Answer
Verified
478.5k+ views
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.
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.
Recently Updated Pages
How many sigma and pi bonds are present in HCequiv class 11 chemistry CBSE
Mark and label the given geoinformation on the outline class 11 social science CBSE
When people say No pun intended what does that mea class 8 english CBSE
Name the states which share their boundary with Indias class 9 social science CBSE
Give an account of the Northern Plains of India class 9 social science CBSE
Change the following sentences into negative and interrogative class 10 english CBSE
Trending doubts
Fill the blanks with the suitable prepositions 1 The class 9 english CBSE
The Equation xxx + 2 is Satisfied when x is Equal to Class 10 Maths
In Indian rupees 1 trillion is equal to how many c class 8 maths CBSE
Which are the Top 10 Largest Countries of the World?
How do you graph the function fx 4x class 9 maths CBSE
Give 10 examples for herbs , shrubs , climbers , creepers
Difference Between Plant Cell and Animal Cell
Difference between Prokaryotic cell and Eukaryotic class 11 biology CBSE
Why is there a time difference of about 5 hours between class 10 social science CBSE