Question

If $\cos x + {\cos ^2}x = 1$ , then the value of ${\sin ^4}x + {\sin ^6}x$ is
a) $- 1 + \sqrt 5$
b) $\dfrac{{ - 1 - \sqrt 5 }}{2}$
c) $\dfrac{{1 - \sqrt 5 }}{2}$
d) $\dfrac{{ - 1 + \sqrt 5 }}{2}$

Answer 1

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.