
</abn>

 

[Legend-X]

You guys are amazing thanks so much!
Actually one question, could you explain what you meant by
noticing that cos2x = 1-tan^2x / 1+tan^2x ?
My class never went over t formula so I'm really bad with those ... and my textbook doesn't explain much either.

+updated geography question. It was somehow missing the given facts.

 

[blaze.bluetane]

Haha thats good, gotta love missing trig formulae

The basic idea of 't' formulae is that if you take the trig double angle results, you can derive expressions for the sine and cosine ratios in terms of tan(A/2), where A is an angle

sin(2A)=2.sin(A).cos(A)

sin(A)=2.sin(A/2).cos(A/2)

sin(A)=[2.sin(A/2).cos(A/2)] / [cos^2(A/2)+sin^2(A/2)]

This is because of the pythagorean identity,

cos^2(A)+sin^2(A)=1

Now if you take this equation

sin(A)=[2.sin(A/2).cos(A/2)] / [cos^2(A/2)+sin^2(A/2)]

and divide numerator and denominator by cos^2(A/2), you will yield the result;

sin(A)=[2.tan.(A/2)] / [1+tan^2(A/2)]

Which is actually the 't' formula for sine...

We take 't=tan(A/2)' and thus,

sin(A)=[2.t] / [1+t^2]

Similarly, when we take the cosine double angle result, and divide cos^2(A/2) through the numerator and denominator, we will yield the result;

cos(A)=[1-t^2]/[1 +t^2]

Hence in your situation, we let A=2x

cos(2x)=[1-tan^2(x)]/[1 +tan^2(x)]

 

[blaze.bluetane]

This is the geometry [not geography lol] proof...
You take the two similar triangles and use the fact that their corresponding similar sides are in proportion to one another... algebraic manipulation and tada!

[BTW I rotated the triangle through about 90 degrees anticlockwise (don't ask why lol)]

The way I solve this is by noticing that there is no 'd' value in the solution. Therefore it must be eliminated in the equations to form the solution

obviously I am missing the labels on the bottom right triangle (ABN)...oops
AB=c and BN=d

 

[Legend-X]

hahaha oops geometry not geography lol
omgosh you are a genius, clear explanation and nice diagrams <3
I finally understand the t-formula, thanks!!

The way I solve this is by noticing that there is no 'd' value in the solution. Therefore it must be eliminated in the equations to form the solution

Oh that's clever. I thought they just put it to confuse us or something haha. I was sooo stuck on that d.

Btw for t-formula do you always divide by the cos^2(x/2)? and is that because it gives you the tan result?

 

[blaze.bluetane]

The trig formulae are a set of results that can be used to solve equations or integrate or solve proofs etc..
You don't really need to be able to prove them but for the sake of maths...

cos(2A)= cos^2(A) - sin^2(A)

cos(2A)= [cos^2(A) - sin^2(A)] / [cos^2(A) + sin^2(A)]

Now divide by the numerator and denominator by cos^2(A)

cos(2A)= [1-t^2] / [1+t^2]

Dividing by the cos^2 allows; [all of the sin^2 to form tan^2], [all of the cos^2 to form 1] and [all of the sincos combos to form tan (sine 't' result)]

It is just another formula to remember because it allow you to express several trig functions into tan (think solving equations, factorising, integrating etc)

cos(2A)= [1-t^2] / [1+t^2]
sin(2A)= [2t] / [1+t^2]
tan(2A)= [2t] / [1-t^2]

bb

 

[jet]

There is another way to derive t-results:
\\

I personally prefer this method, because if you forget them you could easily remember the triangle and visualise it/draw it in an exam.

 

[blaze.bluetane]

lol. yes thats right.

CSSA Trial 2002
Q1 and 3

I thought I remembered these questions from somewhere. Haha