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Question #915b6

1 Answer

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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