Author Topic: Trigonometric Identities (Read 3308 times) Tweet Share

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.

TrueTears · « **Reply #1 on:** July 20, 2010, 05:23:46 pm »

sin(75) = sin(30+45)

now using compound angle formulas.

cltf · « **Reply #2 on:** July 20, 2010, 07:54:21 pm »

Oh I see, thank you!

cltf · « **Reply #3 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.

brightsky · « **Reply #4 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.

brightsky · « **Reply #5 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

brightsky · « **Reply #6 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.

luken93 · « **Reply #7 on:** July 23, 2010, 09:22:30 pm »

Ok I've got one:

Prove that:
$\frac{tan \theta}{tan \Phi} = \frac{tan \theta + cot \Phi}{cot \theta + tan \Phi}$

cltf · « **Reply #8 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?

brightsky · « **Reply #9 on:** July 23, 2010, 09:33:24 pm »

brightsky · « **Reply #10 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.

luken93 · « **Reply #11 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?

brightsky · « **Reply #12 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.

98.40_for_sure · « **Reply #13 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

brightsky · « **Reply #14 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.

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