Question

If we have a trigonometric identity as $\cos x+{{\cos }^{2}}x=1$ , the find 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. $\dfrac{-1+\sqrt{5}}{2}$
E. $1-\sqrt{5}$

Answer 1

To find ${{\sin }^{4}}x+{{\sin }^{6}}x$ , let us write $\cos x+{{\cos }^{2}}x=1$ in the form of a quadratic equation. That is, ${{\cos }^{2}}x+\cos x-1=0$ . Let us find the roots using the formula $x=\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}$ . Here, $x=\cos x$ . Hence, $\cos x=\dfrac{-1\pm \sqrt{1-\left( 4\times 1\times -1 \right)}}{2\times 1}$ . On solving this, we will get $\cos x=\dfrac{-1+\sqrt{5}}{2},\dfrac{-1-\sqrt{5}}{2}$ . We can see that $\dfrac{-1-\sqrt{5}}{2}$ is not within the range of $\cos x$ , that is, [-1,1]. Hence, this value is neglected. We will solve $\cos x+{{\cos }^{2}}x=1$ by writing this in the form $\cos x=1-{{\cos }^{2}}x$ . On solving this, we will get $\cos x={{\sin }^{2}}x$ . Now, we will write ${{\sin }^{4}}x+{{\sin }^{6}}x$ as ${{\left( {{\sin }^{2}}x \right)}^{2}}+{{\left( {{\sin }^{2}}x \right)}^{3}}$ . By substituting the above values in this, we will get the required answer.

Complete step-by-step solution:
We have to find ${{\sin }^{4}}x+{{\sin }^{6}}x$ . It is given that $\cos x+{{\cos }^{2}}x=1$ .
Let us take 1 to the LHS. We will get
$\cos x+{{\cos }^{2}}x-1=0$
Let us rearrange this in the form of a quadratic equation.
$\Rightarrow {{\cos }^{2}}x+\cos x-1=0$
We know that for a quadratic equation $a{{x}^{2}}+bx+c=0$ , $x=\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}$
Here, $x=\cos x$ . Hence,
$\cos x=\dfrac{-1\pm \sqrt{1-\left( 4\times 1\times -1 \right)}}{2\times 1}$
Let us simplify the terms inside the root. We will get
$\begin{aligned} & \cos x=\dfrac{-1\pm \sqrt{1+4}}{2} \\\ & \Rightarrow \cos x=\dfrac{-1\pm \sqrt{5}}{2} \\\ & \Rightarrow \cos x=\dfrac{-1+\sqrt{5}}{2},\dfrac{-1-\sqrt{5}}{2} \\\ \end{aligned}$
We know the range of $\cos x$ is [-1,1] . The value of $\dfrac{-1+\sqrt{5}}{2}=0.61$ and $\dfrac{-1-\sqrt{5}}{2}=-1.61$
We can see that $\dfrac{-1-\sqrt{5}}{2}$ is not within [-1,1]. Hence, this value is neglected. Thus,
$\cos x=\dfrac{-1+\sqrt{5}}{2}...(i)$
Now, let us again consider $\cos x+{{\cos }^{2}}x=1$
Let us take ${{\cos }^{2}}x$ to RHS. We will get
$\cos x=1-{{\cos }^{2}}x...(ii)$
We know that
$\begin{aligned} & {{\cos }^{2}}x+{{\sin }^{2}}x=1 \\\ & \Rightarrow {{\cos }^{2}}x=1-{{\sin }^{2}}x \\\ \end{aligned}$
Hence, we can write (ii) as
$\cos x={{\sin }^{2}}x...(iii)$
Now, let’s consider ${{\sin }^{4}}x+{{\sin }^{6}}x$ .
We can write the above equation as shown below using the rule ${{\left( {{a}^{m}} \right)}^{n}}={{a}^{mn}}$
${{\left( {{\sin }^{2}}x \right)}^{2}}+{{\left( {{\sin }^{2}}x \right)}^{3}}$
From (iii), we can write the above equation as
$\begin{aligned} & {{\left( \cos x \right)}^{2}}+{{\left( \cos x \right)}^{3}} \\\ & ={{\cos }^{2}}x+{{\cos }^{3}}x \\\ \end{aligned}$
Let us take ${{\cos }^{2}}x$ outside.
${{\cos }^{2}}x\left( 1+\cos x \right)$
Let’s substitute the value of $\cos x$ from (i). We will get
${{\left( \dfrac{-1+\sqrt{5}}{2} \right)}^{2}}\left( 1+\dfrac{-1+\sqrt{5}}{2} \right)$
Let us solve this.
$\begin{aligned} & {{\left( \dfrac{-1+\sqrt{5}}{2} \right)}^{2}}\left( \dfrac{2-1+\sqrt{5}}{2} \right) \\\ & \Rightarrow {{\left( \dfrac{-1+\sqrt{5}}{2} \right)}^{2}}\left( \dfrac{1+\sqrt{5}}{2} \right) \\\ \end{aligned}$
We can write the above equation as
$\left( \dfrac{-1+\sqrt{5}}{2} \right)\left( \dfrac{-1+\sqrt{5}}{2} \right)\left( \dfrac{1+\sqrt{5}}{2} \right)$
We can rewrite the second and third terms as
$\left( \dfrac{-1+\sqrt{5}}{2} \right)\left( \dfrac{\sqrt{5}-1}{2} \right)\left( \dfrac{\sqrt{5}+1}{2} \right)$
Let’s combine second and third terms.
$\left( \dfrac{-1+\sqrt{5}}{2} \right)\left[ \dfrac{\left( \sqrt{5}-1 \right)\left( \sqrt{5}+1 \right)}{4} \right]$
We know that ${{a}^{2}}-{{b}^{2}}=\left( a+b \right)\left( a-b \right)$ . Hence,

& \left( \dfrac{-1+\sqrt{5}}{2} \right)\left[ \dfrac{{{\left( \sqrt{5} \right)}^{2}}-{{1}^{2}}}{4} \right] \\\ & \Rightarrow \left( \dfrac{-1+\sqrt{5}}{2} \right)\left[ \dfrac{5-1}{4} \right] \\\ & \Rightarrow \left( \dfrac{-1+\sqrt{5}}{2} \right)\left[ \dfrac{4}{4} \right] \\\ \end{aligned}$$ Let us simplify this further. We will get $$\begin{aligned} & \left( \dfrac{-1+\sqrt{5}}{2} \right)1 \\\ & =\dfrac{-1+\sqrt{5}}{2} \\\ \end{aligned}$$ **Hence, the correct option is D.** **Note:** You may take ${{\cos }^{2}}x$ as the value of a in the equation ${{\cos }^{2}}x+\cos x-1=0$ and $\cos x$ as the value of b to substitute in $x=\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}$ . This will lead to the wrong solution. It is important to check whether the value of $\cos x$ falls within its range. Also, rules of exponents must be known very well.