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VCE Stuff => VCE Mathematics => VCE Mathematics/Science/Technology => VCE Subjects + Help => VCE Specialist Mathematics => Topic started by: cltf on July 20, 2010, 05:23:05 pm

Title: Trigonometric Identities
Post by: cltf on July 20, 2010, 05:23:05 pm: Using the expansion of $\sin (A+B)$ prove that $\sin (75^{\circ})= \frac{\sqrt{6}+\sqrt{2}}{4}$

then again using trig identities find Cos(75)

I am at a loss.
Title: Re: Trigonometric Identities
Post by: TrueTears on July 20, 2010, 05:23:46 pm: sin(75) = sin(30+45)

now using compound angle formulas.
Title: Re: Trigonometric Identities
Post by: cltf on July 20, 2010, 07:54:21 pm: Oh I see, thank you!
Title: Re: Trigonometric Identities
Post by: cltf on July 23, 2010, 07:39:01 pm: got more! yay!

Prove the following:
$\frac{\sin(x)+ 1 - \cos(x)}{\sin(x)-1+\cos(x)}=\frac{1+\tan(\frac{x}{2})}{1-\tan(\frac{x}{2})}$

$\frac{\sin(x)+\sin(\frac{\pi}{2}-x)+1}{\sin(x)-\sin(\frac{\pi}{2}-x)+1}= \cot(\frac{x}{2})$

$\cot(x+y)=\frac{\cot(x)\cot(y)-1}{\cot(x)+\cot(y)}$

that's all for now. Got more to come.
Title: Re: Trigonometric Identities
Post by: brightsky on July 23, 2010, 08:03:47 pm: We know that $\tan\left(\frac{x}{2}\right) = \frac{1 - \cos(x)}{\sin(x)}$ .

Sub that into the LHS:

$\frac{1 + \tan\left(\frac{x}{2}\right)}{1 - \tan\left(\frac{x}{2}\right)} = \frac{1 + \frac{1 - \cos(x)}{\sin(x)}}{1 - \frac{1 - \cos(x)}{\sin(x)}}$

Multiply $\sin(x)$ to the denominator and the numerator we have:

$\frac{\sin(x) + 1 - \cos(x)}{\sin(x) - 1 + \cos(x)}$ as required.
Title: Re: Trigonometric Identities
Post by: brightsky on July 23, 2010, 08:15:07 pm: Alternate the RHS: $\cot\left(\frac{x}{2}\right) = \frac{1}{\tan\left(\frac{x}{2}\right)} = \frac{1}{\frac{1 - \cos(x)}{\sin(x)}} = \frac{\sin(x)}{1 - \cos(x)}$

Alternate the LHS: $\frac{\sin(x) + \sin\left(\frac{\pi}{2} - x\right) + 1}{\sin(x) - \sin\left(\frac{\pi}{2} - x \right) + 1} = \frac{\sin(x) + \cos(x) + 1}{\sin(x) - \cos(x) + 1} = \frac{\sin(x) (1 + \cot(x) + \csc(x))}{(1 - \cos(x))(1 + \cot(x) + \csc(x))} = \frac{\sin(x)}{1 - \cos(x)}$

Hence the original equation is true.

EDIT: cosec -> csc
Title: Re: Trigonometric Identities
Post by: brightsky on July 23, 2010, 09:02:31 pm: First we acknowledge that $\tan(x + y) = \frac{\tan(x) + \tan(y)}{1 - \tan(x) \tan(y)}$

$\frac{\cot(x) \cot(y) - 1}{\cot(x) + \cot(y)}$

$= \frac{\frac{1}{\tan(x)\tan(y)} - 1}{\frac{1}{\tan(x)} + \frac{1}{\tan(y)}}$

$= \frac{\frac{1 - \tan(x)\tan(y)}{\tan(x)\tan(y)}}{\frac{\tan(y) + \tan(x)}{\tan(x)\tan(y)}}$

$= \frac{1 - \tan(x)\tan(y)}{\tan(y)+\tan(x)}$

$= \frac{1}{\tan(x + y)}$

$= \cot (x+y)$ as required.
Title: Re: Trigonometric Identities
Post by: luken93 on July 23, 2010, 09:22:30 pm: Ok I've got one:

Prove that:
$\frac{tan \theta}{tan \Phi} = \frac{tan \theta + cot \Phi}{cot \theta + tan \Phi}$
Title: Re: Trigonometric Identities
Post by: cltf on July 23, 2010, 09:28:56 pm: Quote from: brightsky on July 23, 2010, 08:15:07 pm

Alternate the LHS: $\frac{\sin(x) + \sin\left(\frac{\pi}{2} - x\right) + 1}{\sin(x) - \sin\left(\frac{\pi}{2} - x \right) + 1} = \frac{\sin(x) + \cos(x) + 1}{\sin(x) - \cos(x) + 1} = \frac{\sin(x) (1 + \cot(x) + \cosec(x))}{(1 - \cos(x))(1 + \cot(x) + \cosec(x))} = \frac{\sin(x)}{1 - \cos(x)}$

I don't understand the 3rd step where did the (1+cot(x)+(x)) come from?
Title: Re: Trigonometric Identities
Post by: brightsky on July 23, 2010, 09:33:24 pm: $\frac{\tan(a) + \cot(b)}{\tan(b) + \cot(a)}$

$= \frac{\tan(a) + \frac{1}{\tan(b)}}{\tan(b) + \frac{1}{\tan(a)}}$

$= \frac{\frac{\tan(a)\tan(b) + 1}{\tan(b)}}{\frac{\tan(a)\tan(b) + 1}{\tan(a)}}$

$= \frac{\tan(a)}{\tan(b)}$
Title: Re: Trigonometric Identities
Post by: brightsky on July 23, 2010, 09:35:54 pm: Quote from: cltf on July 23, 2010, 09:28:56 pm
Quote from: brightsky on July 23, 2010, 08:15:07 pm

Alternate the LHS: $\frac{\sin(x) + \sin\left(\frac{\pi}{2} - x\right) + 1}{\sin(x) - \sin\left(\frac{\pi}{2} - x \right) + 1} = \frac{\sin(x) + \cos(x) + 1}{\sin(x) - \cos(x) + 1} = \frac{\sin(x) (1 + \cot(x) + \cosec(x))}{(1 - \cos(x))(1 + \cot(x) + \cosec(x))} = \frac{\sin(x)}{1 - \cos(x)}$

I don't understand the 3rd step where did the (1+cot(x)+(x)) come from?

