Question

Evaluate the following
$4\left( {{\sin }^{4}}{{60}^{\circ }}+{{\cos }^{4}}{{30}^{\circ }} \right)-3\left( {{\tan }^{2}}{{60}^{\circ }}-{{\tan }^{2}}{{45}^{\circ }} \right)+5{{\cos }^{2}}{{45}^{\circ }}$

Answer 1

To evaluate the given expression we need to know the values of $\sin {{60}^{\circ }}$ , $\cos {{30}^{\circ }}$ , $\cos {{45}^{\circ }}$ , $\tan {{45}^{\circ }}$ , $\tan {{60}^{\circ }}$ which are present in the expression. And substitute them in the equation and solve to get the final answer. The trigonometric angles are given in degrees.

Complete step-by-step solution:
Let us consider the given expression as the variable $x$
It is given by
$x=4\left( {{\sin }^{4}}{{60}^{\circ }}+{{\cos }^{4}}{{30}^{\circ }} \right)-3\left( {{\tan }^{2}}{{60}^{\circ }}-{{\tan }^{2}}{{45}^{\circ }} \right)+5{{\cos }^{2}}{{45}^{\circ }}$
The values of $\sin {{60}^{\circ }}$ , $\cos {{30}^{\circ }}$ , $\cos {{45}^{\circ }}$ , $\tan {{45}^{\circ }}$ , $\tan {{60}^{\circ }}$ can be given by
$\sin {{60}^{\circ }}=\dfrac{\sqrt{3}}{2}$
$\cos {{30}^{\circ }}=\dfrac{\sqrt{3}}{2}$
$\cos {{45}^{\circ }}=\dfrac{1}{\sqrt{2}}$
$\tan {{45}^{\circ }}=1$
$\tan {{60}^{\circ }}=\sqrt{3}$
Substitute the above values in the expression represented by $x$
$x=4\left( {{\sin }^{4}}{{60}^{\circ }}+{{\cos }^{4}}{{30}^{\circ }} \right)-3\left( {{\tan }^{2}}{{60}^{\circ }}-{{\tan }^{2}}{{45}^{\circ }} \right)+5{{\cos }^{2}}{{45}^{\circ }}$
$\Rightarrow x=4\left( {{\left( \dfrac{\sqrt{3}}{2} \right)}^{4}}+{{\left( \dfrac{\sqrt{3}}{2} \right)}^{4}} \right)-3\left( {{\left( \sqrt{3} \right)}^{2}}-{{1}^{2}} \right)+5{{\left( \dfrac{1}{\sqrt{2}} \right)}^{2}}$
Now solve the above expression to get the value of $x$
$\Rightarrow x=4\left( 2\times {{\left( {{\left( \dfrac{\sqrt{3}}{2} \right)}^{2}} \right)}^{2}} \right)-3\left( 3-1 \right)+5{{\left( \dfrac{1}{\sqrt{2}} \right)}^{2}}$
$\Rightarrow x=8\left( {{\left( \dfrac{3}{4} \right)}^{2}} \right)-3\left( 2 \right)+5\left( \dfrac{1}{2} \right)$
$\Rightarrow x=8\left( \dfrac{9}{16} \right)-6+\dfrac{5}{2}$
$\Rightarrow x=\dfrac{9}{2}-6+\dfrac{5}{2}$
Do the LCM of denominator and solve the expression
$\Rightarrow x=\dfrac{9+5-12}{2}$
$\Rightarrow x=\dfrac{14-12}{2}$
$\Rightarrow x=\dfrac{2}{2}$
$\Rightarrow x=1$
Hence the value of the given trigonometric expression is $x=1$

Additional information: Trigonometry is related to the relations between the angles and sides of triangles. However, the applications of trigonometry are broadened in a meantime. There are mainly six trigonometric functions namely, sine, cosine, tangent, cotangent, secant, cosecant. Trigonometry deals with the various relations of these functions. The functions cosine and secant are inverse of each other, sine and cosecant are inverses and tan and cotangent are inverse to each other.

Note: The values of sine and cosine are complementary to each other. It is given by the relation $\cos x=\sin \left( 90-x \right)$ . Similarly, secant and cosecant are complementary to each other and for tangent and cotangent also similar relation exists. This is the reason why the value of $\sin {{60}^{\circ }}$ is equal to the value of $\cos {{30}^{\circ }}$ . The representation of angles can be of either degrees or radians. When expressed in radians mostly the angles are of the form $\theta =\dfrac{\pi }{x}$