Answer
Verified
376.5k+ views
Hint: Here in this question we solve quadratic equations in terms of cosx and find the value of it. Using the relation between cos and sin functions we further evaluate the value of given expression which is in terms of sin.
Complete step by step solution:
In the trigonometry we have six trigonometry ratios namely sine , cosine, tangent, cosecant, secant and cotangent. These are abbreviated as sin, cos, tan, csc, sec and cot. The 3 trigonometry ratios are reciprocal of the other trigonometry ratios. Here cosine is the reciprocal of the sine. The secant is the reciprocal of the cosine. The cotangent is the reciprocal of the tangent.
Now consider the given equation
\[\cos x + {\cos ^2}x = 1\]------ (1)
This can be written as
\[ \Rightarrow {\cos ^2}x + \cos x - 1 = 0\]
The above equation is in the form of quadratic equation in \[\cos x\]
By using the formula
\[ \Rightarrow \cos x = \dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}\]
Here a = 1, b = 1 and c = -1
\[ \Rightarrow \cos x = \dfrac{{ - 1 \pm \sqrt {{1^2} - 4(1)( - 1)} }}{{2(1)}}\]
On simplifying we have
\[ \Rightarrow \cos x = \dfrac{{ - 1 \pm \sqrt {1 + 4} }}{2}\]
\[ \Rightarrow \cos x = \dfrac{{ - 1 \pm \sqrt 5 }}{2}\]
Since \[\cos x \ne \dfrac{{ - 1 - \sqrt 5 }}{2}\]
Therefore
\[ \Rightarrow \cos x = \dfrac{{ - 1 + \sqrt 5 }}{2}\] --- (2)
Now consider the equation (1) we have
\[ \Rightarrow \cos x + {\cos ^2}x = 1\]
Take \[{\cos ^2}x\] to RHS we have
\[ \Rightarrow \cos x = 1 - {\cos ^2}x\]
From the trigonometric identity we have \[{\sin ^2}x = 1 - {\cos ^2}x\]
On substituting these we have
\[ \Rightarrow \cos x = {\sin ^2}x\]---- (3)
Now we consider
\[ \Rightarrow {\sin ^4}x + {\sin ^6}x\]
This can be written as
\[ \Rightarrow {\left( {{{\sin }^2}x} \right)^2} + {\left( {{{\sin }^2}x} \right)^3}\]------ (4)
Substituting (3) in (4) we have
\[ \Rightarrow {\cos ^2}x + {\cos ^3}x\]
Take \[{\cos ^2}x\] as a common and it is written as
\[ \Rightarrow {\cos ^2}x(1 + \cos x)\]
On substituting (2) to above equation we have
\[ \Rightarrow {\left( {\dfrac{{ - 1 + \sqrt 5 }}{2}} \right)^2}\left( {1 + \dfrac{{ - 1 + \sqrt 5 }}{2}} \right)\]
On simplifying we have
\[ \Rightarrow \dfrac{{{{\left( { - 1 + \sqrt 5 } \right)}^2}}}{4}\left( {\dfrac{{1 + \sqrt 5 }}{2}} \right)\]
On squaring we have
\[ \Rightarrow \dfrac{{1 + 5 - 2\sqrt 5 }}{4}\left( {\dfrac{{1 + \sqrt 5 }}{2}} \right)\]
On simplifying the first term we have
\[ \Rightarrow \dfrac{{6 - 2\sqrt 5 }}{4}\left( {\dfrac{{1 + \sqrt 5 }}{2}} \right)\]
\[ \Rightarrow \dfrac{{3 - \sqrt 5 }}{2}\left( {\dfrac{{1 + \sqrt 5 }}{2}} \right)\]
On multiplying two terms we have
\[ \Rightarrow \dfrac{1}{4}\left( {3 + 3\sqrt 5 - \sqrt 5 - 5} \right)\]
On simplifying the above inequality we have
\[ \Rightarrow \dfrac{1}{4}\left( { - 2 + 2\sqrt 5 } \right)\]
On further simplifying we have
\[ \Rightarrow \dfrac{{ - 1 + \sqrt 5 }}{2}\]
So, the correct answer is “Option d”.
Note: The question involves the trigonometric functions and we have to find the value trigonometric function given in terms of sin. Students might get confused as to how to find expressions given in terms of sin when we get value in terms of cos, but remember sin and cos are related in many ways by different identities.
