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Find the value of \[{{\cos }^{6}}x-4{{\cos }^{4}}x+8{{\cos }^{2}}x\] , If \[\sin x+{{\sin}^ {2}}x+{{\sin }^{3}}x=1\],
(a) 0
(b) 2
(c) 4
(d) 8

Answer
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Hint: To find the value of given expression, use the identity relating \[\cos x\] and \[\sin x\] as \[{{\sin }^{2}}x+{{\cos }^{2}}x=1\]. Rearrange the terms and solve the expression using trigonometric identities to calculate the value of the given expression.

Complete step-by-step answer:
We know that \[\sin x+{{\sin }^{2}}x+{{\sin }^{3}}x=1\]. We have to calculate the value of \[{{\cos }^{6}}x-4{{\cos }^{4}}x+8{{\cos }^{2}}x\].
We will use the identity \[{{\sin }^{2}}x+{{\cos }^{2}}x=1\] relating \[\cos x\] and \[\sin x\].
We can rewrite the identity \[{{\sin }^{2}}x+{{\cos }^{2}}x=1\] as \[\cos x=\sqrt{1-{{\sin }^{2}}x}\].
Substituting the equation \[\cos x=\sqrt{1-{{\sin }^{2}}x}\] in the expression \[{{\cos }^{6}}x-4{{\cos }^{4}}x+8{{\cos }^{2}}x\], we have \[{{\left( \sqrt{1-{{\sin }^{2}}x} \right)}^{6}}-4{{\left( \sqrt{1-{{\sin }^{2}}x} \right)}^{4}}+8{{\left( \sqrt{1-{{\sin }^{2}}x} \right)}^{2}}\].
Further simplifying the above expression, we have \[{{\left( 1-{{\sin }^{2}}x \right)}^{3}}-4{{\left( 1-{{\sin }^{2}}x \right)}^{2}}+8\left( 1-{{\sin }^{2}}x \right)\].
We know that \[{{\left( a+b \right)}^{3}}={{a}^{3}}+3{{a}^{2}}b+3a{{b}^{2}}+{{b}^{3}}\] and \[{{\left( a+b \right)}^{2}}={{a}^{2}}+2ab+{{b}^{2}}\].
Thus, expanding the expression \[{{\left( 1-{{\sin }^{2}}x \right)}^{3}}-4{{\left( 1-{{\sin }^{2}}x \right)}^{2}}+8\left( 1-{{\sin }^{2}}x \right)\], we have \[1-3{{\sin }^{2}}x+3{{\sin }^{4}}x-{{\sin }^{6}}x-4-4{{\sin }^{4}}x+8{{\sin }^{2}}x+8-8{{\sin }^{2}}x\].
So, we have \[{{\cos }^{6}}x-4{{\cos }^{4}}x+8{{\cos }^{2}}x=-{{\sin }^{6}}x-{{\sin }^{4}}x-3{{\sin }^{2}}x+5.....\left( 1 \right)\].
We can rewrite the expression \[\sin x+{{\sin }^{2}}x+{{\sin }^{3}}x=1\] as \[\sin x+{{\sin }^{3}}x=1-{{\sin }^{2}}x\].
Squaring the equation \[\sin x+{{\sin }^{3}}x=1-{{\sin }^{2}}x\] on both sides, we have \[{{\left( \sin x+{{\sin }^{3}}x \right)}^{2}}={{\left( 1-{{\sin }^{2}}x \right)}^{2}}\].
We know that \[{{\left( a+b \right)}^{2}}={{a}^{2}}+2ab+{{b}^{2}}\].
Thus, we have \[{{\sin }^{2}}x+2{{\sin }^{4}}x+{{\sin }^{6}}x=1-2{{\sin }^{2}}x+{{\sin }^{4}}x\].
Further simplifying the above expression, we have \[-{{\sin }^{6}}x-{{\sin }^{4}}x-3{{\sin }^{2}}x=-1.....\left( 2 \right)\].
Substituting the value of equation (2) in equation (1), we have \[{{\cos }^{6}}x-4{{\cos }^{4}}x+8{{\cos }^{2}}x=5-1=4\].
Hence, the value of expression \[{{\cos }^{6}}x-4{{\cos }^{4}}x+8{{\cos }^{2}}x\] is 4, which is option (c).

Note: Trigonometric functions are real functions which relate any angle of a right angled triangle to the ratios of any two of its sides. The widely used trigonometric functions are sine, cosine and tangent. However, we can also use their reciprocals, i.e., cosecant, secant and cotangent. We can use geometric definitions to express the value of these functions on various angles using unit circle (circle with radius 1). We also write these trigonometric functions as infinite series or as solutions to differential equations. Thus, allowing us to expand the domain of these functions from the real line to the complex plane. One should be careful while using the trigonometric identities and rearranging the terms to convert from one trigonometric function to the other one.