Trig identitys

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February 15th 2011, 05:50 PM
Kromletch

Trig identitys

My book lacks answers and i suffer much confusion in this. So here goes, but first quick question. How do i produce signs like pi or theta on here like some of the experts do, along with representing fractions as a fraction as opposed to for example 2x/4

I misspelled Identities also in title im aware of this :P

anyway

tan x csc x cos x =1

tan=sin x/cos x
csc x= ? says something like csc(-x)=-csc x
cos x= 1/sec x

so i get

(sin x/cos x)(csc x)(1/sec x)=1 and am stuck here, the csc identity confuses me
February 15th 2011, 05:58 PM
pickslides

Quote:

Originally Posted by Kromletch

tan x csc x cos x =1

tan=sin x/cos x

Looks like a good start

$\displaystyle \tan x \csc x \cos x$

$\displaystyle \frac{\sin x}{\cos x} \csc x \cos x$

$\displaystyle \sin x \csc x$

$(\displaystyle \csc x =\frac{1}{\sin x})$

...
February 15th 2011, 06:21 PM
harish21

To type symbols and fractions and other notations, learn Latex
February 15th 2011, 09:14 PM
Kromletch

ok what about (4sin^2 x-1)/(2sin x+1)=2sin x-1

would it be

(2sin x-1)/(x+1) =2sin x-1

this is where i get lost again as im not sure how to work the problem. seems like i could cancel out the 2sin x-1 from each side but it seems illegal
February 15th 2011, 09:28 PM
topsquark

Quote:

Originally Posted by Kromletch

ok what about (4sin^2 x-1)/(2sin x+1)=2sin x-1

would it be

(2sin x-1)/(x+1) =2sin x-1

this is where i get lost again as im not sure how to work the problem. seems like i could cancel out the 2sin x-1 from each side but it seems illegal

First, I'm not sure where you got the x + 1 in the denominator.

Typically you want to work on only one side of the (suspected) identity. In this case
$\displaystyle \frac{4~sin^2(x) - 1}{2~sin(x) + 1}$

Note that the numerator is in the form $\displaystyle a^2 - b^2 = (a - b)(a + b)$ . Can you take it from here?

-Dan
February 15th 2011, 11:43 PM
Kromletch

Lets see. ( 2sin(x)-1)(2sin(x)+1)

then i can cancel out 2sin(x)+1 which leaves 2sin(x)-1=2sin(x)-1

which equals 1.

And that is the idea of an identity ?
February 16th 2011, 02:42 AM
Archie Meade

Quote:

Originally Posted by Kromletch

Lets see. ( 2sin(x)-1)(2sin(x)+1)

then i can cancel out 2sin(x)+1 which leaves 2sin(x)-1=2sin(x)-1

which equals 1.

And that is the idea of an identity ?

$\displaystyle\frac{4sin^2x-1}{2sinx+1}=2sinx-1$

The idea of the identity is that the above expression is equal for all x.
Hence, instead of using the fraction on the left in calculations,
the right side, which is simpler, may be used instead.

You are asked to show they are equal.
If they are equal, then the denominator must divide into the numerator,
hence the numerator should have the denominator as a factor.
The 2nd factor should be the part on the right.
That's the clue to work it through.

$4sin^2x-1=(2sinx+1)(2sinx-1)\;\;?$

$4sin^2x-1=2sinx(2sinx-1)+(1)(2sinx-1)\;\;?$

$4sin^2x-1=4sin^2x-2sinx+2sinx-1\;\;?$

Yes, they are equal.
Hence, the fraction may be replaced with

$2sinx-1$
February 16th 2011, 05:39 AM
Kromletch

Hmm im starting to understand, ive ran into yet again another different situation.

So (sinx-cosx)(sinx+cosx)=1-2cosx^2
Multiply them
sinx^2+sinxcosx-sinxcosx-cosx^2=1-2cosx^2
for
sinx^2-cosx^2=1-2cosx^2

and am again unsure of where to go from here, im looking in my book at pyhagrean identities seems similar
February 16th 2011, 07:02 AM
Archie Meade

Quote:

Originally Posted by Kromletch

Hmm im starting to understand, ive ran into yet again another different situation.

So (sinx-cosx)(sinx+cosx)=1-2cosx^2
Multiply them
sinx^2+sinxcosx-sinxcosx-cosx^2=1-2cosx^2
for
sinx^2-cosx^2=1-2cosx^2

and am again unsure of where to go from here, im looking in my book at pyhagrean identities seems similar

You made good progress with that.

$sin^2x-cos^2x=1-2cos^2x\;\;?$

There are a few ways to continue..

You have "difference of squares" on the left, but you have already expanded that out.

Notice that we have 1 on the right, so we could use the identity

$sin^2x+cos^2x=1,$ which is a Pythagorean Identity, as you suspected...

This gives

$sin^2x-cos^2x=sin^2x+cos^2x-2cos^2x\;\;?$

$cos^2x-2cos^2x=(1-2)cos^2x=-cos^2x$

$sin^2x-cos^2x=sin^2x-cos^2x\;\;?$

True