Question

I need some help with 'Verifying Trigonometric Identities':

13. (sinx + tanx) / (1 + cosx) = tanx

15. cosx = (1 - tan(x/2)^2)/(1 tan(x/2)^2)

Answer 1

\[\frac{\sin(x)+\frac{\sin(x)}{\cos(x)}}{1+\cos(x)}\]

Answer 2

multiply top and bottom by cos(x) to get rid of that compound fraction as a first step

Answer 3

\[\frac{ \sin \theta \cos \theta + \sin \theta}{ \cos \theta + \cos ^{2} \theta }\]

This is what I come up with, I'm not sure what to do from here..

Answer 4

Ok good stuff I wouldn't have multiplied the bottom all out but that is fine...
We are going to factor sin(x) on top and cos(x) on bottom
you should see something that cancels

Answer 5

\[\frac{\sin(x)(\cos(x)+1)}{\cos(x)(1+\cos(x))}\]

Answer 6

Awesome, thanks! Any chance you could point me in the right direction for the second one?

Answer 7

I'm assuming there is a type-o in the second one

Answer 8

Is that a wrong assumption?

Answer 9

\[\cos x = \frac{ 1 - \tan ^{2} \frac{ x }{ 2 } }{ 1 + \tan ^{2} \frac{ x }{ 2 }}\]

Answer 10

I usually prefer things in terms of sin and cos so I don't have to remember extra identities

Answer 11

but if you know tan(x/2) 's identity we can use that here

Answer 12

do you know the half angle identity for tan?

Answer 13

It is fine if you don't like I said I prefer writing things in terms of sin and cos

Answer 14

\[\frac{ \sin \alpha }{ 1 + \cos \alpha } = \tan \frac{ \alpha }{ 2 }\]

Answer 15

The upcoming test is over all of the trigonometric identities so it's best if I use them all.

Answer 16

Yaknow?

Answer 17

let's do this way
and let's see how hard it is the other way
this way might actually be easier...
\[\frac{1-\frac{\sin^2(\frac{x}{2})}{\cos^2(\frac{x}{2})}}{1+\frac{\sin^2(\frac{x}{2})}{\cos^2(\frac{x}{2})}}\]

Answer 18

try multiplying top and bottom by cos^2(x/2) to get rid of the compound fraction first

Answer 19

Alright, I got this:
\[\frac{ \cos ^{2} \frac{ x }{ 2 } - \sin ^{2}\frac{ x }{ 2 }}{ \cos ^{2} \frac{ x }{ 2 } + \sin ^{2}\frac{ x }{ 2 } }\]

Answer 20

what does the bottom equal ?

Answer 21

It's equal to 1

Answer 22

yes \[\sin^2(\theta)+\cos^2(\theta)=1 \]

Answer 23

do you know what the top equals?

Answer 24

I don't think so

Answer 25

\[\cos^2(\theta)-\sin^2(\theta)=\cos(2 \theta) \\ \text{ so if } \theta=\frac{x}{2} \text{ we have } \\ \cos^2(\frac{x}{2})-\sin^2(\frac{x}{2})=\cos(2 \cdot \frac{x}{2})=\cos(x)\]

Answer 26

I will show you the other way if you like now

Answer 27

or if you have a question about this way let me know

Answer 28

Dude, if you're not too busy, that would be awesome. Either way you've already helped me a ton!

Answer 29

\[\frac{1-\tan^2(\frac{x}{2})}{1+\tan^2(\frac{x}{2})} =\frac{1-(\frac{\sin(x)}{1+\cos(x)})^2}{1+(\frac{\sin(x)}{1+\cos(x)})^2} \\ =\frac{(1+\cos(x))^2}{(1+\cos(x))^2} \cdot \frac{1-(\frac{\sin(x)}{1+\cos(x)})^2}{1+(\frac{\sin(x)}{1+\cos(x)})^2} =\frac{(1+\cos(x))^2-\sin^2(x)}{(1+\cos(x))^2+\sin^2(x)}\]
While you look at that I will typing more

Answer 30

\[\frac{1+2\cos(x)+\cos^2(x)-\sin^2(x)}{1+2\cos(x)+\cos^2(x)+\sin^2(x)}=\frac{1+2\cos(x)+\cos^2(x)-(1-\cos^2(x))}{1+2\cos(x)+1}\]

Answer 31

\[\frac{1+2\cos(x)+\cos^2(x)-1+\cos^2(x)}{2+2\cos(x)}=\frac{2\cos(x)+2\cos^2(x)}{2+\cos(x)} \\ =\frac{\cos(x)(2+\cos(x))}{2+\cos(x)}=\cos(x)\]

Answer 32

so a little more longer but still works

Answer 33

Wow...Thank you so much!

Answer 34

np

Answer 35

good luck on your test

Answer 36

Thank you~!