Thread: Okay, last topic for today.. it's about expressing a sec as a sin..

1. Okay, last topic for today.. it's about expressing a sec as a sin..

using the fundamental identities. Man, this stuff is easy yet for some reason my head doesn't understand it at all..

So I have to convert sec, into sin which I thought was simple but I must be over thinking once again.

$secx, sinx$

Process:
$1 + tan^2 = sec^2$ (One of the fundamental identities.)
$sec^2 = 1 + tan^2$
$sec^2 = 1 + sin^2 / cos^2$ (Changing from tangent to sin/cos)

Now here I get a little lost cause I dunno what to do with that cos. I tried doing something like..

$sec^2 = cos^2 + sin^2 / cos^2$ (Multiplied 1 by the common denominator)

Dunno where exactly I went wrong or even if I did it correctly in the first place. Meh, any help on this easy problem would put my mind to ease and help me understand how to do the rest. I just want to learn how to do it correctly ><.

EDIT:
I forgot to add that the answer is secx = 1 / √(1 - sin^2). Also, is there a way to change from cos to sin? o-o

2. First, can you convert $\cos x$ into terms of $\sin x$?

Once you do that, all you have to do is take the reciprocal of your answer because $\sec x = \frac{1}{\cos x}$

3. Hello, Megaman!

We can write all the trig functions in terms of $\sin\theta.$

We already know that: . $\boxed{\csc\theta \:=\:\frac{1}{\sin\theta}}$

From: . $\sin^2\theta + \cos^2\theta \:=\:1$, we get: . $\cos^2\!\theta \:=\:1 - \sin^2\!\theta \quad\Rightarrow\quad\boxed{ \cos\theta \:=\:\sqrt{1-\sin^2\!\theta}}$

. . And we have: . $\sec\theta \:=\:\frac{1}{\cos\theta} \quad\Rightarrow\quad \boxed{\sec\theta \:=\:\frac{1}{\sqrt{1-\sin^2\!\theta}}}$

Since $\tan\theta \,=\,\frac{\sin\theta}{\cos\theta}$, we have: . $\boxed{\tan\theta \:=\:\frac{\sin\theta}{\sqrt{1-\sin^2\!\theta}}}$

. . And we have: . $\cot\theta \:=\:\frac{1}{\tan\theta} \quad\Rightarrow\quad \boxed{\cot\theta \:=\:\frac{\sqrt{1-\sin^2\!\theta}}{\sin\theta}}$

Now we have the entire list in terms of $\sin\theta$

. . $\begin{array}{cccccccc}\sin\theta &=& \sin\theta & \quad & \csc\theta &=& \dfrac{1}{\sin\theta} \\ \\
\cos\theta &=& \sqrt{1-\sin^2\!\theta} & & \sec\theta & = &\dfrac{1}{\sqrt{1-\sin^2\!\theta }} \\ \\
\tan\theta &=& \dfrac{\sin\theta}{\sqrt{1-\sin^2\!\theta}} & & \cot\theta &=& \dfrac{\sqrt{1-\sin^2\!\theta}}{\sin\theta} \end{array}$

I've joked with my classes that, if we were ever shipwrecked
and our survival depended on doing some serious trigonometry,
but the sea water had seeped into our calculator so that
the only trig key that works is the $\boxed{\text{sin}}$ key . . . hey, not to worry!