Courses
Courses for Kids
Free study material
Offline Centres
More
Store

# The value of ${\cos ^2}{1^ \circ } + {\cos ^2}{2^ \circ } + {\cos ^2}{3^ \circ } + ....... + {\cos ^2}{90^ \circ } =$ (a) $0$ (b) $1$ (c) $45$ (d) $\dfrac{{89}}{2}$

Last updated date: 11th Sep 2024
Total views: 366.6k
Views today: 4.66k
Verified
366.6k+ views
Hint: The given problem requires us to simplify and find the value of the given trigonometric expression. The question requires thorough knowledge of trigonometric functions, formulae and identities. The question describes the wide ranging applications of trigonometric identities and formulae. We must keep in mind the trigonometric identities while solving such questions.

In the given question, we are required to evaluate the value of the trigonometric summation expression ${\cos ^2}{1^ \circ } + {\cos ^2}{2^ \circ } + {\cos ^2}{3^ \circ } + ....... + {\cos ^2}{90^ \circ }$ using the basic concepts of trigonometry, formulae and identities.
We are given the summation of squares of cosine functions with angles ranging from ${1^ \circ }$ to ${90^ \circ }$ .
Now, we know that the trigonometric functions sine and cosine are complementary of each other.
Writing the middlemost term and last few terms of the summation, we get,
$\Rightarrow {\cos ^2}{1^ \circ } + {\cos ^2}{2^ \circ } + {\cos ^2}{3^ \circ } + ...... + {\cos ^2}{45^ \circ } + ......{\cos ^2}{87^ \circ } + {\cos ^2}{88^ \circ } + {\cos ^2}{89^ \circ } + {\cos ^2}{90^ \circ }$
Now, expressing the angles of later half of terms as the compliments of the angle of first half of the terms in the summation, we get,
$\Rightarrow {\cos ^2}{1^ \circ } + {\cos ^2}{2^ \circ } + {\cos ^2}{3^ \circ } + ..... + {\cos ^2}{45^ \circ } + ......{\cos ^2}\left( {{{90}^ \circ } - {3^ \circ }} \right) + {\cos ^2}\left( {{{90}^ \circ } - {2^ \circ }} \right) + {\cos ^2}\left( {{{90}^ \circ } - {1^ \circ }} \right) + {\cos ^2}{90^ \circ }$ Now, we can use the trigonometric formula $\cos x = \sin \left( {{{90}^ \circ } - x} \right)$ in order to simplify the trigonometric expression given to us.
$\Rightarrow {\cos ^2}{1^ \circ } + {\cos ^2}{2^ \circ } + {\cos ^2}{3^ \circ } + ...... + {\cos ^2}{45^ \circ } + ......{\sin ^2}\left( {{3^ \circ }} \right) + {\sin ^2}\left( {{2^ \circ }} \right) + {\sin ^2}\left( {{1^ \circ }} \right) + {\cos ^2}{90^ \circ }$
Now, grouping the sine and cosine terms with same angles. Similarly, the rest of the terms would also follow the same pattern except the term ${\cos ^2}{45^ \circ }$ and ${\cos ^2}{90^ \circ }$ . So, we get,
$\Rightarrow \left( {{{\cos }^2}{1^ \circ } + {{\sin }^2}{1^ \circ }} \right) + \left( {{{\cos }^2}{2^ \circ } + {{\sin }^2}{2^ \circ }} \right) + \left( {{{\cos }^2}{3^ \circ } + {{\sin }^2}{3^ \circ }} \right) + ......\left( {{{\cos }^2}{{44}^ \circ } + {{\sin }^2}{{44}^ \circ }} \right) + {\cos ^2}{45^ \circ } + {\cos ^2}{90^ \circ }$
Now, we can use the trigonometric identity ${\sin ^2}x + {\cos ^2}x = 1$ in the expression. Hence, we get,
$\Rightarrow \left( 1 \right) + \left( 1 \right) + \left( 1 \right) + ......\left( 1 \right) + {\cos ^2}{45^ \circ } + {\cos ^2}{90^ \circ }$
So, adding all the ones in the summation and substituting the value of $\cos {90^ \circ }$ as zero and $\cos {45^ \circ }$ as $\dfrac{1}{{\sqrt 2 }}$ , we get,
$\Rightarrow 44 + {\left( {\dfrac{1}{{\sqrt 2 }}} \right)^2} + {\left( 0 \right)^2}$
Simplifying the expression, we get,
$\Rightarrow 44 + \dfrac{1}{2}$
$\Rightarrow \dfrac{{89}}{2}$
So, we get the value of the expression ${\cos ^2}{1^ \circ } + {\cos ^2}{2^ \circ } + {\cos ^2}{3^ \circ } + ....... + {\cos ^2}{90^ \circ }$ as $\dfrac{{89}}{2}$ .
Hence, option (d) is correct.
So, the correct answer is “Option d”.

Note: All the trigonometric ratios can be converted into each other using the simple trigonometric identities listed above. The given problem involves the use of trigonometric formulae and identities. Such questions require thorough knowledge of trigonometric conversions and ratios. Algebraic operations and rules like transposition rule come into significant use while solving such problems.