Pre-Calc question

J

John292

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Hello, My name is John and this is me a 54 year old engineer from the USA trying to help my son with his precalc homework! I am honored to to find a site in GB. Your engineering work is always the best!

The problem is 1+cos (y) / [ 1+ sec (y) ] and the answer is to be in sin or cos

The answer is cos (y) per the book

I tried 1/ [1+sec(y)] + cos(y)/ [1+sec(y)]

and then I get lost trying to rid of the + sign
 
\(\displaystyle \frac{1+cos(y)}{1+sec(y)}\)

Use the identity \(\displaystyle \frac{1}{cos(y)}=sec(y)\)

\(\displaystyle \frac{1+cos(y)}{1+\frac{1}{cos(y)}}\)

Cross multiply the bottom as you would any fraction:

\(\displaystyle \frac{1+cos(y)}{\frac{cos(y)+1}{cos(y)}}\)

\(\displaystyle \frac{1+cos(y)}{1}\cdot \frac{cos(y)}{1+cos(y)}\)

See the cancellations?. We're left with

\(\displaystyle cos(y)\)
 
Thank you very much for your help! You put me in the right direction.

I worked thru the next problem.

(sec[sup:1z99rjfc]2[/sup:1z99rjfc] -1)/(sec[sup:1z99rjfc]2[/sup:1z99rjfc]) Trying to learn this editor

[sec[sup:1z99rjfc]2[/sup:1z99rjfc](x)-1]/[sec[sup:1z99rjfc]2[/sup:1z99rjfc](x)]

subed in 1/cos[sup:1z99rjfc]2[/sup:1z99rjfc](x) = sec[sup:1z99rjfc]2[/sup:1z99rjfc](x) and did the flip multiply

cos[sup:1z99rjfc]2[/sup:1z99rjfc](x)[sec[sup:1z99rjfc]2[/sup:1z99rjfc](x)-1] = cos[sup:1z99rjfc]2[/sup:1z99rjfc](x) sec[sup:1z99rjfc]2[/sup:1z99rjfc](x) - cos[sup:1z99rjfc]2[/sup:1z99rjfc](x)

then subed in 1 = sec(x) cos(x) = sec[sup:1z99rjfc]2[/sup:1z99rjfc](x) cos[sup:1z99rjfc]2[/sup:1z99rjfc](x)

and get 1-cos[sup:1z99rjfc]2[/sup:1z99rjfc](x)

subing in sin[sup:1z99rjfc]2[/sup:1z99rjfc](x) = 1-cos[sup:1z99rjfc]2[/sup:1z99rjfc](x)

and the answer is sin[sup:1z99rjfc]2[/sup:1z99rjfc](x)

So I sub and I flip and get it to work! The correct wording would be good.

If you can help with next one I would be appreciative,

[1+csc(x)] / [cos(x)+cot(x)] and the answer is sec(x)

Good day!
 
Make the subs so everything is in terms of cos and sin and you'll see it.

\(\displaystyle csc(x)=\frac{1}{sin(x)}, \;\ cot(x)=\frac{cos(x)}{sin(x)}\)
 
Yes that did work out. Thank you for the lead.

So the question that comes to my mind........

How do you know where to start? Is there a rule?

Is it sort of simular to playing chess with Trig Identities?
 
Yeah, one gets better at them from doing them so much.

The thing is to get everything in terms of cos and sin.

There are plenty of identities to do that.

They are all in your sons pre-calc book.
 
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