Prelim 2016 Maths Help Thread (1 Viewer)

[InteGrand]

I have the worked our solution and they did some weird shit...
final answer: 3tan(θ)-tan^2(θ)/1-3tan^2(θ)

That is the answer in terms of tan(θ). They probably just typo'ed the Q. and meant in terms of tan(θ).

 

[Green Yoda]

That is the answer in terms of tan(θ). They probably just typo'ed the Q. and meant in terms of tan(θ).

oh lol, I was half way through and I just checked the solution if I was going correctly and they started with something really weird..I'll finish it off and see what I get.

 

[InteGrand]

oh lol, I was half way through and I just checked the solution if I was going correctly and they started with something really weird..I'll finish it off and see what I get.

What is the weird thing they started with?

 

[Green Yoda]

3sin(θ)-4sin^3(θ)/4cos^3(θ)-3cos(θ)
The first step...

 

[leehuan]

Incidentally,

sin(3x)=3sin(x)-4sin³x

As for cos

 

[InteGrand]

3sin(θ)-4sin^3(θ)/4cos^3(θ)-3cos(θ)
The first step...

They assumed the results for sin(3θ) and cos(3θ) in terms of sin(θ) and cos(θ). Were these earlier parts of the Q.? They aren't formulas you're expected to memorise, but you should be able to derive them.

It's probably easier to do this Q. by just expanding tan(2θ + θ) though. And to save writing, just introduce an abbreviation like t ≡ tan(θ).

 

[Green Yoda]

They assumed the results for sin(3θ) and cos(3θ) in terms of sin(θ) and cos(θ). Were these earlier parts of the Q.? They aren't formulas you're expected to memorise, but you should be able to derive them.

Yes part a and b was to find sin(2θ) and cos(2θ)
But would it still work If I did it normally??

 

[InteGrand]

Yes part a and b was to find sin(2θ) and cos(2θ)

You mean 3θ, right? If so, that explains their method used.

But would it still work If I did it normally??

Yeah, it's probably faster in fact.

 

[Green Yoda]

oh yeah sorry 3θ

 

[Green Yoda]

Yup doing it the normal way was indeed easier

 

[leehuan]

 

[Green Yoda]

help:
simplify sin^2(50)+sin^2(40)

 

[jathu123]

help:
simplify sin^2(50)+sin^2(40)

Since sin(θ) = cos(90-θ), you can change the sin^2(50) into cos^2(90-50) = cos^2(40).
now you have cos^2(40)+sin^2(40) which is = 1 (by the Pythagorean trig identity)

 

[InteGrand]

help:
simplify sin^2(50)+sin^2(40)

= 1 by Pythagorean trig. identity, because sin^2(40 deg) = cos^2(50 deg) (complementary angles identity).

 

[Green Yoda]

= 1 by Pythagorean trig. identity, because sin^2(40 deg) = cos^2(50 deg) (complementary angles identity).

ohh yeah does it also work with sin^2(θ)?
I though it only worked with just θ

 

[InteGrand]

ohh yeah does it also work with sin^2(θ)?
I though it only worked with just θ

Yes it works:
sin^2 x = (sin(x))^2 = (cos(90 deg - x))^2 = cos^2 (90 deg - x).

 

[leehuan]

Basically the important thing to note in what InteGrand said is literally the fact that:

sin²(x)=(sin(x))²

So you can literally chuck the identity under the square.

As an example, similarly:
cos⁴(x)=(1-sin²(x))²

 

[Green Yoda]

help
simplify sin(x)cos(x)cos(2x)

My working out so far
1/2sin(2x)cos(2x)

 

[InteGrand]

help
simplify sin(x)cos(x)cos(2x)

My working out so far
1/2sin(2x)cos(2x)

Apply that rule again and it becomes (1/4)*sin(4x).

 

[Green Yoda]

Apply that rule again and it becomes (1/4)*sin(4x).

Apply what rule?