1. ## Trigonometry Addition Formula (Tangent)

Hello , I'm Joshua. This is my first time here and my first post. I came from Philippines and I am studying at Computer College University. My course is B.S. Computer Engineering. A freshman. I came here because, I think it might help me.

Okay here's my problem.

When I was at highschool, math are too easy for me. Specially, Highschool Algebra, Trigo's. But then after I entered the college life. I cant understand ALL math subjects, there are too numbers for me to put it in my brain. We have two math subjects, first is College Algebra, and the second is Trigonometry. I failed them both at Prelim . And this MidTerm I was hoping I could pass it. But after the MidTerm examination I failed it again. I cant believe that this is happening to me. Every night, I am almost crying, because I cant take it anymore. It is really hard.

I felt stupid, pathetic. Even how many times it is discussed I cant understand it.

Please... Help me! I NEED to advance to second year, well... not just second year but to Graduate also! I love my mother, she's the only one who pay my tuition fees, and all payments for school. Im so ashamed to myself.

Help me...

to avoid from off topic..

I'll put Formulas and Examples. Kindly illustrate them how to use this formulas to the givens, and how to answer them one by one.(Actually, There are a lot of things that I am confuse, not just this.)

Formulas:

double-angle formulas:

source formulas: Trigonometric Addition Formulas -- from Wolfram MathWorld

Given:
1) cos2x= 1-tan²x / 1+tan²x

2) cosΘ
/1-sinΘ = 1+tanΘ/² / 1-tanΘ/²

3)cosΘ / 1-sinΘ = 1+1cosΘ+sinΘ / 1+cosΘ / 1+cosΘ-sinΘ / 1+cosΘ

I am sorry if there is wrong grammars. Thank you!

2. Originally Posted by Roach
Hello , I'm Joshua. This is my first time here and my first post. I came from Philippines and I am studying at Computer College University. My course is B.S. Computer Engineering. A freshman. I came here because, I think it might help me.

Okay here's my problem.

When I was at highschool, math are too easy for me. Specially, Highschool Algebra, Trigo's. But then after I entered the college life. I cant understand ALL math subjects, there are too numbers for me to put it in my brain. We have two math subjects, first is College Algebra, and the second is Trigonometry. I failed them both at Prelim . And this MidTerm I was hoping I could pass it. But after the MidTerm examination I failed it again. I cant believe that this is happening to me. Every night, I am almost crying, because I cant take it anymore. It is really hard.

I felt stupid, pathetic. Even how many times it is discussed I cant understand it.

Please... Help me! I NEED to advance to second year, well... not just second year but to Graduate also! I love my mother, she's the only one who pay my tuition fees, and all payments for school. Im so ashamed to myself.

Help me...

to avoid from off topic..

I'll put Formulas and Examples. Kindly illustrate them how to use this formulas to the givens, and how to answer them one by one.(Actually, There are a lot of things that I am confuse, not just this.)

Formulas:

double-angle formulas:

source formulas: Trigonometric Addition Formulas -- from Wolfram MathWorld

Given:
1) cos2x= 1-tan²x / 1+tan²x

2) cosΘ /1-sinΘ = 1+tanΘ/² / 1-tanΘ/²

3)cosΘ / 1-sinΘ = 1+1cosΘ+sinΘ / 1+cosΘ / 1+cosΘ-sinΘ / 1+cosΘ

I am sorry if there is wrong grammars. Thank you!

I'm assuming that you're asked to verify the identities you've been given...

1) $\displaystyle \frac{1 - \tan^2{x}}{1 + \tan^2{x}} = \frac{1 - \frac{\sin^2{x}}{\cos^2{x}}}{1 + \frac{\sin^2{x}}{\cos^2{x}}}$

$\displaystyle = \frac{\frac{\cos^2{x}}{\cos^2{x}} - \frac{\sin^2{x}}{\cos^2{x}}}{\frac{\cos^2{x}}{\cos ^2{x}} + \frac{\sin^2{x}}{\cos^2{x}}}$

$\displaystyle = \frac{\frac{\cos^2{x} - \sin^2{x}}{\cos^2{x}}}{\frac{\cos^2{x} + \sin^2{x}}{\cos^2{x}}}$

$\displaystyle = \frac{\frac{\cos{(2x)}}{\cos^2{x}}}{\frac{1}{\cos^ 2{x}}}$

$\displaystyle = \frac{\cos{(2x)}}{1}$

$\displaystyle = \cos{(2x)}$.

3. Hello, Joshua!

Welcome aboard!

Here's the first one.
I think there is a typo in the second one.

