Fundamental Identities

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sin(A+B)=sin(A)cos(B)+cos(A)sin(B) proof - geometrical

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1 of 2 videos by Tiago Hands

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:

    •The reciprocal identities
    •The pythagorean identities
    •The quotient identities

    They are all shown in the following image:

    https://academics.utep.edu/Portals/1788/CALCULUS%20MATERIAL/5_1%20USING%20OF%20FUNDAMENTALS%20IDENTITIES.pdf

    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^5-x^3+2x# is an odd function since

    #g(-x)=(-x)^5-(-x)^3+2(-x)=-x^5+x^3-2x=-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#

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