Fundamental Identities
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Key Questions

Those are formulas that you can derive from the pitagoras triangle.
For example:
#c^2= a^2+b^2#
Expressed in sin and cos you have:
#1 = cos^2(theta) + sin^2(theta)#
This is one identity, then you play a bit with this formula and get another identity.Divide by
#cos^2(theta)# :#1/(cos^2(theta))  (cos^2(theta))/(cos^2(theta)) = (sin^2(theta))/(cos^2(theta))# #sec^2(theta)  1 = tg^2(theta)# And keep going.

"The fundamental trigonometric identities" are the basic identities:
â€¢The reciprocal identities
â€¢The pythagorean identities
â€¢The quotient identitiesThey are all shown in the following image:
When it comes down to simplifying with these identities, we must use combinations of these identities to reduce a much more complex expression to its simplest form.
Here are a few examples I have prepared:
a) Simplify:
# tanx/cscx xx secx# Apply the quotient identity
#tantheta = sintheta/costheta# and the reciprocal identities#csctheta = 1/sintheta# and#sectheta = 1/costheta# .#=(sinx/cosx)/(1/sinx) xx 1/cosx# #=sinx/cosx xx sinx/1 xx 1/cosx# #=sin^2x/cos^2x# Reapplying the quotient identity, in reverse form:
#=tan^2x# b) Simplify:
#(cscbeta  sin beta)/cscbeta# Apply the reciprocal identity
#cscbeta = 1/sinbeta# :#=(1/sinbeta  sin beta)/(1/sinbeta)# Put the denominator on a common denominator:
#=(1/sinbeta  sin^2beta/sinbeta)/(1/sinbeta)# Rearrange the pythagorean identity
#cos^2theta + sin^2theta = 1# , solving for#cos^2theta# :#cos^2theta = 1  sin^2theta# #=(cos^2beta/sinbeta)/(1/sinbeta)# #=cos^2beta/sinbeta xx sin beta/1# #=cos^2beta# c) Simplify:
#sinx/cosx + cosx/(1 + sinx)# :Once again, put on a common denominator:
#=(sinx(1 + sinx))/(cosx(1 + sinx)) + (cosx(cosx))/(cosx(1 + sinx))# Multiply out:
#=(sinx + sin^2x + cos^2x)/(cosx(1 + sinx))# Applying the pythagorean identity
#cos^2x + sin^2x = 1# :#=(sinx + 1)/(cosx(1 + sinx))# Cancelling out the
#sinx + 1# since it appears both in the numerator and in the denominator.#=cancel(sinx + 1)/(cosx(cancel(sinx + 1))# #=1/cosx# Applying the reciprocal identity
#1/costheta = sectheta# #=secx# Finally, on a last note, I know that here in Canada, British Columbia more specifically, these identities are given on a formula sheet, but I don't know what it's like elsewhere. In any event, many students, me included, memorize these identities because they're that important to mathematics. I would highly recommend memorization.
Practice exercises:
Simplify the following expressions:
a)
#cosalpha + tan alphasinalpha# b)
#cscx/sinx  cotx/tanx# c)
#sin^4theta  cos^4theta# d)
#(tan beta + cot beta)/csc^2beta# Hopefully this helps, and good luck!

Even & Odd Functions
A function
#f(x)# is said to be#{("even if "f(x)=f(x)),("odd if "f(x)=f(x)):}# Note that the graph of an even function is symmetric about the
#y# axis, and the graph of an odd function is symmetric about the origin.
Examples
#f(x)=x^4+3x^2+5# is an even function since#f(x)=(x)^4+(x)^2+5=x^4+3x^2+5=f(x)# #g(x)=x^5x^3+2x# is an odd function since#g(x)=(x)^5(x)^3+2(x)=x^5+x^32x=f(x)#
I hope that this was helpful.

Divide the fundamental identity
# sin^2x + cos^2x = 1# by#sin^2x# or#cos^2x# to derive the other two:#sin^2x/sin^2x + cos^2x/sin^2x = 1/sin^2x# #1 + cot^2x = csc^2x# #sin^2x/cos^2x + cos^2x/cos^2x = 1/cos^2x# #tan^2x + 1 = sec^2x#