1) Prove: .$\displaystyle \cos2x\:=\:\frac{1-\tan^2x}{1+\tan^2x}$
The right side is: . $\displaystyle \frac{1-\tan^2x}{1 + \tan^2x} \;=\; \frac{1-\dfrac{\sin^2x}{\cos^2x}}{1 + \dfrac{\sin^2x}{\cos^2x}}$

Multiply by $\displaystyle \frac{\cos^2x}{\cos^2x}\!:\quad \frac{\overbrace{\cos^2x - \sin^2x}^{\text{This is }\cos2x}}{\underbrace{\cos^2x + \sin^2x}_{\text{This is 1}}} \;=\;\frac{\cos2x}{1} \;=\;\cos2x$

4. Originally Posted by Roach
Hello , I'm Joshua. This is my first time here and my first post. I came from Philippines and I am studying at Computer College University. My course is B.S. Computer Engineering. A freshman. I came here because, I think it might help me.

Okay here's my problem.

When I was at highschool, math are too easy for me. Specially, Highschool Algebra, Trigo's. But then after I entered the college life. I cant understand ALL math subjects, there are too numbers for me to put it in my brain. We have two math subjects, first is College Algebra, and the second is Trigonometry. I failed them both at Prelim . And this MidTerm I was hoping I could pass it. But after the MidTerm examination I failed it again. I cant believe that this is happening to me. Every night, I am almost crying, because I cant take it anymore. It is really hard.

I felt stupid, pathetic. Even how many times it is discussed I cant understand it.

Please... Help me! I NEED to advance to second year, well... not just second year but to Graduate also! I love my mother, she's the only one who pay my tuition fees, and all payments for school. Im so ashamed to myself.

Help me...

to avoid from off topic..

I'll put Formulas and Examples. Kindly illustrate them how to use this formulas to the givens, and how to answer them one by one.(Actually, There are a lot of things that I am confuse, not just this.)

Formulas:

double-angle formulas:

source formulas: Trigonometric Addition Formulas -- from Wolfram MathWorld

Given:
1) cos2x= 1-tan²x / 1+tan²x

2) cosΘ /1-sinΘ = 1+tanΘ/² / 1-tanΘ/²

3)cosΘ / 1-sinΘ = 1+1cosΘ+sinΘ / 1+cosΘ / 1+cosΘ-sinΘ / 1+cosΘ

I am sorry if there is wrong grammars. Thank you!

i would use x for convenience .
(b)
$\displaystyle \frac{\cos x}{1-\sin x} = \frac{\cos^2 \frac{x}{2}-\sin^2 \frac{x}{2}}{\cos^2 \frac{x}{2}-2\sin \frac{x}{2}\cos \frac{x}{2}+\sin^2 \frac{x}{2}}$

$\displaystyle \frac{(\cos \frac{x}{2}+\sin \frac{x}{2})(\cos \frac{x}{2}-\sin \frac{x}{2})}{(\cos \frac{x}{2}-\sin \frac{x}{2})^2}$

$\displaystyle \frac{\frac{\cos \frac{x}{2}+\sin \frac{x}{2}}{\cos \frac{x}{2}}}{\frac{\cos \frac{x}{2}-\sin \frac{x}{2}}{\cos \frac{x}{2}}}$

$\displaystyle \frac{1+\tan \frac{x}{2}}{1-\tan \frac{x}{2}}$

5. Im confuse, Im sorry for being dumb... Kindly illustrate them how to use the formulas to the givens, and how to answer them step by step, I mean how to answer the given I gave. And about Latex, what is it?

6. We HAVE used the formulas given and shown you step by step.

As for the LaTeX, use the LaTeX help forum on this site.

7. Hello Joshua
Originally Posted by Roach
Im confuse, Im sorry for being dumb... Kindly illustrate them how to use the formulas to the givens, and how to answer them step by step, I mean how to answer the given I gave. And about Latex, what is it?

You'll need to study the answers carefully, and you'll find which formulae are used at each stage.

I guess the writers of the answers assumed that you knew or could work out for yourself the following formulae:

$\displaystyle \tan x = \frac{\sin x}{\cos x}$

$\displaystyle \sin^2x+\cos^2x = 1$ or $\displaystyle \sin^2\tfrac12x+\cos^2\tfrac12x=1$, or ... etc.

Also when you replace $\displaystyle 2\alpha$ by $\displaystyle x$ in the formulae for $\displaystyle \cos2\alpha$, you get equivalent formulae that express $\displaystyle \cos x$ in terms of $\displaystyle \tfrac12x$:

$\displaystyle \cos x =\cos^2\tfrac12x - \sin^2\tfrac12x$

$\displaystyle =2\cos^2\tfrac12x-1$

$\displaystyle =1-2\sin^2\tfrac12x$

Have another careful look through the proofs that Soroban and mathaddict gave you. You should be able to see where each step comes from now. If you can't, tell us exactly where the problem is, and we'll see if we can explain further.

