Question

The value of ${\cos ^2}45^\circ - {\sin ^2}15^\circ $ is
A. $\dfrac{{\sqrt 3 - 1}}{{2 - \sqrt 2 }}$
B. $\dfrac{{\sqrt 3 + 1}}{{2\sqrt 2 }}$
C. $\dfrac{{\sqrt 3 }}{{\sqrt 2 }}$
D. $\dfrac{{\sqrt 3 }}{4}$

Answer 1

Hint: We can expand $\sin 15^\circ $ as $\sin \left( {45^\circ - 30^\circ } \right)$ . Then we can simplify it using the identity, $\sin \left( {A - B} \right) = \sin A\cos B - \cos A\sin B$ . Then we can substitute this in the given expression. We can then give the values for \[\cos 45^\circ ,\sin 45^\circ ,\cos 30^\circ \] and \[\sin 30^\circ \] . After further simplification, we will get the required solution.

Complete step-by-step answer:
We need to find the value of ${\cos ^2}45^\circ - {\sin ^2}15^\circ $
We can take the 2nd term of the expression, $\sin 15^\circ $
We can write 15 as $15 = 45 - 30$
 $ \Rightarrow \sin 15^\circ = \sin \left( {45^\circ - 30^\circ } \right)$
 We know that $\sin \left( {A - B} \right) = \sin A\cos B - \cos A\sin B$ . On applying this, we get,
 $ \Rightarrow \sin 15^\circ = \sin 45^\circ \cos 30^\circ - \cos 45^\circ \sin 30^\circ $ .
We know that \[\cos 45^\circ = \dfrac{1}{{\sqrt 2 }}\] , \[\cos 30^\circ = \dfrac{{\sqrt 3 }}{2}\] , \[\sin 45^\circ = \dfrac{1}{{\sqrt 2 }}\] and \[\sin 30^\circ = \dfrac{1}{2}\] . On substituting these in the above equation, we get,
 $ \Rightarrow \sin 15^\circ = \dfrac{1}{{\sqrt 2 }} \times \dfrac{{\sqrt 3 }}{2} - \dfrac{1}{{\sqrt 2 }} \times \dfrac{1}{2}$
On simplification, we get,
 $ \Rightarrow \sin 15^\circ = \dfrac{{\sqrt 3 - 1}}{{2\sqrt 2 }}$
Now we can take the square.
 $ \Rightarrow {\sin ^2}15^\circ = {\left( {\dfrac{{\sqrt 3 - 1}}{{2\sqrt 2 }}} \right)^2}$
 \[ \Rightarrow {\sin ^2}15^\circ = \dfrac{{{{\left( {\sqrt 3 - 1} \right)}^2}}}{{{{\left( {2\sqrt 2 } \right)}^2}}}\]
We know that ${\left( {a - b} \right)^2} = {a^2} + {b^2} - 2ab$ . On applying this in numerator, we get,
 $ \Rightarrow {\sin ^2}15^\circ = \dfrac{{1 + 3 - 2\sqrt 3 }}{8}$
On simplification we get,
 $ \Rightarrow {\sin ^2}15^\circ = \dfrac{{4 - 2\sqrt 3 }}{8}$
On dividing the numerator and denominator by 2, we get,
 $ \Rightarrow {\sin ^2}15^\circ = \dfrac{{2 - \sqrt 3 }}{4}$ … (1)
Now we need to find the value of ${\cos ^2}45^\circ $ .
We know that \[\cos 45^\circ = \dfrac{1}{{\sqrt 2 }}\]
On squaring on both sides, we get,
 \[{\cos ^2}45^\circ = \dfrac{1}{2}\] .. (2)
We have the expression ${\cos ^2}45^\circ - {\sin ^2}15^\circ $
Let $I = {\cos ^2}45^\circ - {\sin ^2}15^\circ $
We can substitute equation (1) and (2) to the above expression.
 $ \Rightarrow I = \dfrac{1}{2} - \dfrac{{2 - \sqrt 3 }}{4}$
We can multiply and divide the 1st term with 2 to make the denominators equal.
 $ \Rightarrow I = \dfrac{2}{4} - \dfrac{{2 - \sqrt 3 }}{4}$
As the denominators are equal, we can add the numerators.
 $ \Rightarrow I = \dfrac{{2 - 2 + \sqrt 3 }}{4}$
On simplification, we get,
 $ \Rightarrow I = \dfrac{{\sqrt 3 }}{4}$
Therefore, the value of ${\cos ^2}45^\circ - {\sin ^2}15^\circ $ is $\dfrac{{\sqrt 3 }}{4}$
So, the correct answer is option D.

