Question

Prove that: ${{\cos }^{2}}45{}^\circ -{{\sin }^{2}}15{}^\circ =\dfrac{\sqrt{3}}{4}$.

Answer 1

Hint: For solving this problem, first we convert the angle of sine in terms of 45 and 30 degrees by using subtraction. From the standard table of trigonometric functions, we know the value of sin 45 and 30 but not sin 15. Now, we expand the sine term by using formula sin (A-B). After simplification, wait to put the values from the table of trigonometric functions to obtain the final result.

Complete step-by-step answer:
Some of the useful trigonometric formula used in solving this problem:
sin (A - B) = sin A cos B - cos A sin B
The specific value of functions of sin and cos which are useful for this problem can be illustrated as:
\[\begin{align}
  & \sin 30{}^\circ =\dfrac{1}{2} \\
 & \sin 45{}^\circ =\dfrac{1}{\sqrt{2}} \\
 & \cos 30{}^\circ =\dfrac{\sqrt{3}}{2} \\
 & \cos 45{}^\circ =\dfrac{1}{\sqrt{2}} \\
\end{align}\]

According to the problem statement, we consider the left-hand side of the equation for proving equivalence of both sides. First, we convert the ${{\sin }^{2}}15{}^\circ \text{ to si}{{\text{n}}^{2}}\left( 45-30 \right){}^\circ $. Now, by using the above expansion, we can rewrite the left-hand side as
$\begin{align}
  & \Rightarrow {{\cos }^{2}}45{}^\circ -\text{si}{{\text{n}}^{2}}\left( 45-30 \right){}^\circ \\
 & \Rightarrow {{\cos }^{2}}45{}^\circ -{{\left( \sin 45{}^\circ \cos 30{}^\circ -\cos 45{}^\circ \sin 30{}^\circ \right)}^{2}} \\
\end{align}$
Now, we expand the terms by using the algebraic identity $\left( a-b \right)={{a}^{2}}+{{b}^{2}}-2ab$ to get
\[\Rightarrow {{\cos }^{2}}45{}^\circ -\left[ \left( {{\sin }^{2}}45{}^\circ \right)\left( {{\cos }^{2}}30{}^\circ \right)+\left( {{\cos }^{2}}45{}^\circ \right)\left( {{\sin }^{2}}30{}^\circ \right)-2\left( \sin 45{}^\circ \right)\left( \cos 30{}^\circ \right)\left( \cos 45{}^\circ \right)\left( \sin 30{}^\circ \right) \right]\]
Now, putting the values from the table, we get
$\begin{align}
  & \Rightarrow {{\left( \dfrac{1}{\sqrt{2}} \right)}^{2}}-\left[ {{\left( \dfrac{1}{\sqrt{2}} \right)}^{2}}{{\left( \dfrac{\sqrt{3}}{2} \right)}^{2}}+{{\left( \dfrac{1}{\sqrt{2}} \right)}^{2}}{{\left( \dfrac{1}{2} \right)}^{2}}-2\left( \dfrac{1}{\sqrt{2}} \right)\left( \dfrac{\sqrt{3}}{2} \right)\left( \dfrac{1}{\sqrt{2}} \right)\left( \dfrac{1}{2} \right) \right] \\
 & \Rightarrow \dfrac{1}{2}-\left[ \dfrac{1}{2}\times \dfrac{3}{4}+\dfrac{1}{2}\times \dfrac{1}{4}-2\times \dfrac{\sqrt{3}}{8} \right] \\
 & \Rightarrow \dfrac{1}{2}-\left[ \dfrac{3}{8}+\dfrac{1}{8}-\dfrac{\sqrt{3}}{4} \right] \\
 & \Rightarrow \dfrac{1}{2}-\left[ \dfrac{4}{8}-\dfrac{\sqrt{3}}{4} \right] \\
 & \Rightarrow \dfrac{1}{2}-\dfrac{1}{2}+\dfrac{\sqrt{3}}{4} \\
 & \Rightarrow \dfrac{\sqrt{3}}{4} \\
\end{align}$
Hence, we proved the equivalence of both sides by considering the expression of the left side.

Note: Students must remember the trigonometric table and the trigonometric formulas associated with different functions. This problem can be alternatively solved by using the identity ${{\cos }^{2}}X-{{\sin }^{2}}Y=\cos \left( X+Y \right)\cos \left( X-Y \right)$. Now, putting X = 45 and Y = 15, we can easily obtain the same result.