Lukecrayonz
sin^2(x1)/cos(x)



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Lukecrayonz
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@amistre64

amistre64
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might need to know alittle more about the question

Lukecrayonz
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It just says to simplify @amistre64

amistre64
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already seems "simple" to me :/

amistre64
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you can try to shuffle some things about i spose; sin^2 = 1cos^2
cos(a+b) = cosa cosb  sina sinb and stuff

amistre64
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cos(x) = cos(x) ... just some basic trig identities and properties that might be useful

Lukecrayonz
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@amistre64 would you like the possible answers?

Lukecrayonz
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Also quick question on another problem I'm stuck on.

amistre64
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options would be able to narrow it down :)

Lukecrayonz
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@amistre64 You there?

Lukecrayonz
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@saifoo.khan

Lukecrayonz
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A. sinx
B. cosx
C. Sinx
D. cosx

saifoo.khan
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Where are you stuck? @Lukecrayonz

Lukecrayonz
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@saifoo.khan don't even know where to start :'(((


Lukecrayonz
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The one Sin(AB)?

saifoo.khan
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Yep.

Lukecrayonz
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How.. That makes no sense to me?

Lukecrayonz
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If its sin^2 how would I use Sin(AB)?

saifoo.khan
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Ohhh. :o

Lukecrayonz
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hmm?

Lukecrayonz
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I think I got it

saifoo.khan
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I guess, @satellite73 or @amistre64 will help you here. :/

Lukecrayonz
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sin^2x1=cos^2x
cos(x)=cosx
so the answer is cosx

saifoo.khan
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Wait. How?

Lukecrayonz
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cos^2x/cosx=cosx?

saifoo.khan
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Step 1.

Lukecrayonz
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cos(x)=cosx

Lukecrayonz
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Pythagorean identities sin^2x+cos^2x=1

saifoo.khan
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\[\frac{\cos^2(x1) +1}{\cos(x)}\]Then we should get this?

amistre64
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shuffling trigs is a pain :)

saifoo.khan
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No pain, no gain. ;)

amistre64
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do you need to show work?