Note: Alternate solution is given by,
We need to find the value of ${\cos ^2}45^\circ - {\sin ^2}15^\circ $
We can use the identity, \[{a^2} - {b^2} = \left( {a + b} \right)\left( {a - b} \right)\]
 \[ \Rightarrow {\cos ^2}45^\circ - {\sin ^2}15^\circ = \left( {\cos 45^\circ + \sin 15^\circ } \right)\left( {\cos 45^\circ - \sin 15^\circ } \right)\] …. (a)
We can write 15 as $15 = 45 - 30$
 $ \Rightarrow \sin 15^\circ = \sin \left( {45^\circ - 30^\circ } \right)$
 We know that $\sin \left( {A - B} \right) = \sin A\cos B - \cos A\sin B$ . On applying this, we get,
 $ \Rightarrow \sin 15^\circ = \sin 45^\circ \cos 30^\circ - \cos 45^\circ \sin 30^\circ $ .
We know that \[\cos 45^\circ = \dfrac{1}{{\sqrt 2 }}\] , \[\cos 30^\circ = \dfrac{{\sqrt 3 }}{2}\] , \[\sin 45^\circ = \dfrac{1}{{\sqrt 2 }}\] and \[\sin 30^\circ = \dfrac{1}{2}\] . On substituting these in the above equation, we get,
 $ \Rightarrow \sin 15^\circ = \dfrac{1}{{\sqrt 2 }} \times \dfrac{{\sqrt 3 }}{2} - \dfrac{1}{{\sqrt 2 }} \times \dfrac{1}{2}$
On simplification, we get,
 $ \Rightarrow \sin 15^\circ = \dfrac{{\sqrt 3 - 1}}{{2\sqrt 2 }}$
We know that \[\cos 45^\circ = \dfrac{1}{{\sqrt 2 }}\] .
Substituting these in equation (a), we get,
 \[ \Rightarrow {\cos ^2}45^\circ - {\sin ^2}15^\circ = \left( {\dfrac{1}{{\sqrt 2 }} + \dfrac{{\sqrt 3 - 1}}{{2\sqrt 2 }}} \right)\left( {\dfrac{1}{{\sqrt 2 }} - \dfrac{{\sqrt 3 - 1}}{{2\sqrt 2 }}} \right)\]
On taking LCM of terms inside brackets we get,
 \[ \Rightarrow {\cos ^2}45^\circ - {\sin ^2}15^\circ = \left( {\dfrac{2}{{2\sqrt 2 }} + \dfrac{{\sqrt 3 - 1}}{{2\sqrt 2 }}} \right)\left( {\dfrac{2}{{2\sqrt 2 }} - \dfrac{{\sqrt 3 - 1}}{{2\sqrt 2 }}} \right)\]
On adding we get,
 \[ \Rightarrow {\cos ^2}45^\circ - {\sin ^2}15^\circ = \left( {\dfrac{{2 + \left( {\sqrt 3 - 1} \right)}}{{2\sqrt 2 }}} \right)\left( {\dfrac{{2 - \left( {\sqrt 3 - 1} \right)}}{{2\sqrt 2 }}} \right)\]
On simplification we get,
 \[ \Rightarrow {\cos ^2}45^\circ - {\sin ^2}15^\circ = \dfrac{{\left( {2 - \left( {\sqrt 3 - 1} \right)} \right)\left( {2 + \left( {\sqrt 3 - 1} \right)} \right)}}{8}\]
We know that \[{a^2} - {b^2} = \left( {a + b} \right)\left( {a - b} \right)\] , using this we get,
 \[ \Rightarrow {\cos ^2}45^\circ - {\sin ^2}15^\circ = \dfrac{{{2^2} - {{\left( {\sqrt 3 - 1} \right)}^2}}}{8}\]
On simplification we get,
 \[ \Rightarrow {\cos ^2}45^\circ - {\sin ^2}15^\circ = \dfrac{{4 - \left( {4 - 2\sqrt 3 } \right)}}{8}\]
Hence, we have,
 \[ \Rightarrow {\cos ^2}45^\circ - {\sin ^2}15^\circ = \dfrac{{2\sqrt 3 }}{8}\]
On dividing the numerator and denominator by 2 we get,
 \[ \Rightarrow {\cos ^2}45^\circ - {\sin ^2}15^\circ = \dfrac{{\sqrt 3 }}{4}\]
Therefore, the value of ${\cos ^2}45^\circ - {\sin ^2}15^\circ $ is $\dfrac{{\sqrt 3 }}{4}$ .