Trigonometry

[pamann]

Verify that the equation is an identity:

2cos^2 B– 1 = 1 - tan^2 B / 1 + tan^2 B :?

My answer: cos2A = -1

Is this correct?[/quote]

 

[Dinky_Bell]

Basically all that you need to do is choose a number and place it as B

B=?

2cos^2 B– 1 = 1 - tan^2 B / 1 + tan^2 B


If the two sides (either side of the '=' sign) are equal you proved it as an identity...if they aren't then you've proved it as a non-identity

 

[Daniel_Feldman]

Dinky, your reasoning is not correct. Just because the equation may be true for one value of b does not mean that it wil be true for all values of B, and vice versa.


The fact that the directions say "verify that this is an identity" means that the statement is true. Here's how we prove it. Note the general property that if a=b, then b=a. This allows us to begin with either side of the equation and verify the other side. What we are not allowed to do is ot work with both sides simultaneously.

2cos^2 B– 1 = 1 - tan^2 B / 1 + tan^2 B

I'll start with the left side and work to get to the right side:

1 - tan^2 B / 1 + tan^2 B

1+tan^2 B=sec^2 B=1/cos^2 B

so

1 - tan^2 B / 1 + tan^2 B=(1-tan^2 B)/sec^2 B=(1-tan^2 B)(cos^2 B)

tan^2 B=sin^2 B/cos^2 B, so

(1-tan^2 B)(cos^2 B)=(1-(sin^2 B/cos^2 B))(cos^2 B)

Multiplying through

(1-(sin^2 B/cos^2 B))(cos^2 B)=cos^2 B-sin^2 B


cos^2 B-sin^2 B=2cos^2 B-1 (Recall the double angle formula for cosines: cos(2x)=2cos^2 x-1=1-2sin^2 x=cos^2 x-sin^2 x)


Got it?

 

[Dinky_Bell]

it didn't know you could do it that way... my teacher told us to just choose a radian of the circle (ex: pi/2) and see if it works....


I guess these boards dont just help the ppl who post but the ones who pay attention as well...thankx

 

[Daniel_Feldman]

You could do that for one value, but can't conclude that is works for everything.


Think of it with a more basic equation:

2x+3=13


You could test, say, x=3, and find out that it does not work. But would that allow you to conclude that 2x+3 never equals 13? Certainly not!