Sorry, Latex didn't display properly. Fixed it up now. :)
Title: Re: Trigonometric Identities
Post by: luken93 on July 23, 2010, 09:52:55 pm: Quote from: brightsky on July 23, 2010, 09:33:24 pm
$\frac{\tan(a) + \cot(b)}{\tan(b) + \cot(a)}$

$= \frac{\tan(a) + \frac{1}{\tan(b)}}{\tan(b) + \frac{1}{\tan(a)}}$

$= \frac{\frac{\tan(a)\tan(b) + 1}{\tan(b)}}{\frac{\tan(a)\tan(b) + 1}{\tan(a)}}$

$= \frac{\tan(a)}{\tan(b)}$
is there a way to do it starting from LHS or is it fine to do it from RHS?
Several of my teachers were having a fight as to whether you should do it from either or LHS?
Or is it simply a matter of reversing the steps you took?
Title: Re: Trigonometric Identities
Post by: brightsky on July 23, 2010, 09:54:22 pm: Yep, just a matter of reversing steps. Both LHS and RHS approaches are legitimate. Your only task is to prove that LHS = RHS.
Title: Re: Trigonometric Identities
Post by: 98.40_for_sure on July 23, 2010, 09:56:27 pm: Can you... half solve LHS and RHS?

I've always solved fully from one side, rather than lil bits from both sides until they are equal
Title: Re: Trigonometric Identities
Post by: brightsky on July 23, 2010, 10:07:03 pm: Quote from: 99.95_for_sure on July 23, 2010, 09:56:27 pm
Can you... half solve LHS and RHS?

I've always solved fully from one side, rather than lil bits from both sides until they are equal

I don't see why you can't.
Title: Re: Trigonometric Identities
Post by: QuantumJG on July 23, 2010, 10:12:56 pm: Quote from: 99.95_for_sure on July 23, 2010, 09:56:27 pm
Can you... half solve LHS and RHS?

I've always solved fully from one side, rather than lil bits from both sides until they are equal

~~This is true.~~ That is the right way.

With a proof you should either start at the RHS or the LHS (whatever side looks the best to tackle) and then show that after manipulation it comes out to being what the other side states.

These proofs are probably the nicest you will ever see.

EDIT: I noticed ambiguity in my post.
Title: Re: Trigonometric Identities
Post by: cltf on July 23, 2010, 11:01:32 pm: Quote from: brightsky on July 23, 2010, 08:15:07 pm
Alternate the RHS: $\cot\left(\frac{x}{2}\right) = \frac{1}{\tan\left(\frac{x}{2}\right)} = \frac{1}{\frac{1 - \cos(x)}{\sin(x)}} = \frac{\sin(x)}{1 - \cos(x)}$

Alternate the LHS: $\frac{\sin(x) + \sin\left(\frac{\pi}{2} - x\right) + 1}{\sin(x) - \sin\left(\frac{\pi}{2} - x \right) + 1} = \frac{\sin(x) + \cos(x) + 1}{\sin(x) - \cos(x) + 1} = \frac{\sin(x) (1 + \cot(x) + \csc(x))}{(1 - \cos(x))(1 + \cot(x) + \csc(x))} = \frac{\sin(x)}{1 - \cos(x)}$

Hence the original equation is true.

EDIT: cosec -> csc

To be precise I am confused about where the Cosec and Cot come from.
Title: Re: Trigonometric Identities
Post by: brightsky on July 23, 2010, 11:06:31 pm: Factorise the $\sin(x)$ out.

$\sin(x) + \cos(x) + 1 = \sin(x) \left(1 + \frac{1}{\tan(x)} + \frac{1}{\sin(x)}\right) = \sin(x) (1 + \cot(x) + \csc(x))$
Title: Re: Trigonometric Identities
Post by: 98.40_for_sure on July 23, 2010, 11:19:17 pm: Lol, i doubt any "normal" skilled maths people would even think of factorizing it like that :P
Title: Re: Trigonometric Identities
Post by: cltf on July 23, 2010, 11:24:24 pm: did you factorize $(1-\cos(x))$ out of the denominator? or was it $\sin (x)$ because in neither of the workings do i get $(1+\cot (x) + \csc(x)?$

and one more equation

prove:

$\frac{1-\tan(x)\tan(2x)}{1+\tan(x)\tan(2x)}= 4\cos^{2}(x)-3$

last one!!! out of the 56!
Title: Re: Trigonometric Identities
Post by: brightsky on July 23, 2010, 11:45:32 pm: $\sin(x) (1 + \cot(x) + \csc(x)) = \sin(x) + \cos(x) + 1$

$(1 - \cos(x)) (1 + \cot(x) + \csc(x)) = 1 + \cot(x) + \csc(x) - \cos(x) - \cot(x)\cos(x) - \cos(x) \csc(x) = 1 + \sin(x) - \cos(x)$
Title: Re: Trigonometric Identities
Post by: bar0029 on August 13, 2010, 08:58:00 pm: trig identities = love hahah