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If $\text{sinA}\ \text{+}\ \text{si}{{\text{n}}^{2}}\text{A+si}{{\text{n}}^{3}}\text{A=1,}$ then find the value of ${{\cos }^{6}}\text{A}\ -\ 4{{\cos }^{4}}\text{A}\ \text{+}\ \text{8co}{{\text{s}}^{2}}\text{A}$is:-A). 1.B). 2.C). 3.D). 4

Last updated date: 24th Jun 2024
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Hint: Try using the equation given to us,$\text{sinA}\ \text{+}\ \text{si}{{\text{n}}^{2}}\text{A+si}{{\text{n}}^{3}}\text{A=1,}$you can use the basic trigonometry property, ${{\sin }^{2}}\text{A}\ \text{+}\ \text{co}{{\text{s}}^{2}}\text{A}\ \text{=}\ \text{1,}$ which will be helpful in getting to the solution.

We need to find the value of ${{\cos }^{6}}\text{A}\ \text{- 4co}{{\text{s}}^{4}}\text{A}\ \text{+}\ \text{8co}{{\text{s}}^{2}}\text{A}\text{.}$ Let us use the equation given to us.
$\Rightarrow \ \sin \text{A}\ \text{+}\ \text{si}{{\text{n}}^{2}}\text{A}\ \text{+}\ \text{si}{{\text{n}}^{3}}\text{A}\ \text{=1}$
Taking ${{\sin }^{2}}\text{A}$ to the right hand side,
$\Rightarrow \ \text{sinA}\ \text{+}\ \text{si}{{\text{n}}^{\text{3}}}\text{A}\ \text{=}\ \text{1}\ \text{-}\ \text{si}{{\text{n}}^{\text{2}}}\text{A}$
$\Rightarrow \ \text{sinA}\left( 1+\text{si}{{\text{n}}^{\text{2}}}\text{A} \right)=\ \text{co}{{\text{s}}^{\text{2}}}\text{A}$
Here we can substitute the value of $\text{si}{{\text{n}}^{\text{2}}}\text{A}\ \ \ \text{as}\ \ \ \ \text{1-}\ \text{co}{{\text{s}}^{\text{2}}}\text{A}$
Therefore we will get,
$\text{sinA}\ \left( 1+1-\text{co}{{\text{s}}^{\text{2}}}\text{A} \right)=\ \text{co}{{\text{s}}^{\text{2}}}\text{A}$
$\text{sinA}\left( 2-\text{co}{{\text{s}}^{\text{2}}}\text{A} \right)={{\cos }^{2}}\left( \text{A} \right)$
Now squatting on both the sides,
$\text{si}{{\text{n}}^{\text{2}}}\text{A}{{\left( 2-\text{co}{{\text{s}}^{\text{2}}}\text{A} \right)}^{2}}\ =\ {{\cos }^{4}}A$
Again we can replace $\text{si}{{\text{n}}^{\text{2}}}\text{A}\ \text{by}\ \left( 1-\text{co}{{\text{s}}^{\text{2}}}\text{A} \right)$
$\Rightarrow \ \left( 1-\text{co}{{\text{s}}^{\text{2}}}\text{A} \right)\left( 4+\text{co}{{\text{s}}^{\text{4}}}\text{A}\ \text{-}\ \text{4co}{{\text{s}}^{\text{2}}}\text{A} \right)=\ \text{co}{{\text{s}}^{\text{4}}}\text{A}$
$\Rightarrow \ \text{4+co}{{\text{s}}^{\text{4}}}\text{A}\ \text{-}\ \text{4co}{{\text{s}}^{\text{2}}}\text{A}\ \text{-}\ \text{4co}{{\text{s}}^{\text{2}}}\text{A}\ \text{-}\ \text{co}{{\text{s}}^{\text{6}}}\text{A}\ \text{+}\ \text{4cos}{{\text{A}}^{\text{4}}}\text{A}\ \text{=}\ \text{co}{{\text{s}}^{\text{4}}}\text{A}$
Taking all the $\cos$ terms on are side,
$\text{co}{{\text{s}}^{\text{4}}}\text{A}\ \text{-}\ \text{co}{{\text{s}}^{\text{4}}}\text{A}\ \text{+4co}{{\text{s}}^{\text{2}}}\text{A}\ \text{+}\ \text{4co}{{\text{s}}^{\text{2}}}\text{A}\ \text{+}\ \text{co}{{\text{s}}^{\text{6}}}\text{A}\ \text{-4co}{{\text{s}}^{\text{4}}}\text{A}\ \text{=}\ \text{4}$
$\Rightarrow \ \text{co}{{\text{s}}^{\text{6}}}\text{A}\ \text{-}\ \text{4co}{{\text{s}}^{\text{4}}}\text{A}\ \text{+}\ \text{8co}{{\text{s}}^{\text{2}}}\text{A}\ \text{=}\ \text{4}$

Hence, our answer will be 4.

Note: In such questions rearrangement of the given equation always leads us to the desired equation. Carefully verify each step before moving into the next step. Additionally some basic trigonometric ideates to keep in mind would be,
${{\sin }^{2}}\theta \ +{{\cos }^{2}}\theta =1$
${{\tan }^{2}}\theta +1={{\sec }^{2}}\theta$
${{\cot }^{2}}\theta +1=\text{cose}{{\text{c}}^{\text{2}}}\theta$.