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 Proofs: Derivative Trig Functions (Math | Calculus | Derivatives | Table Of | Trig Functions)
 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)

## Proofs of Derivative of Trig Functions

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)       Q.E.D.

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/2)
cos(x) = sin(x + PI/2)
= sin(u) * (x + PI/2) (Set u = x + PI/2)
= cos(u) * 1 = cos(x + PI/2) = -sin(x)       Q.E.D.

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)       Q.E.D.

Reciprocals

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)       Q.E.D.

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