# Trig Identity

1. Dec 2, 2009

### mcca408

1. The problem statement, all variables and given/known data

verify the following identity:

Sec(x)Sin2(x)
______________________ = 1 - cos(x)

1 + sec(x)

2. Relevant equations
sec(x)=1/cos(x)
sin2(x)=1-cos2(x)

3. The attempt at a solution
I never know how to start off these problems. I have to take the left side and show that it equals the right by doing trig identities. I try several steps and keep going through an endless loop. I believe I must start of by multiplying 1-sec(x)/1-sec(x)
That way i get

(1-sec(x))(sec(x)sin2(x))
__________________________________

1 - sec2(x)

I'm not sure if i'm starting off correctly

1. The problem statement, all variables and given/known data

2. Relevant equations

3. The attempt at a solution

2. Dec 2, 2009

### Staff: Mentor

It would probably be easier to start by replacing the sec(x) terms on the left side with 1/cos(x).

3. Dec 2, 2009

### mcca408

Sec(x)Sin2(x)
______________________ = 1 - cos(x)

1 + sec(x)

ok using the identity sec(x)=1/cos(x)

I get

1/cos(x) * sin2(x)
__________________________

1 + 1/cos(x)

and that =

sin2(x)/cos(x)
______________________

1 + 1/cos(x)

multiply top and bottom by cos(x)

Sin2(x) / 2

Did i do anything wrong?

Thanks

4. Dec 2, 2009

### Staff: Mentor

Yes. cos(x)*(1 + 1/cos(x)) is not equal to 2. Try again.

5. Dec 2, 2009

### DrMath

I think a trick in handling trigo qns is to convert all to sine and cosine functions.
since sec, csc and even tan functions can be derived from sine and cosine.

Do remember this to help in your future sch works as well.

6. Dec 2, 2009

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7. Dec 2, 2009

### Anakin_k

Turn the denominator into $$\frac{cos(x)+1}{cos(x)}$$.

So that would look like $$\frac{\frac{sin^{2}(x)}{cos(x)}}{\frac{cos(x)+1}{cos(x)}}.$$

What icystrike showed is the faster way but since you're already this far, you can try what I suggested.