Question

The value of $2\left( {{{\sin }^6}\theta + {{\cos }^6}\theta } \right) - 3\left( {{{\sin }^4}\theta + {{\cos }^4}\theta } \right) + 1$ is
A. 2
B. 0
C. 4
D. 6

Answer 1

Here in this question, we have to find the exact value of a given trigonometric function. we get the required value. For this, first we rewrite the trigonometric ratio in given function as by power of power rule of exponent then expand the function using a algebraic identities like${\left( {a + b} \right)^3} = {a^3} + {b^3} + 3ab\left( {a + b} \right)$and ${\left( {a + b} \right)^2} = {a^2} + {b^2} + 2ab$. Then further simplify by using trigonometric identities to get the required solution.

Complete step by step answer:
A function of an angle expressed as the ratio of two of the sides of a right triangle that contains that angle; the sine, cosine, tangent, cotangent, secant, or cosecant known as trigonometric function Also called circular function.Consider the given question:
$2\left( {{{\sin }^6}\theta + {{\cos }^6}\theta } \right) - 3\left( {{{\sin }^4}\theta + {{\cos }^4}\theta } \right) + 1$ --------(1)
Rewrite the each trigonometric ratio by power of power rule of exponent i.e., ${\left( {{a^m}} \right)^n} = {a^{mn}}$, then
$\Rightarrow \,\,\,2\left( {{{\left( {{{\sin }^2}\theta } \right)}^3} + {{\left( {{{\cos }^2}\theta } \right)}^3}} \right) - 3\left( {{{\left( {{{\sin }^2}\theta } \right)}^2} + {{\left( {{{\cos }^2}\theta } \right)}^2}} \right) + 1$ -----(2)

We know the algebraic identities,
${\left( {a + b} \right)^3} = {a^3} + {b^3} + 3ab\left( {a + b} \right) \Rightarrow {a^3} + {b^3} = {\left( {a + b} \right)^3} - 3ab\left( {a + b} \right)$
${\left( {a + b} \right)^2} = {a^2} + {b^2} + 2ab \Rightarrow {a^2} + {b^2} = {\left( {a + b} \right)^2} - 2ab$
Here, $a = \sin \theta$ and $b = \cos \theta$, then
On applying the algebraic identities in equation (2), then we have $\Rightarrow \,\,\,2\left( {{{\left( {{{\sin }^2}\theta + {{\cos }^2}\theta } \right)}^3} - 3{{\sin }^2}\theta \cdot {{\cos }^2}\theta \left( {{{\sin }^2}\theta + {{\cos }^2}\theta } \right)} \right) - 3\left( {{{\left( {{{\sin }^2}\theta + {{\cos }^2}\theta } \right)}^2} - 2{{\sin }^2}\theta \cdot {{\cos }^2}\theta } \right) + 1$

As we know the standard trigonometric identity: ${\sin ^2}\theta + {\cos ^2}\theta = 1$, then the above equation becomes
$\Rightarrow \,\,\,2\left( {{{\left( 1 \right)}^3} - 3{{\sin }^2}\theta \cdot {{\cos }^2}\theta \left( 1 \right)} \right) - 3\left( {{{\left( 1 \right)}^2} - 2{{\sin }^2}\theta \cdot {{\cos }^2}\theta } \right) + 1$
$\Rightarrow \,\,\,2\left( {1 - 3{{\sin }^2}\theta \cdot {{\cos }^2}\theta } \right) - 3\left( {1 - 2{{\sin }^2}\theta \cdot {{\cos }^2}\theta } \right) + 1$
$\Rightarrow \,\,\,2 - 6{\sin ^2}\theta \cdot {\cos ^2}\theta - 3 + 6{\sin ^2}\theta \cdot {\cos ^2}\theta + 1$
$\Rightarrow \,\,\,3 - 6{\sin ^2}\theta \cdot {\cos ^2}\theta - 3 + 6{\sin ^2}\theta \cdot {\cos ^2}\theta$
On simplification we get
$\therefore \,\,\,\,0$
Hence, the required value is 0.

Therefore, option B is the correct answer.

Note: When solving the trigonometry-based questions, we have to know the definitions of all six trigonometric ratios sine, cosine, tangent, secant, cosecant and cotangent and know the standard three trigonometric identities and student should know the basic algebraic identities and choose it depending upon the question. The use of algebraic identities makes the solution easy.