simplifying and proving identities

• Jul 16th 2008, 06:08 PM
Dave19
simplifying and proving identities
degrees=(d)theta=(t) alpha=(a) beta=(B) pie(p)squared=(s)
it will make much more sense if you use these.
comment:just do the ones you know you don't have to do all of them.
also write it out on paper in the right way not my way it will make it easier and i just need the work i am given the answers.

5)prove: tan(t)-cot(t)/tan(t)+cot(t)=2sin(s)(t)-1

6)prove:1-cos(t)/sin(t)=sin(t)/1+cos(t)

7)prove:1-tan(s)(t)/1+tan(s)(t)=1-2sin(s)(t)
• Jul 16th 2008, 08:18 PM
Mathstud28
Quote:

Originally Posted by Dave19
degrees=(d)theta=(t) alpha=(a) beta=(B) pie(p)squared=(s)
it will make much more sense if you use these.
comment:just do the ones you know you don't have to do all of them.
also write it out on paper in the right way not my way it will make it easier and i just need the work i am given the answers.

5)prove: tan(t)-cot(t)/tan(t)+cot(t)=2sin(s)(t)-1

6)prove:1-cos(t)/sin(t)=sin(t)/1+cos(t)

7)prove:1-tan(s)(t)/1+tan(s)(t)=1-2sin(s)(t)

3)

$\cos\left(\frac{3\pi}{2}+x\right)$

$\text{We know the addition formula to be}~~\sin\left(A+B\right)=\sin\left(A\right)\cos\l eft(B\right)+\cos\left(A\right)\sin\left(B\right)$

$\Rightarrow\sin\left(\frac{3\pi}{2}+x\right)=\sin\ left(\frac{3\pi}{2}\right)\cos(x)+\cos\left(\frac{ 3\pi}{2}\right)\sin(x)$

$=-\cos(x)+0\sin(x)$

$=-\cos(x)$

$\therefore\quad\boxed{\sin\left(\frac{3\pi}{2}+x\r ight)=-\cos(x)}$
• Jul 16th 2008, 08:42 PM
Chris L T521
Quote:

Originally Posted by Dave19
degrees=(d)theta=(t) alpha=(a) beta=(B) pie(p)squared=(s)
it will make much more sense if you use these.
comment:just do the ones you know you don't have to do all of them.
also write it out on paper in the right way not my way it will make it easier and i just need the work i am given the answers.

5)prove: tan(t)-cot(t)/tan(t)+cot(t)=2sin(s)(t)-1

6)prove:1-cos(t)/sin(t)=sin(t)/1+cos(t)

7)prove:1-tan(s)(t)/1+tan(s)(t)=1-2sin(s)(t)

note:I need these done for tomorrow morning so its either now or never

$(1)$

$\cos^2x+\csc^2x+\cot^2x+\sin^2x$

$\text{We need to use the Pythagorean identities:}$

$\sin^2x=1-\cos^2x \ \text{and} \ \cot^2x=\csc^2x-1$

$\text{Thus, } \cos^2x+\csc^2x+\underbrace{\csc^2x-1}_{\cot^2x}+\underbrace{1-\cos^2x}_{\sin^2x}=\color{red}\boxed{2\csc^2x}$

Does this make sense?

--Chris
• Jul 16th 2008, 11:56 PM
earboth
Quote:

Originally Posted by Dave19
degrees=(d)theta=(t) alpha=(a) beta=(B) pie(p)squared=(s)
it will make much more sense if you use these.
comment:just do the ones you know you don't have to do all of them.
...

...

$\cos^2(\theta)+\csc^2(\theta)+\cot^2(\theta)+\sin^ 2(\theta) = \underbrace{\cos^2(\theta)+\sin^2(\theta)}_{\text{ = 1}} +\csc^2(\theta)+\cot^2(\theta)$

$1+\csc^2(\theta)+\cot^2(\theta)=1+\frac1{\sin^2(\t heta)} + \frac{\cos^2(\theta)}{\sin^2(\theta)} = \frac{\sin^2(\theta)}{\sin^2(\theta)} +\frac1{\sin^2(\theta)} + \frac{\cos^2(\theta)}{\sin^2(\theta)}$

$\frac{\sin^2(\theta)}{\sin^2(\theta)} +\frac1{\sin^2(\theta)} + \frac{\cos^2(\theta)}{\sin^2(\theta)}=\frac{\sin^2 (\theta) + 1 + \cos^2(\theta)}{\sin^2(\theta)} = \frac{\underbrace{\sin^2(\theta) + \cos^2(\theta)}_{\text{= 1}} + 1 }{\sin^2(\theta)}$

$\frac2{\sin^2(\theta)}= 2 \cdot \left(\frac1{\sin(\theta)}\right)^2 = 2\csc^2(\theta)$
• Jul 17th 2008, 12:07 AM
Moo
I always transform into sin, cos and tan, because it's much easier :p
$\sec(t)=\frac{1}{\cos(t)}$

$\csc(t)=\frac{1}{\sin(t)}$

$\cot(t)=\frac{\cos(t)}{\sin(t)}$

and it's PI, not pie.

