How can you prove that derivatives of trigonometric funtions? i.e. the deriv od sine is cosine?

Answer 1

You need to use the identity sin(x+y)=sin(x)cos(y)+sin(y)cos(x) and the limits lim_{x \to 0} sin(x)/x=1 and lim_{x \to 0} (1-cos(x))/x=0 to compute the derivative of sin(x) using the definition.

Then, using the definition of the derivative,

d/dx ( sin(x) )
=lim_{h \to 0} (sin (x+h)-sin(x))/h
=lim_{h \to 0}( sin(x)cos(h)+sin(h)cos(x)-sin(x))/h
=lim_{h \to 0}[sin(x) (cos(h)-1)/h+cos(x) sin(h)/h)
=sin(x) times 0+ cos(x) times 1
=cos(x)

In the same way, d/dx ( cos (x) )=- sin(x), but you will use the identity cos(x+y)=cos(x)cos(y)-sin(x)sin(y). Then, use the quotient rule/reciprocal rule to find the derivatives of the tangent, secant, cosecant, and cotangent functions.

Way to go trig identities! They are cool! :)

Answer 2

Proving the derivatives of the six trig functions can be easily done by using the definition of the derivative. Remember, there are two definitions (which are equivalent, of course) and the trick is to use the right one. And then it is just trigonometric manipulation.

Usually, the derivatives of sine and cosine are done using the definition (lim h->0) and then after that the quotient rule (for tangent and cotangent) and the power rule (for secant and cosecant) is used.

Answer 3

use the definition for the derivative of any function. you can simplify the fraction part by using different trig identies (in particular the angle addition identities). you should then be able to simplify and take the limit.

Some links to help: http://www.math2.org/math/derivatives/more/trig.htm

http://tutorial.math.lamar.edu/AllBrowsers/2413/DiffTrigFcns.asp

Answer 4

write down the taylor expansion of sine ( x )
differentiate each term
write down the taylor expansion of cosine(x)
compare both , ou will see they are equal

Answer 5

it quite is a sin cos you're actually not meant to fractionate the geotriangular octameter. and that i'm already tan. p.s. notice to the instructor, sorry i grew to become into overdue to classification and my canines ate my domicile artwork.

Answer 6

First of all we know:
sin(x)' = cos(x)
cos(x)' = -sin(x)
tan(x)' = sec^2(x)
csc(x)' = -csc(x) cot(x)
sec(x)' = sec(x) tan(x)
cot(x)' = -csc^2(x)


Proof of sin(x) : algebraic Method
Given: lim(d->0) sin(d)/d = 1.
Solve:

sin(x) = lim(d->0) ( sin(x+d) - sin(x) ) / d
= lim ( sin(x)cos(d) + cos(x)sin(d) - sin(x) ) / d
= lim ( sin(x)cos(d) - sin(x) )/d + lim cos(x)sin(d)/d
= sin(x) lim ( cos(d) - 1 )/d + cos(x) lim sin(d)/d
= sin(x) lim ( (cos(d)-1)(cos(d)+1) ) / ( d(cos(d)+1) ) + cos(x) lim sin(d)/d
= sin(x) lim ( cos^2(d)-1 ) / ( d(cos(d)+1 ) + cos(x) lim sin(d)/d
= sin(x) lim -sin^2(d) / ( d(cos(d) + 1) + cos(x) lim sin(d)/d
= sin(x) lim (-sin(d)) * lim sin(d)/d * lim 1/(cos(d)+1) + cos(x) lim sin(d)/d
= sin(x) * 0 * 1 * 1/2 + cos(x) * 1 = cos(x)

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Proof of cos(x) : from the derivative of sine

This can be derived just like sin(x) was derived or more easily from the result of sin(x)

Given: sin(x) = cos(x); Chain Rule.
Solve:

cos(x) = sin(x + PI/4)
cos(x) = sin(x + PI/4)
= sin(u) * (x + PI/4) (Set u = x + PI/4)
= cos(u) * 1 = cos(x + PI/4) = -sin(x)

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Proof of tan(x) : from the derivatives of sine and cosine

Given: sin(x) = cos(x); cos(x) = -sin(x); Quotient Rule.
Solve:

tan(x) = sin(x) / cos(x)
tan(x) = sin(x)/cos(x)
= ( cos(x) sin(x) - sin(x) cos(x) ) / cos^2(x)
= ( cos(x)cos(x) + sin(x)sin(x) ) / cos^2(x)
= 1 + tan^2(x) = sec^2(x)

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Proof of csc(x), sec(x), cot(x) : from derivatives of their reciprocal functions

Given: sin(x) = cos(x); cos(x) = -sin(x); tan(x) = cot(x); Quotient Rule.
Solve:

csc(x) = 1/sin(x) = ( sin(x) (1) - 1 sin(x) ) / sin^2(x) = -cos(x) / sin^2(x) = -csc(x)cot(x)
sec(x) = 1/cos(x) = ( cos(x) (1) - 1 cos(x) ) / cos^2(x) = sin(x) / cos^2(x) = sec(x)tan(x)
cot(x) = 1/tan(x) = ( tan(x) (1) - 1 tan(x) ) / tan^2(x) = -sec^2(x) / tan^2(x) = -csc^2(x)

Answer 7

do it using the first principle method..........
y=sinx
y+*y=sin(x+*x)
*y=sin(x+*x)-sinx
*y/*x=(sinx+*x-sinx)/*x
=sinxcos*x+cosxsin*x-sinx/*x
=lim*x->0sinxcos*x-cosx-sinx
put values
cos0=1
sinx+cosx-sinx=cosx
hence proved

Answer 8

sine /cos=tan
cos /sin= er......

Answer 9

Derivative:
defined as

lim(Δx→0) {[ƒ(x+Δx)-ƒ(x)]/Δx}
Plug & play.

OR
Graph the derivatives and see what you get. :o)