I need to reduce trig identities to the simplest form. i was wondering if you could take me further.
(secx)(cscx)-tanx
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cotx(1/secx)+sinx
i changed the top to (1/cosx)(1/sinx)-(sinx/cosx)
Thats as far as i can do it
I need to reduce trig identities to the simplest form. i was wondering if you could take me further.
(secx)(cscx)-tanx
__________________
cotx(1/secx)+sinx
i changed the top to (1/cosx)(1/sinx)-(sinx/cosx)
Thats as far as i can do it
Keep going ... Convert everything into terms of sinx and cosx and combine the fractions by finding the common denominator. For example in the numerator:
$\displaystyle \sec x \csc x - \tan x$
$\displaystyle = \frac{1}{\cos x \sin x} - \frac{\sin x}{\cos x}$
$\displaystyle = \frac{1}{\cos x \sin x} - \frac{\sin^{2} x}{\cos x \sin x}$
$\displaystyle = \frac{\overbrace{1 - \sin^{2} x}^{\mbox{Look familiar?}}}{\cos x \sin x}$
Hello, Airjunkie!
Get a common denominator and combine the fractions.Reduce to the simplest form: .$\displaystyle \frac{\sec x\csc x -\tan x}{\cot x\!\cdot\!\frac{1}{\sec x}+\sin x}$
i changed the top to: .$\displaystyle \frac{1}{\cos x}\!\cdot\!\frac{1}{\sin x} -\frac{\sin x}{\cos x}$ . . . . Good!
. . $\displaystyle \frac{1}{\cos x\sin x} - \frac{\sin x}{\cos x}\cdot\frac{\sin x}{\sin x} \;\;=\;\;\frac{1}{\cos x\sin x} - \frac{\sin^2\!x}{\cos x\sin x} $
. . $\displaystyle = \;\;\frac{1-\sin^2\!x}{\cos x\sin x} \;\;=\;\;\frac{\cos^2\!x}{\cos x\sin x} \;\;=\;\;\frac{\cos x}{\sin x}$
The denominator is: .$\displaystyle \frac{\cos x}{\sin x}\!\cdot\!\frac{\cos x}{1} + \sin x \;\;=\;\;\frac{\cos^2\!x}{\sin x} + \frac{\sin^2\!x}{\sin x}$
. . $\displaystyle =\;\;\frac{\overbrace{\cos^2\!x + \sin^2\!x}^{\text{This is 1}}}{\sin x} \;\;=\;\;\frac{1}{\sin x}$
The problem becomes: . $\displaystyle \frac{\dfrac{\cos x}{\sin x}}{\dfrac{1}{\sin x}} \;\;=\;\;\boxed{\cos x}$
Let's concentrate piece by piece.
1.$\displaystyle \frac{\cos \left(\frac{\pi}{2} - \theta\right)}{sin^{2}\left(\frac{\pi}{2} - \theta\right) + cos^{2}\left(\frac{\pi}{2} - \theta\right)}$
First off, you should immediately recognize the identity found in the denominator. As for the numerator, if you know your sin
and cos graphs like the back of your hand:
$\displaystyle \cos \left(\frac{\pi}{2} - \theta \right) = \overbrace{\cos \left(\theta - \frac{\pi}{2}\right)}^{ \cos (- x ) = \cos x }$
which is just y = cos x shifted to the right by pi/2. What would that be?
2. $\displaystyle \frac{\frac{1}{\frac{1}{\sin\theta}}}{\sec \theta} = \frac{\sin \theta}{\frac{1}{\cos \theta}} = \mbox{etc.}$
3. $\displaystyle - \frac{1}{\cot (\pi + \theta)} = - \tan(\theta + \pi)$
Ignoring the negative sign for now, if you shift the graph y = tanx to the left by pi, what do you get?
4.
$\displaystyle \sin^{2}(- \theta) + \cos^{2}(\theta) = (\sin \left(-\theta\right))^{2} + \cos^{2}(\theta) = \underbrace{(-\sin (\theta))^{2}}_{\sin(-\theta) = -\sin (\theta)} + \cos^{2} (\theta) = \sin^{2}\theta + \cos^{2} \theta$
Im not trying to be mean but unless i see steps it cant help me.
By the way this isnt really for a grade, but she said these will be on the tests. So id really like to study and memorize the rules with them. I need to see the whole thing. Ill try to look over it again