Question #915b6

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Jan 17, 2016

Use the identity #sin^2theta+cos^2theta=1# and raise it to powers that would make the equation contain the desired trigonometric functions.

Explanation:

  • First: solving #(sin^2theta+cos^2theta)^3=(1)^3#
    #(sin^2theta+cos^2theta)(sin^4theta+2sin^2theta*cos^2theta+cos^4theta)=1#

#(sin^6theta+2sin^4theta*cos^2theta+sin^2theta*cos^4theta)+(cos^2theta*sin^4theta+2sin^2theta*cos^4theta+cos^6theta)=1#

#(sin^6theta+cos^6theta)+[(2sin^4theta*cos^2theta+2sin^2theta*cos^4theta)+(sin^2theta*cos^4theta+cos^2theta*sin^4theta)]=1#

#(sin^6theta+cos^6theta)+[sin^2theta*cos^2theta*((2sin^2theta+2cos^2theta)+(cos^2theta+sin^2theta))]=1#

#(sin^6theta+cos^6theta)+[sin^2theta*cos^2theta*((2)+(1))]=1#
#(sin^6theta+cos^6theta)+3[sin^2theta*cos^2theta]=1#

#:.sin^6theta+cos^6theta=1-3[sin^2theta*cos^2theta]#

  • Using a the same method starting with #(sin^2theta+cos^2theta)^2=(1)^2#
    you get:
    #sin^4theta+cos^4theta=1-2[sin^2theta*cos^2theta]#

  • The left side of the equation #=#
    #2(sin^6theta+cos^6theta)-3(sin^4theta+cos^4theta)+1#
    #=2(1-3sin^2theta*cos^2theta)-3(1-2sin^2theta*cos^2theta)+1#
    #=2cancel(-6sin^2theta*cos^2theta)-3+cancel(6sin^2theta*cos^2theta)+1#
    #=2-3+1 =0#

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