Relating Trigonometric Functions

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The Reciprocal, Quotient, and Pythagorean Identities
9:33 — by James S.

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Key Questions

  • The reciprocal functions are as follows:

    #sin(a) * csc(a) = 1#

    #cos(a) * sec(a) = 1#

    #tan(a) * cot(a) = 1#

  • In below proof ^2 denotes square.......
    By reciprocal identities-
    sin x= 1/cosec x and cos x= 1/sec x
    Also by pythoragas identities-
    sin^2 x + cos^2 x=1 ...................... -eqn 1
    1 + tan^2 x= sec^2 x ..................... -eqn 2

    Now putting sec x= 1/cos x in eqn 2
    1 + tan^2 x= 1/ cos^2 x
    i.e, tan^2 x= -1 + 1/cos^2 x
    i.e, tan^2 x= (1 - cos^2 x)/ cos^2 x
    i.e, tan^2 x= sin^2 x/ cos^2 x...................... using eqn 1
    i.e, tan x= sin x/cos x
    Similarly we can prove cot x= cosx/sinx

    And hence tan x= sin x/cos x and cot x= cosx/sinx are the quotient identities......

  • In function of trigonometry the elements of first set which have relation with second elements are domain and elements of second set which have relation with first set are range. BUt in function there is co-domain als which can be defined as all the elements of secon set .

  • It means to determine if the value of a trigonometric function is positive or negative; for example, since #sin({3pi}/2)=-1<0#, its sign is negative, and since #cos(-pi/3)=1/2>0#, its sign is positive.


    I hope that this was helpful.

  • The Pythagorean identity is:

    #color(red)(sin^2x+cos^2x=1#

    However, it does not have to apply to just sine and cosine.

    To find the form of the Pythagorean identity with the other trigonometric identities, divide the original identity by sine and cosine.

    SINE:

    #(sin^2x+cos^2x=1)/sin^2x#

    This gives:

    #sin^2x/sin^2x+cos^2x/sin^2x=1/sin^2x#

    Which equals

    #color(red)(1+cot^2x=csc^2x#

    To find the other identity:

    COSINE:

    #(sin^2x+cos^2x=1)/cos^2x#

    This gives:

    #sin^2x/cos^2x+cos^2x/cos^2x=1/cos^2x#

    Which equals

    #color(red)(tan^2x+1=sec^2x#

    These identities can all be algebraically manipulated to prove many things:

    #{(sin^2x=1-cos^2x),(cos^2x=1-sin^2x):}#

    #{(tan^2x=sec^2x-1),(cot^2x=csc^2x-1):}#

Questions

  • Barney V. answered · 1 week ago