Author Topic: Trigonometric Identities (Read 3483 times) Tweet Share

QuantumJG · « **Reply #15 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.

cltf · « **Reply #16 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.

brightsky · « **Reply #17 on:** July 23, 2010, 11:06:31 pm »

Factorise the $\sin(x)$ out.

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

98.40_for_sure · « **Reply #18 on:** July 23, 2010, 11:19:17 pm »

Lol, i doubt any "normal" skilled maths people would even think of factorizing it like that

cltf · « **Reply #19 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!

brightsky · « **Reply #20 on:** July 23, 2010, 11:45:32 pm »

bar0029 · « **Reply #21 on:** August 13, 2010, 08:58:00 pm »

trig identities = love hahah

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Author Topic: Trigonometric Identities (Read 3483 times) Tweet Share

QuantumJG

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brightsky

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98.40_for_sure

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cltf

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brightsky

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bar0029

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