calculus

[jas397]

prove each identity
a)(sin(x))(sec(x))=(tan(x))
b)(cos^2(x))-(sin^2(x))=1-(2sin^2(x))
c)sec(x)-cos(x)=(sin(x))(tan(x))
d)(1-sin(t)/cos(t))=(cos(t)/1+sin(t))
e)(sec(t)-1/1-cos(t))=sec(t)
f)(1+tanx/sinx)-secx=csc(x)
g)tan(x)+cot(x)=(1/cos(x))(sin(x))

 

[soroban]

Hello, jas397!

Don't you know ANY of the basic trig facts?

$\displaystyle [1]\;\sin x\,=\,\frac{1}{\csc x},\;[2]\;\csc x = \frac{1}{\sin x}$

$\displaystyle [3]\;\cos x\,=\,\frac{1}{\sec x},\;[4]\;\sec x\,=\,\frac{1}{\cos x}$

$\displaystyle [5]\;\tan x\,=\,\frac{1}{\cot x},\;[6]\;\cot x\,=\,\frac{1}{\tan x}$

$\displaystyle [7]\;\tan x\,=\,\frac{\sin x}{\cos x},\;[8]\;\cot x\,=\,\frac{\cos x}{\sin x}$

\(\displaystyle [9]\;\sin^2x\,+\,\cos^2x\:=\:1,\;[10]\;\tan^2x\,+\,1\:=\:\sec^2x,\;[11]\;\cot^2x\,+\,1\:=\:\csc^2^2x\)

Prove each identity.

$\displaystyle a)\;\sin(x)\cdot\sec(x)\,=\,\tan(x)$

We have: $\displaystyle \,\sin(x)\cdot\sec(x)$

From [4], this becomes: $\displaystyle \,\sin(x)\cdot\frac{1}{\cos x}\:= \:\frac{\sin(x)}{\cos(x)}$

From [7], we have: \(\displaystyle \.\frac{\sin(x)}{\cos(x)}\:=\:\tan(x)\)

$\displaystyle b)\;\cos^2(x)\,-\,sin^2(x)\:=\:1\,-\,2\cdot\sin^2(x)$

The left side is: $\displaystyle \,\cos^2(x)\,-\,\sin^2(x)$

$\displaystyle \;\;\;$From [9], we have: $\displaystyle \sin^2x\,+\,\cos^2x\:=\:1\;\;\Rightarrow\;\;cos^2x\:=\:1\,-\,\sin^2x$

Substitute: $\displaystyle \,(1\,-\,\sin^2x)\,-\,\sin^2x\:=\:1\,-\,2\sin^2x$


See how it's done?

 

[jas397]

thanx soroban u really helped, noooooot

 

[stapel]

thanx soroban u really helped, noooooot

He's done two of them for you. Aren't you going to do anything on these? Or are you waiting for Soroban to do every last bit of every last one?

Eliz.

 

[happy]

thanx soroban u really helped, noooooot

That was so rude of you to say, jas! Change your ways and then come back here.

 

[jas397]

i was jus jokin he relly helped thnx

 

[Matt]

Why is this a "sticky"....?

It looks like regular members, when creating threads, have the option of making them into stickies. "Why is this a sticky?" indeed. I can only think of two purposes in allowing non-moderators to create stickies:

(1) to make desperate students appear even more desperate when seeking help, which would only provoke us tutors to lash out at them more :wink:

(2) to help trolls distract genuine students and well-intentioned tutors by allowing them (the trolls) to post flamebait as "must-see" stickies

Honestly though, how would creating stickies be useful to us (non-moderators)? All I see is a lot of room for abuse.

I think the moderators should remove this option ASAP.

 

[stapel]

I think the moderators should remove this option ASAP.

Maybe you could post this thought in the "Administration Issues" forum...?

Eliz.

 

[Gene]

It has been my policy to remove the sticky-ness as soon as I see it. I assume that it is built in to phpBB. Otherwise only Ted should have that button.
-----------------
Gene

 

[Matt]

It turns out that Ted accidentally set the "sticky" option for all registered users in the Calculus board only, instead of just for moderators/admins. He fixed it now.