Here is the expression:
$$2(\sin^6x + \cos^6x) - 3(\sin^4 x + \cos^4 x) + 1$$
The exercise is to evaluate it. In my text book the answer is $0$ I tried to factor the expression, but it got me nowhere.
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Jean-Claude Arbaut
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Planet_Earth
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2What is the expression equal to? Right now it's just an expression, not an equation; there's no meaningful way to solve it. – DylanSp Mar 23 '16 at 13:14
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what do you want to achieve? – user190080 Mar 23 '16 at 13:14
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The exercise is to evaluate the expression, the answer in my textbook is 0. – Planet_Earth Mar 23 '16 at 13:16
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Evaluate it for what value of $x$? – DylanSp Mar 23 '16 at 13:16
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@DylanSp: I expect that the result follows for all values $x$... – abiessu Mar 23 '16 at 13:16
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so you actually want to investigate the function $f(x)=2(\sin^6x + \cos^6x) - 3(\sin^4 x + \cos^4 x) + 1$ – user190080 Mar 23 '16 at 13:17
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I will translate exactly the words from my textbook, since it isn't in eng. The value of the expression (from the answer) is? – Planet_Earth Mar 23 '16 at 13:17
5 Answers
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$$\sin^4 x+\cos^4 x=(\sin^2 x+\cos^2 x)^2-2\sin^2x\cos^2x=1-2\sin^2x\cos^2x$$ $$\sin^6 x+\cos^6 x=(\sin^2 x+\cos^2 x)^3-3\sin^2 x\cos^2 x(\sin^2 x+\cos^2 x)=1-3\sin^2x\cos^2x$$
Thus your expression simplifies to
$$2-6\sin^2x\cos^2x-3+6\sin^2x\cos^2x+1$$
Which is $0$, for all $x$.
Jean-Claude Arbaut
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This is it, I will give you best answer when I can(~7 mins). Thank you. – Planet_Earth Mar 23 '16 at 13:19
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$$2(\sin^6x +\cos^6x) - 3(\sin^4x+\cos^4x)+1=$$ $$=2(\sin^2x +\cos^2x)(\sin^4x-\sin^2x \cos^2 x+\cos^4x) - 3(\sin^4x+\cos^4x)+1=$$ $$=2(\sin^4x-\sin^2x \cos^2 x+\cos^4x) - 3(\sin^4x+\cos^4x)+1=$$ $$=-2\sin^2x \cos^2 x -\sin^4x-\cos^4x+1=$$ $$-(\sin^2x +\cos^2x)^2+1=-1+1=0$$
Roman83
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Let $t=\sin^2(x)$.
$$2(t^3+(1-t)^3)-3(t^2+(1-t)^2)+1=6t^2-6t+2-6t^2+6t-3+1=0.$$
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$$ \begin{align} 1 &=\left(\sin^2(x)+\cos^2(x)\right)^3\\ &=\sin^6(x)+3\sin^4(x)\cos^2(x)+3\sin^2(x)\cos^4(x)+\cos^6(x)\\ &=\sin^6(x)+3\sin^2(x)\cos^2(x)+\cos^6(x)\tag{1}\\\\ 1 &=\left(\sin^2(x)+\cos^2(x)\right)^2\\ &=\sin^4(x)+2\sin^2(x)\cos^2(x)+\cos^4(x)\tag{2} \end{align} $$ Subtracting $3$ times $(2)$ from $2$ times $(1)$ gives $-1$. Therefore, $$ 2\left(\sin^6(x)+\cos^6(x)\right)-3\left(\sin^4(x)+\cos^4(x)\right)+1=0\tag{3} $$
robjohn
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Consider $a^3+b^3=(a+b)^3-3ab(a+b)$ and $a^2+b^2=(a+b)^2-2ab$. For $a=\sin^2x$ and $b=\cos^2x$ we have $$ 2(\sin^6x+\cos^6x)-3(\sin^4x+\cos^4x)+1= 2(a+b)^3-6ab(a+b)-3(a+b)^2+6ab+1 $$ However, $a+b=\sin^2x+\cos^2x=1$, so the expression simplifies to $$ 2-6ab-3+6ab+1=0 $$
egreg
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