1. ## Basic trig simplification

I am having the HARDEST TIME simplifying this into cos y

(1+cos y)/(1+sec y)

When I do it, I always end up with cos^2 y..

Thanks guys, appreciate it.

2. Originally Posted by shiznid12
I am having the HARDEST TIME simplifying this into cos y

(1+cos y)/(1+sec y)

When I do it, I always end up with cos^2 y..

Thanks guys, appreciate it.
sec(y)=1/cos(y). Combine the denominator into one term, then the cancellation should be clear. If not, post your steps and we'll take a look.

3. I understand that... but how would it cancel out

(1+cos y)/(1+1/cos y)

I understand the 1's would/could cancel, but then you still have:

cos y/(1/cos y)

And the only way I can imagine you would get rid of the denom is multiply the cos y uptop.

I'm clearly doing something wrong, and this is the only problem i've had any issues with thus far.

4. Originally Posted by shiznid12
I am having the HARDEST TIME simplifying this into cos y

(1+cos y)/(1+sec y)

When I do it, I always end up with cos^2 y..

Thanks guys, appreciate it.
$\displaystyle \frac{1+\cos{y}}{1 + \sec{y}} \cdot \frac{\cos{y}}{\cos{y}} =\frac{(1+\cos{y})\cos{y}}{\cos{y} + 1} =\cos{y}$

5. Is there an explanation for why you multiply top and bottom by cos y?

That would be very useful knowledge for my test tomorrow.

Better yet, why don't you multiple up the 1 + 1/cos also because it's part of the denominator too?

6. Originally Posted by shiznid12
Is there an explanation for why you multiply top and bottom by cos y?

experience, really ... I saw (1 + secy) in the denominator ... multiplying by cosy makes the denominator match a factor in the numerator (1 + cosy)

Better yet, why don't you multiple up the 1 + 1/cos also because it's part of the denominator too?

what would that accomplish? ... the idea here is to simplify, not make more complex.
here is probably the most basic method of simplification ...

$\displaystyle \frac{1+\cos{y}}{1+\sec{y}} =$

$\displaystyle \frac{1+\cos{y}}{1 + \frac{1}{\cos{y}}} =$

$\displaystyle \frac{1+\cos{y}}{\frac{\cos{y}}{\cos{y}} + \frac{1}{\cos{y}}} =$

$\displaystyle \frac{1+\cos{y}}{\frac{\cos{y}+1}{\cos{y}}} =$

$\displaystyle \frac{1+\cos{y}}{1} \cdot \frac{\cos{y}}{\cos{y}+1} = \cos{y}$

...