Question

${\huge{\boxed{ \mathcal{\orange{Hello!}}}}}$
$\large\mathcal\red{Please\:solve\:this..✍}$
PROVE;
$\small\mathfrak\orange{sin⁸ø\:-\:cos⁸ø\:=\:(sin²ø\:-\:cos²ø)}$
​

Answer 1

Correct Statement is

To prove :-

$\sf \:  sin⁸ø\:-\:cos⁸ø\:=(\:sin²ø\:-\:cos²ø)(1 - 2 {sin}^{2} ø {cos}^{2} ø)$

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$\begin{gathered}\Large{\bold{{\underline{Formula \: Used \::}}}} \end{gathered}$

$\bf \:  ⟼ {x}^{2} - {y}^{2} = (x + y)(x - y)$

$\bf \:  ⟼ {x}^{2} + {y}^{2} = {(x + y)}^{2} - 2xy$

$\bf \:  ⟼ {sin}^{2} ø + {cos}^{2} ø = 1$

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Proof :-

Proof :- Consider LHS :-

$\bf \:  sin⁸ø\:-\:cos⁸ø\:$

$\bf \:  = { ({sin}^{4}ø) }^{2} - { ({cos}^{4}ø )}^{2}$

$\sf \:  = ( {sin}^{4} ø + {cos}^{4} ø)( {sin}^{4} ø - {cos}^{4} ø)$

$\sf \:  = \bigg( { ({sin}^{2}ø) }^{2} + { ({cos}^{2}ø )}^{2} \bigg )\bigg( { ({sin}^{2}ø) }^{2} - { ({cos}^{2}ø )}^{2} \bigg )$

$\sf \:  = \bigg(\bigg( { {sin}^{2}ø } + { {cos}^{2}ø \bigg )}^{2} - 2 {sin}^{2} ø {cos}^{2} ø \bigg)( {sin}^{2} ø + {cos}^{2} ø)( {sin}^{2} ø - {cos}^{2} ø)$

$\sf \:  = (1 - 2 {sin}^{2} ø {cos}^{2} ø)( {sin}^{2} ø - {cos}^{2} ø)$

$\large{\boxed{\boxed{\bf{Hence, Proved}}}}$

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$\large {\bf \: \: \: \: \: \: \: \: \: \: \: \: ⟼ Explore \: more } ✍$

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Additional Information:-

Additional Information:- Relationship between sides and T ratios

sin θ = Opposite Side/Hypotenuse

cos θ = Adjacent Side/Hypotenuse

tan θ = Opposite Side/Adjacent Side

sec θ = Hypotenuse/Adjacent Side

cosec θ = Hypotenuse/Opposite Side

cot θ = Adjacent Side/Opposite Side

Reciprocal Identities

cosec θ = 1/sin θ

sec θ = 1/cos θ

cot θ = 1/tan θ

sin θ = 1/cosec θ

cos θ = 1/sec θ

tan θ = 1/cot θ

Co-function Identities

sin (90°−x) = cos x

cos (90°−x) = sin x

tan (90°−x) = cot x

cot (90°−x) = tan x

sec (90°−x) = cosec x

cosec (90°−x) = sec x

Fundamental Trigonometric Identities

sin²θ + cos²θ = 1

sec²θ - tan²θ = 1

cosec²θ - cot²θ = 1

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$</p><p>\begin{gathered}\begin{gathered}\begin{gathered}\begin{gathered}\bf Trigonometry\: Table \\ \begin{gathered}\begin{gathered}\begin{gathered}\begin{gathered}\boxed{\boxed{\begin{array}{ |c |c|c|c|c|c|} \bf\angle A & \bf{0}^{ \circ} & \bf{30}^{ \circ} & \bf{45}^{ \circ} & \bf{60}^{ \circ} & \bf{90}^{ \circ} \\ \\ \rm sin A & 0 & \dfrac{1}{2}& \dfrac{1}{ \sqrt{2} } & \dfrac{ \sqrt{3}}{2} &1 \\ \\ \rm cos \: A & 1 & \dfrac{ \sqrt{3} }{2}& \dfrac{1}{ \sqrt{2} } & \dfrac{1}{2} &0 \\ \\ \rm tan A & 0 & \dfrac{1}{ \sqrt{3} }&1 & \sqrt{3} & \rm \infty \\ \\ \rm cosec A & \rm \infty & 2& \sqrt{2} & \dfrac{2}{ \sqrt{3} } &1 \\ \\ \rm sec A & 1 & \dfrac{2}{ \sqrt{3} }& \sqrt{2} & 2 & \rm \infty \\ \\ \rm cot A & \rm \infty & \sqrt{3} & 1 & \dfrac{1}{ \sqrt{3} } & 0\end{array}}}\end{gathered}\end{gathered}\end{gathered} \end{gathered}\end{gathered}\end{gathered}\end{gathered}\end{gathered}$