Question

Evaluate: \[{\sin ^2}90^\circ \cdot {\cos ^2}45^\circ + 4{\tan ^2}30^\circ + \dfrac{1}{2}{\sin ^2}a - 2{\cos ^2}90^\circ + \dfrac{1}{{24}}\].

Answer 1

Hint:
Here, we need to evaluate the given expression. We will substitute the values of the trigonometric ratios of specific angles in the given expression. Then, we will simplify the expression to get the required answer.

Complete step by step solution:
We will use the values of the trigonometric ratios of specific angles to simplify the given expression.
The trigonometric ratios used in the expression are \[\sin 90^\circ \], \[\cos 45^\circ \], \[\tan 30^\circ \], \[\sin a\], and \[\cos 90^\circ \].
We know that the sine of angle measuring \[90^\circ \] is equal to 1, the cosine of angle measuring \[45^\circ \] is equal to \[\dfrac{1}{{\sqrt 2 }}\], the tangent of angle measuring \[30^\circ \] is equal to \[\dfrac{1}{{\sqrt 3 }}\], and the cosine of angle measuring \[90^\circ \] is equal to 0.
Thus, we have \[\sin 90^\circ = 1\], \[\cos 45^\circ = \dfrac{1}{{\sqrt 2 }}\], \[\tan 30^\circ = \dfrac{1}{{\sqrt 3 }}\], and \[\cos 90^\circ = 0\].
We will substitute these values in the given expression.
Substituting \[\sin 90^\circ = 1\], \[\cos 45^\circ = \dfrac{1}{{\sqrt 2 }}\], \[\tan 30^\circ = \dfrac{1}{{\sqrt 3 }}\], and \[\cos 90^\circ = 0\] in \[{\sin ^2}90^\circ \cdot {\cos ^2}45^\circ + 4{\tan ^2}30^\circ + \dfrac{1}{2}{\sin ^2}a - 2{\cos ^2}90^\circ + \dfrac{1}{{24}}\], we get
\[ \Rightarrow {\left( 1 \right)^2}{\left( {\dfrac{1}{{\sqrt 2 }}} \right)^2} + 4{\left( {\dfrac{1}{{\sqrt 3 }}} \right)^2} + \dfrac{1}{2}{\sin ^2}a - 2{\left( 0 \right)^2} + \dfrac{1}{{24}}\]
Applying the exponents on the bases, we get
\[ \Rightarrow 1 \cdot \dfrac{1}{2} + 4 \cdot \dfrac{1}{3} + \dfrac{1}{2}{\sin ^2}a - 2 \cdot 0 + \dfrac{1}{{24}}\]
Multiplying the terms of the expression, we get
\[\begin{array}{l} \Rightarrow \dfrac{1}{2} + \dfrac{4}{3} + \dfrac{1}{2}{\sin ^2}a - 0 + \dfrac{1}{{24}}\\ \Rightarrow \dfrac{1}{2} + \dfrac{4}{3} + \dfrac{1}{{24}} + \dfrac{1}{2}{\sin ^2}a\end{array}\]
The L.C.M. of the denominators 2, 3, and 24 is 24.
Rewriting the terms of the expression with a denominator of 24, we get
\[ \Rightarrow \dfrac{{12}}{{24}} + \dfrac{{32}}{{24}} + \dfrac{1}{{24}} + \dfrac{1}{2}{\sin ^2}a\]
Adding the terms of the expression, we get
\[\begin{array}{l} \Rightarrow \dfrac{{12 + 32 + 1}}{{24}} + \dfrac{1}{2}{\sin ^2}a\\ \Rightarrow \dfrac{{45}}{{24}} + \dfrac{1}{2}{\sin ^2}a\end{array}\]
We can observe that both 45 and 24 are a multiple of 3.
Thus, we can simplify the fraction further.
Simplifying the expression, we get
\[ \Rightarrow \dfrac{{15}}{8} + \dfrac{1}{2}{\sin ^2}a\]
Thus, we get
\[ \Rightarrow {\sin ^2}90^\circ \cdot {\cos ^2}45^\circ + 4{\tan ^2}30^\circ + \dfrac{1}{2}{\sin ^2}a - 2{\cos ^2}90^\circ + \dfrac{1}{{24}} = \dfrac{{15}}{8} + \dfrac{1}{2}{\sin ^2}a\]

\[\therefore\] The value of the given expression is \[\dfrac{{15}}{8} + \dfrac{1}{2}{\sin ^2}a\].

Note:
You should remember the trigonometric ratios of the standard angles of \[0^\circ \], \[30^\circ \], \[45^\circ \], \[60^\circ \], and \[90^\circ \]. Also, we need to remember that \[{\sin ^2}x\] means \[{\left( {\sin x} \right)^2}\]. A common mistake we can make is to substitute \[\sin 90^\circ = 1\], \[\cos 45^\circ = \dfrac{1}{{\sqrt 2 }}\], \[\tan 30^\circ = \dfrac{1}{{\sqrt 3 }}\], and \[\cos 90^\circ = 0\] in the given expression and writing \[\left( 1 \right)\left( {\dfrac{1}{{\sqrt 2 }}} \right) + 4\left( {\dfrac{1}{{\sqrt 3 }}} \right) + \dfrac{1}{2}{\sin ^2}a - 2\left( 0 \right) + \dfrac{1}{{24}}\]. This will give an incorrect answer.