Complete step by step solution:
In the trigonometry we have six trigonometry ratios namely sine , cosine, tangent, cosecant, secant and cotangent. These are abbreviated as sin, cos, tan, csc, sec and cot. The 3 trigonometry ratios are reciprocal of the other trigonometry ratios. Here cosine is the reciprocal of the sine. The secant is the reciprocal of the cosine. The cotangent is the reciprocal of the tangent.
Now consider the given equation
\[\cos x + {\cos ^2}x = 1\]------ (1)
This can be written as
\[ \Rightarrow {\cos ^2}x + \cos x - 1 = 0\]
The above equation is in the form of quadratic equation in \[\cos x\]
By using the formula
\[ \Rightarrow \cos x = \dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}\]
Here a = 1, b = 1 and c = -1
\[ \Rightarrow \cos x = \dfrac{{ - 1 \pm \sqrt {{1^2} - 4(1)( - 1)} }}{{2(1)}}\]
On simplifying we have
\[ \Rightarrow \cos x = \dfrac{{ - 1 \pm \sqrt {1 + 4} }}{2}\]
\[ \Rightarrow \cos x = \dfrac{{ - 1 \pm \sqrt 5 }}{2}\]
Since \[\cos x \ne \dfrac{{ - 1 - \sqrt 5 }}{2}\]
Therefore
\[ \Rightarrow \cos x = \dfrac{{ - 1 + \sqrt 5 }}{2}\] --- (2)
Now consider the equation (1) we have
\[ \Rightarrow \cos x + {\cos ^2}x = 1\]
Take \[{\cos ^2}x\] to RHS we have
\[ \Rightarrow \cos x = 1 - {\cos ^2}x\]
From the trigonometric identity we have \[{\sin ^2}x = 1 - {\cos ^2}x\]
On substituting these we have
\[ \Rightarrow \cos x = {\sin ^2}x\]---- (3)
Now we consider
\[ \Rightarrow {\sin ^4}x + {\sin ^6}x\]
This can be written as
\[ \Rightarrow {\left( {{{\sin }^2}x} \right)^2} + {\left( {{{\sin }^2}x} \right)^3}\]------ (4)
Substituting (3) in (4) we have
\[ \Rightarrow {\cos ^2}x + {\cos ^3}x\]
Take \[{\cos ^2}x\] as a common and it is written as
\[ \Rightarrow {\cos ^2}x(1 + \cos x)\]
On substituting (2) to above equation we have
\[ \Rightarrow {\left( {\dfrac{{ - 1 + \sqrt 5 }}{2}} \right)^2}\left( {1 + \dfrac{{ - 1 + \sqrt 5 }}{2}} \right)\]
On simplifying we have
\[ \Rightarrow \dfrac{{{{\left( { - 1 + \sqrt 5 } \right)}^2}}}{4}\left( {\dfrac{{1 + \sqrt 5 }}{2}} \right)\]
On squaring we have
\[ \Rightarrow \dfrac{{1 + 5 - 2\sqrt 5 }}{4}\left( {\dfrac{{1 + \sqrt 5 }}{2}} \right)\]
On simplifying the first term we have
\[ \Rightarrow \dfrac{{6 - 2\sqrt 5 }}{4}\left( {\dfrac{{1 + \sqrt 5 }}{2}} \right)\]
\[ \Rightarrow \dfrac{{3 - \sqrt 5 }}{2}\left( {\dfrac{{1 + \sqrt 5 }}{2}} \right)\]
On multiplying two terms we have
\[ \Rightarrow \dfrac{1}{4}\left( {3 + 3\sqrt 5 - \sqrt 5 - 5} \right)\]
On simplifying the above inequality we have
\[ \Rightarrow \dfrac{1}{4}\left( { - 2 + 2\sqrt 5 } \right)\]
On further simplifying we have
\[ \Rightarrow \dfrac{{ - 1 + \sqrt 5 }}{2}\]
So, the correct answer is “Option d”.
Note: The question involves the trigonometric functions and we have to find the value trigonometric function given in terms of sin. Students might get confused as to how to find expressions given in terms of sin when we get value in terms of cos, but remember sin and cos are related in many ways by different identities.
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