8. I can understand a little bit but I have a question.

What happened here?
at given number 2.

$\displaystyle \frac{\cos x}{1-\sin x} = \frac{\cos^2 \frac{x}{2}-\sin^2 \frac{x}{2}}{\cos^2 \frac{x}{2}-2\sin \frac{x}{2}\cos \frac{x}{2}+\sin^2 \frac{x}{2}}$

Why it so happen quickly that the denominator becomes

$\displaystyle {(\cos \frac{x}{2}-\sin \frac{x}{2})^2}$

EDIT:

$\displaystyle \tan x = \frac{\sin2x}{1+\cos2x}$

$\displaystyle 1-\tan x \tan \frac{x}{2} = \sec$

9. Originally Posted by Roach
I can understand a little bit but I have a question.

What happened here?
at given number 2.

$\displaystyle \frac{\cos x}{1-\sin x} = \frac{\cos^2 \frac{x}{2}-\sin^2 \frac{x}{2}}{\cos^2 \frac{x}{2}-2\sin \frac{x}{2}\cos \frac{x}{2}+\sin^2 \frac{x}{2}}$

Why it so happen quickly that the denominator becomes

$\displaystyle {(\cos \frac{x}{2}-\sin \frac{x}{2})^2}$

EDIT:

$\displaystyle \tan x = \frac{\sin2x}{1+\cos2x}$

$\displaystyle 1-\tan x \tan \frac{x}{2} = \sec$
Well it has the form of $\displaystyle a^2-2ab+b^2 = (a-b)^2$

For the first part ,

$\displaystyle \frac{\sin 2x}{1+\cos 2x}=\tan x$

Think of the double angle formula .

From RHS , $\displaystyle \frac{2\sin x\cos x}{\cos^2 x+\sin^2 x+\cos^2 x-\sin^2 x}$

$\displaystyle \frac{2\sin x\cos x}{2\cos^2 x}$

try continuing from here.

10. is there a formula for $\displaystyle \cos ^2 x$

11. Originally Posted by Roach
is there a formula for $\displaystyle \cos ^2 x$
erm , are u referring to your previous question , you can just cancel out the cos .

By the way , $\displaystyle \cos^2x+\sin^2x=1\Rightarrow \cos^2x=1-\sin^2x$

12. 1 is equivalent to $\displaystyle \cos ^2 x$ ?

well... I couldnt find the answer.
$\displaystyle \frac{\sin 2 x}{2\cos ^2 x}$

Im confuse with $\displaystyle 2\cos ^2 x$... oh man..

13. Originally Posted by Roach
1 is equivalent to $\displaystyle \cos ^2 x$ ?

well... I couldnt find the answer.
$\displaystyle \frac {\sin 2 x}{2\cos ^2 x}$

Im confuse with $\displaystyle 2\cos ^2 x$... oh man..
No . Recall this famous identity .
$\displaystyle 1=\cos^2x+\sin^2x$

So just now you were asking $\displaystyle \cos^2x=...$ ? Then just do some algebra work .

Well , i will continue from where i stop ..

$\displaystyle \frac{2\sin x\cos x}{2\cos^2 x}$

now do u see both the cos x in the numerator and denominator ? You can cancel them and you are left with

$\displaystyle \frac{2\sin x}{2\cos x}$--- cancel the 2 as well .. then you are left with .

$\displaystyle \frac{\sin x}{\cos x}$

Then this will be $\displaystyle \tan x$

No . Recall this famous identity .
$\displaystyle 1=\cos^2x+\sin^2x$

So just now you were asking $\displaystyle \cos^2x=...$ ? Then just do some algebra work .

Well , i will continue from where i stop ..

$\displaystyle \frac{2\sin x\cos x}{2\cos^2 x}$

now do u see both the cos x in the numerator and denominator ? You can cancel them and you are left with

$\displaystyle \frac{2\sin x}{2\cos x}$--- cancel the 2 as well .. then you are left with .

$\displaystyle \frac{\sin x}{\cos x}$

Then this will be $\displaystyle \tan x$
OH!! I GET IT!.. Finally... Ahh... Thank you very much !! thank you thank you !..

Finally... what about the second given?
$\displaystyle 1-\tan x \tan \frac{x}{2} = \sec$

15. Wow, the longer i stay in this forum the more i appreciate the helping hands of its members. Thank you MHF members!

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