anonymous
  • anonymous
How can i remember all trig functions and their identities? I mean, its easy to remember what sin, cos, tan are but with cosecant, secant, cotangent, arccosine, arcsine, arctan i cant remember their derivatives and identities. Can you please give some advice? Thank you very much.
OCW Scholar - Single Variable Calculus
  • Stacey Warren - Expert brainly.com
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katieb
  • katieb
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anonymous
  • anonymous
I don't bother to remember others that sin, cos and tan including their inverses. IMHO secant and cosecant are totally useless and cotangent is only needed when tangent approaches infinity. What comes to the derivatives one only must remember that cosine starts as decreasing. So:\[\frac{ d }{ dx }\sin x = \cos x, \frac{ d }{ dx }\cos x = -\sin x\]One can either remember tangent's derivative one or the other way:\[\frac{ d }{ dx }\tan x = 1+\tan^2x=\frac{ 1 }{ \cos^2x }\]But these can be derived from each other:\[\frac{ 1 }{ \cos^2x }=\frac{ (\cos^2x+\sin^2x) }{ \cos^2x }=1+\frac{ \sin^2x }{ \cos^2x }=1+\tan^2x\]The derivatives of the arccusfunctions can be calculated by the identity:\[\frac{ dy }{ dx }=\frac{ 1 }{ \frac{ dx }{ dy } }\]So we can easily see from the equation\[\frac{ d }{ dx }\tan x = 1+\tan^2x\] that\[\frac{ d }{ dx }\arctan x = \frac{ 1 }{1+x^2 }\]and because\[\sin^2x+\cos^2x=1\]that\[\frac{ d }{ dx }\arcsin x=\frac{ 1 }{ \sqrt{1-x^2} }, \frac{ d }{ dx }\arccos x=\frac{ -1 }{ \sqrt{1-x^2} }\]Hope that helped.
anonymous
  • anonymous
It's just a question of practice. Remember that sin^2 x + cos^2 x = 1, then you can easily derive that sec^2 = 1+tan^2 and csc^2 = 1+ cot^2x. That's for the Pythagorean identities. Topi wrote a great explanation of the derivatives.

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