Question

The value of $\cos {65^ \circ }\cos {55^ \circ }\cos {5^ \circ }$ is

Answer 1

The simplest way to solve this problem is to use the basic formula in compound angles
$\cos (A + B) + \cos (A - B) = 2\cos A\cos B$
To use the above formula we need to first multiply and divide $\cos {65^ \circ }\cos {55^ \circ }\cos {5^ \circ }$ with 2 .So now we can apply the above mentioned formula and get ($2\cos {65^ \circ }\cos {55^ \circ }$)$\cos {5^ \circ }$.Then the one in bracket can be written as the form of above mentioned formula . Then we substitute the values of cosine of angle ${65^ \circ } + {55^ \circ } = {120^ \circ }$
Now we multiply $\cos {5^ \circ }$ with the simplified terms in brackets. We get $\cos {5^ \circ }\cos {10^ \circ }$ as one of the terms after multiplication with $\cos {5^ \circ }$.To this term we multiply and divide 2. So that we can spit $2\cos {5^ \circ }\cos {10^ \circ }$ using the above mentioned formula. After spitting two terms get subtracted and we are left with only one term .Lastly, we have to assign the value of the remaining trigonometric value of the angle.

Complete step-by-step solution:
In this question we need to find the value of $\cos {65^ \circ }\cos {55^ \circ }\cos {5^ \circ }$
For solving the question we need following information
$\cos {120^ \circ } = \dfrac{{ - 1}}{2} \\\ \cos {15^ \circ } = \dfrac{{\sqrt 3 + 1}}{{2\sqrt 2 }}$

= \dfrac{{2\cos {{65}^ \circ }\cos {{55}^ \circ }\cos {5^ \circ }}}{2} \\\ = \dfrac{{(2\cos {{65}^ \circ }\cos {{55}^ \circ })\cos {5^ \circ }}}{2} \\\ = \dfrac{{(\cos ({{65}^ \circ } + {{55}^ \circ }) + \cos ({{65}^ \circ } - {{55}^ \circ }))\cos {5^ \circ }}}{2} \\\ = \dfrac{{(\cos ({{120}^ \circ }) + \cos ({{65}^ \circ } - {{55}^ \circ }))\cos {5^ \circ }}}{2} \\\ = \dfrac{{(\cos ({{120}^ \circ }) + \cos ({{10}^ \circ }))\cos {5^ \circ }}}{2} \\\ = \dfrac{{\left( { - \dfrac{1}{2} + \cos ({{10}^ \circ })} \right)\cos {5^ \circ }}}{2} \\\ = \dfrac{{\left( { - \dfrac{1}{2}} \right)\cos {5^ \circ } + \cos ({{10}^ \circ })\cos {5^ \circ }}}{2} \\\ = \dfrac{{\dfrac{{ - \cos {5^ \circ }}}{2} + \cos ({{10}^ \circ })\cos {5^ \circ }}}{2} \\\ = \dfrac{{\dfrac{{ - \cos {5^ \circ }}}{2} + \dfrac{{2\cos ({{10}^ \circ })\cos {5^ \circ }}}{2}}}{2} \\\ = \dfrac{{ - \cos {5^ \circ } + 2\cos ({{10}^ \circ })\cos {5^ \circ }}}{{2 \times 2}} \\\ = \dfrac{{ - \cos {5^ \circ } + (2\cos ({{10}^ \circ })\cos {5^ \circ })}}{4} \\\ = \dfrac{{ - \cos {5^ \circ } + \cos ({{10}^ \circ } + {5^ \circ }) + \cos ({{10}^ \circ } - {5^ \circ })}}{4} \\\ = \dfrac{{ - \cos {5^ \circ } + \cos ({{15}^ \circ }) + \cos ({5^ \circ })}}{4} \\\ = \dfrac{{\cos ({{15}^ \circ }) + \cos ({5^ \circ }) - \cos {5^ \circ }}}{4} \\\ = \dfrac{{\cos ({{15}^ \circ })}}{4} \\\ = \dfrac{{\sqrt 3 + 1}}{{4 \times 2\sqrt 2 }} \\\ = \dfrac{{\sqrt 3 + 1}}{{8\sqrt 2 }} $$ So the value of $\cos {65^ \circ }\cos {55^ \circ }\cos {5^ \circ }$is

\dfrac{{\sqrt 3 + 1}}{{8\sqrt 2 }} $$

Note: In the trigonometric questions like this we generally don’t need the value of each trigonometric term. We can just simplify the expression into trigonometric terms of known angles .In this whole question we used only one basic formula of compound angles. Here we used the formula mentioned in HINT only twice, rest of the problem is just simplification and multiplying and dividing with a constant to convert an expression into a favorable expression which can be simplified using the above mentioned formula.