Question

The value of $4\left( {{\sin }^{4}}{{30}^{\circ }}+{{\cos }^{4}}{{60}^{\circ }} \right)-3\left( {{\cos }^{2}}{{45}^{\circ }}+{{\sin }^{2}}{{90}^{\circ }} \right)$ is
$\left( A \right)\text{ }-\dfrac{1}{2}$
$\left( B \right)\text{ }-4$
$\left( C \right)\text{ 2}$
$\left( D \right)\text{ }\dfrac{1}{2}$

Answer 1

Hint: In this question we have been given with a trigonometric expression which we have to simplify and find its value. We will solve this question by substituting the values of the trigonometric functions and then expanding the terms. we will then simplify the terms to get the required value of the expression.

Complete step by step answer:
We have the expression given to us as:
$\Rightarrow 4\left( {{\sin }^{4}}{{30}^{\circ }}+{{\cos }^{4}}{{60}^{\circ }} \right)-3\left( {{\cos }^{2}}{{45}^{\circ }}+{{\sin }^{2}}{{90}^{\circ }} \right)$
We can see that all the values of the angles in the trigonometric functions is in angles therefore, we will substitute the values.
We know that:
$\Rightarrow \sin {{30}^{\circ }}=\dfrac{1}{2}$
$\Rightarrow \cos {{60}^{\circ }}=\dfrac{1}{2}$
$\Rightarrow \cos {{45}^{\circ }}=\dfrac{1}{\sqrt{2}}$
$\Rightarrow \sin {{90}^{\circ }}=1$
On substituting the values in the expression, we get:
$= 4\left( {{\left( \dfrac{1}{2} \right)}^{4}}+{{\left( \dfrac{1}{2} \right)}^{4}} \right)-3\left( {{\left( \dfrac{1}{\sqrt{2}} \right)}^{2}}+{{1}^{2}} \right)$
Now we can write the expression as:
$= 4\left( \dfrac{1}{{{2}^{4}}}+\dfrac{1}{{{2}^{4}}} \right)-3\left( \dfrac{1}{{{\left( \sqrt{2} \right)}^{2}}}+{{1}^{2}} \right)$
Now we know that ${{2}^{4}}=16$ and ${{\left( \sqrt{2} \right)}^{2}}=2$ therefore, on substituting, we get:
$= 4\left( \dfrac{1}{16}+\dfrac{1}{16} \right)-3\left( \dfrac{1}{2}+1 \right)$
Now on taking the lowest common multiple in both the fractions, we get:
$= 4\left( \dfrac{1+1}{16} \right)-3\left( \dfrac{1+2}{2} \right)$
On adding the terms, we get:
$= 4\left( \dfrac{2}{16} \right)-3\left( \dfrac{3}{2} \right)$
On simplifying the terms, we get:
$= \dfrac{2}{4}-3\left( \dfrac{3}{2} \right)$
On further simplifying and multiplying the terms, we get:
$= \dfrac{1}{2}-\dfrac{9}{2}$
Now since the denominator is the same, we can combine and write the fraction as:
$= \dfrac{1-9}{2}$
On subtracting the terms, we get:
$= \dfrac{-8}{2}$
On simplifying the terms, we get:
$= -4$, which is the required solution.

So, the correct answer is “Option B”.

Note: It is to be remembered that to add two or more fractions, the denominator of both them should be the same, if the denominator is not the same, the lowest common multiple known as L.C.M should be taken. The various trigonometric identities and formulae should be remembered while doing these types of sums. the various Pythagorean identities should also be remembered while doing these types of questions.