Question

If ${{P}_{n}}={{\cos }^{n}}\theta +{{\sin }^{n}}\theta$ then $2{{P}_{6}}-3{{P}_{4}}+1=$?

& \left( a \right)2 \\\ & \left( b \right)3 \\\ & \left( c \right)0 \\\ & \left( d \right)1 \\\ \end{aligned}$$

Answer 1

In order to solve the given problem, we will be substituting the given P value in to $2{{P}_{6}}-3{{P}_{4}}+1$. After substituting the value, we must express it in the form of a trigonometric identity for solving it conveniently. And upon solving it we obtain the required answer.

Complete step by step answer:
Let us have a brief discussion regarding the trigonometric functions. The counter-clockwise angle between the initial arm and the terminal arm of an angle in standard position is called the principal angle. Its value is between ${{0}^{\circ }}$ and ${{360}^{\circ }}$. The relationship between the angles and sides of a triangle are given by the trigonometric functions. The basic trigonometric functions are sine, cosine, tangent, cotangent, secant and cosecant. These are the basic main trigonometric functions used.
Now let us find the value of $2{{P}_{6}}-3{{P}_{4}}+1$ when ${{P}_{n}}={{\cos }^{n}}\theta +{{\sin }^{n}}\theta$
Firstly, let us substitute the values. We get,
$2{{P}_{6}}-3{{P}_{4}}+1=2\left( {{\cos }^{6}}\theta +{{\sin }^{6}}\theta \right)-3\left( {{\cos }^{4}}\theta +{{\sin }^{4}}\theta \right)+1$
Now let us convert them into the form of ${{\sin }^{2}}\theta +{{\cos }^{2}}\theta =1$ so that our calculation would be easier.

& \Rightarrow 2\left( {{\cos }^{6}}\theta +{{\sin }^{6}}\theta \right)-3\left( {{\cos }^{4}}\theta +{{\sin }^{4}}\theta \right)+1 \\\ & \Rightarrow 2\left[{{\left( {{\cos }^{2}}\theta +{{\sin }^{2}}\theta \right)}^{3}}-3{{\cos }^{2}}\theta {{\sin }^{2}}\theta \left( {{\sin }^{2}}\theta +{{\cos }^{2}}\theta \right)\right]-3\left[ \left( {{\sin }^{2}}\theta +{{\cos }^{2}}\theta \right)-2{{\cos }^{2}}\theta {{\sin }^{2}}\theta \right]+1 \\\ & \\\ \end{aligned}$$Now upon solving this, we get $$\begin{aligned} & \Rightarrow 2\left( 1-3{{\cos }^{2}}\theta {{\sin }^{2}}\theta \right)-3\left( 1-2{{\cos }^{2}}\theta {{\sin }^{2}}\theta \right)+1 \\\ & \Rightarrow 2-6{{\cos }^{2}}\theta {{\sin }^{2}}\theta -3+6{{\cos }^{2}}\theta {{\sin }^{2}}\theta +1 \\\ & \Rightarrow 0 \\\ \end{aligned}$$ $$\therefore $$ The value of $$2{{P}_{6}}-3{{P}_{4}}+1$$ when $${{P}_{n}}={{\cos }^{n}}\theta +{{\sin }^{n}}\theta $$ is $$0$$. **So, the correct answer is “Option c”.** **Note:** We must be careful while expanding the functions into the required form as incorrectly expanding the functions is the commonly committed error. We must always try to expand it into the identities whose values we are aware of, for easy calculations. In the above problem, we have considered RHS for solving as given as substitution. But in some of the cases, we must be considering as per given.