Furthermore, I think it's not very useful to give you all the solution... just try to do it with the information and then, tell us if you can't do it ;)

Quote:

Originally Posted by Dave19
degrees=(d)theta=(t) alpha=(a) beta=(B) pie(p)squared=(s)
it will make much more sense if you use these.
comment:just do the ones you know you don't have to do all of them.
also write it out on paper in the right way not my way it will make it easier and i just need the work i am given the answers.

$\frac{1}{\cos^2(t)}+\frac{1}{\sin^2(t)}$
get the common denominator, and use the identity $\cos^2(t)+\sin^2(t)=1$

Quote:

Use $\sin(a+b)=\cos(a)\sin(b)+\cos(b)\sin(a)$

:)

Quote:

Use the same identity as above :

$\sin(2t)=\cos(t)\sin(t)+\dots$

Quote:

5)prove: tan(t)-cot(t)/tan(t)+cot(t)=2sin(s)(t)-1
I guess it is $\frac{\tan(t)-\cot(t)}{\tan(t)+\cot(t)}$ (please use parenthesis ! pemdo rules or something like that...)

transform $\tan=\frac \sin \cos$ and $\cot=\frac \cos \sin$

then get the common denominator on the numerator and the denominator.

Quote:

6)prove:1-cos(t)/sin(t)=sin(t)/1+cos(t)
$\frac{1-\cos(t)}{\sin(t)}=\frac{\sin(t)}{1+\cos(t)}$
or
$1-\frac{\cos(t)}{\sin(t)}=\frac{\sin(t)}{1+\cos(t)}$
?
Do you see the purpose of parenthesis ? :)

Quote:

7)prove:1-tan(s)(t)/1+tan(s)(t)=1-2sin(s)(t)
idem
• Jul 17th 2008, 08:22 AM
Soroban
Hello, Dave19!

Quote:

1) Simplify: . $\cos^2\!\theta+\csc^2\!\theta + \cot^2\!\theta + \sin^2\!\theta$

Answer: . $2\csc^2\!\theta$

We have: . $\underbrace{\sin^2\!\theta + \cos^2\!\theta} + \csc^2\!\theta + \cot^2\!\theta$
. . . . . . . . . $= \;\;1 \;+ \;\csc^2\!\theta \;+\; \overbrace{\csc^2\!\theta - 1} \;\;=\;\;2\csc^2\!\theta$

Quote:

2) Simplify: . $\sec^2\theta+\csc^2\!\theta$

Answer: . $\sec^2\!\theta\csc^2\!\theta$

We have: . $\frac{1}{\cos^2\!\theta} + \frac{1}{\sin^2\!\theta} \;=\;\frac{\overbrace{\sin^2\!\theta + \cos^2\!\theta}^{\text{This is 1}}}{\cos^2\!\theta\sin^2\!\theta}$

. . $=\;\frac{1}{\cos^2\!\theta\sin^2\theta} \;=\;\sec^2\!\theta\csc^2\theta$

Quote:

5) Prove: . $\frac{\tan\theta-\cot\theta}{\tan\theta +\cot\theta} \:=\:2\sin^2\!\theta -1$
We have: . $\frac{\;\dfrac{\sin\theta}{\cos\theta} - \dfrac{\cos\theta}{\sin\theta}\;} {\dfrac{\sin\theta}{\cos\theta} + \dfrac{\cos\theta}{\sin\theta}}$

Multiply top and bottom by $\sin\theta\cos\theta$

. . $\frac{\sin\theta\cos\theta\left(\dfrac{\sin\theta} {\cos\theta} - \dfrac{\cos\theta}{\sin\theta}\right)} {\sin\theta\cos\theta\left(\dfrac{\sin\theta}{\cos \theta} + \dfrac{\cos\theta}{\sin\theta}\right)}
\;\;=\;\;\frac{\sin^2\!\theta-\cos^2\!\theta}{\underbrace{\sin^2\!\theta + \cos^2\theta}_{\text{This is 1}}}$

. . $= \;\;\sin^2\!\theta - \cos^2\!\theta \;\;=\;\;\sin^2\!\theta - (1 - \sin^2\!\theta) \;\;=\;\;2\sin^2\!\theta - 1$

Quote:

6) Prove: . $\frac{1-\cos\theta}{\sin\theta} \:=\:\frac{\sin\theta}{1+\cos\theta}$
Multiply the left side by: $\frac{1+\cos\theta}{1+\cos\theta}$

. $\frac{1-\cos\theta}{\sin\theta}\cdot{\color{blue}\frac{1+\ cos\theta}{1+\cos\theta}} \;=\;\frac{1-\cos^2\!\theta}{\sin\theta\,(1 + \cos\theta)} \;=\;\frac{\sin^2\!\theta}{\sin\theta\,(1 + \cos\theta)} \;=\;\frac{\sin\theta}{1 + \cos\theta}$