Question

If sinθ+cosθ=x prove that sin6θ+cos6θ=4-3(x²-1)²/4

Answer 1

$\textbf{Given:}$

$x=sin\,\theta+cos\,\theta$

$\text{squaring on both sides, we get}$

$\implies\,x^2=(sin\,\theta+cos\,\theta)^2$

$\implies\,x^2=1+2\;sin\,\theta\,cos\,\theta$

$\implies\,x^2-1=2\;sin\,\theta\,cos\,\theta$...........(1)

$\text{Now,}$

$sin^6\theta+cos^6\theta$

$=(sin^2\theta)^3+(cos^2\theta)^3$

$\text{Using,}$

$\boxed{\bf\,a^3+b^3=(a+b)^3-3ab(a+b)}$

$=(sin^2\theta+cos^2\theta)^3-3\,sin^2\theta\,cos^2\theta$

$=(sin^2\theta+cos^2\theta)^3-3\,sin^2\theta\,cos^2\theta(sin^2\theta+cos^2\theta)$

$=1^3-3\,sin^2\theta\,cos^2\theta(1)$

$=1-\frac{3}{4}(4\,sin^2\theta\,cos^2\theta)$

$=1-\frac{3}{4}(2\,sin\theta\,cos\theta)^2$

$=1-\frac{3}{4}(x^2-1)^2$ $\text{Using (1)}$

$=1-\displaystyle\frac{3(x^2-1)^2}{4}$

$=\displaystyle\frac{4-3(x^2-1)^2}{4}$

$\therefore\;\bf\,sin^6\theta+cos^6\theta=\frac{4-3(x^2-1)^2}{4}$

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