Question

Prove that $\cos {10^ \circ }\cos {30^ \circ }\cos {50^ \circ }\cos {70^ \circ } = \dfrac{3}{{16}}$

Answer 1

Here in this question we should know basic values of cosine and some trigonometric identities that are mentioned below:-
$\cos {30^ \circ } = \dfrac{{\sqrt 3 }}{2}$
$\cos {60^ \circ } = \dfrac{1}{2}$
$\cos {45^ \circ } = \dfrac{1}{{\sqrt 2 }}$
$\cos {90^ \circ } = 0$
$\cos {180^ \circ } = 1$
$2\cos A\cos B = \cos (A + B) + \cos (A - B)$

Complete step-by-step answer:
First of all we will see that it any value of cosine is known in the question if yes then we will put that value and then will try to convert the L.H.S equal to R.H.S
$\Rightarrow \cos {10^ \circ }\cos {30^ \circ }\cos {50^ \circ }\cos {70^ \circ }$ (We know the value of$\cos {30^ \circ } = \dfrac{{\sqrt 3 }}{2}$)
$\Rightarrow \cos {10^ \circ } \times \dfrac{{\sqrt 3 }}{2}\cos {50^ \circ }\cos {70^ \circ }$
$\Rightarrow \dfrac{{\sqrt 3 }}{2}\cos {70^ \circ }\cos {50^ \circ }\cos {10^ \circ }$
(Now we will apply identity $2\cos A\cos B = \cos (A + B) + \cos (A - B)$ but for this we have to make that identity by multiplying and dividing the equation by 2)
$\Rightarrow \dfrac{{\sqrt 3 }}{2}\cos {70^ \circ } \times \dfrac{2}{2}\cos {50^ \circ }\cos {10^ \circ }$
$\Rightarrow \dfrac{{\sqrt 3 }}{4}\cos {70^ \circ } \times 2\cos {50^ \circ }\cos {10^ \circ }$
$\Rightarrow \dfrac{{\sqrt 3 }}{4}\cos {70^ \circ } \times [\cos ({50^ \circ } + {10^ \circ }) + \cos ({50^ \circ } - {10^ \circ })]$ (Here$A = {50^ \circ },B = {10^ \circ }$)
$\Rightarrow \dfrac{{\sqrt 3 }}{4}\cos {70^ \circ } \times [\cos ({50^ \circ } + {10^ \circ }) + \cos ({50^ \circ } - {10^ \circ })]$
$\Rightarrow \dfrac{{\sqrt 3 }}{4}\cos {70^ \circ } \times [\cos ({60^ \circ }) + \cos ({40^ \circ })]$
$\Rightarrow \dfrac{{\sqrt 3 }}{4}\cos {70^ \circ } \times [\dfrac{1}{2} + \cos ({40^ \circ })]$ (Putting value of $\cos {60^ \circ } = \dfrac{1}{2}$)
$\Rightarrow \dfrac{{\sqrt 3 }}{4}[\cos {70^ \circ } \times \dfrac{1}{2} + \cos {70^ \circ }\cos {40^ \circ }]$
(Multiplying $\cos {70^ \circ }$inside so that again identity $2\cos A\cos B = \cos (A + B) + \cos (A - B)$can be applied)
$\Rightarrow \dfrac{{\sqrt 3 }}{4}[\dfrac{1}{2}\cos {70^ \circ } + \dfrac{2}{2}\cos {70^ \circ }\cos {40^ \circ }]$
(Again we will multiply and divide the equation with 2 to make identity)
$\Rightarrow \dfrac{{\sqrt 3 }}{{4 \times 2}}[\cos {70^ \circ } + 2\cos {70^ \circ }\cos {40^ \circ }]$ (Taking denominators 2 common)
$\Rightarrow \dfrac{{\sqrt 3 }}{8}[\cos {70^ \circ } + \cos ({70^ \circ } + {40^ \circ }) + \cos ({70^ \circ } - {40^ \circ })]$ (Here$A = {70^ \circ },B = {40^ \circ }$)
$\Rightarrow \dfrac{{\sqrt 3 }}{8}[\cos {70^ \circ } + \cos {110^ \circ } + \cos {30^ \circ }]$
$\Rightarrow \dfrac{{\sqrt 3 }}{8}[\cos {70^ \circ } + \cos {110^ \circ } + \dfrac{{\sqrt 3 }}{2}]$ (Putting value of $\cos {30^ \circ } = \dfrac{{\sqrt 3 }}{2}$)
$\Rightarrow \dfrac{{\sqrt 3 }}{8}[\cos {70^ \circ } - \cos {70^ \circ } + \dfrac{{\sqrt 3 }}{2}]$ ()
$\Rightarrow \dfrac{{\sqrt 3 }}{8} \times \dfrac{{\sqrt 3 }}{2} = \dfrac{3}{{16}}$ (Cancelling equal and opposite terms)
Hence it is proved that $\cos {10^ \circ }\cos {30^ \circ }\cos {50^ \circ }\cos {70^ \circ } = \dfrac{3}{{16}}$

Note: Here in this question students may find $\cos {110^ \circ } = - \cos {70^ \circ }$this step little confusing so below explanation is there:-
Since ${110^ \circ }$angle is greater than ${90^ \circ }$and less than ${180^ \circ }$so it is located in quadrant second where only ‘sine’ terms are positive and ‘cosine’ terms are negative.
$\cos {110^ \circ } = - (\cos {180^ \circ } - {70^ \circ })$
$\therefore \cos {110^ \circ } = - \cos {70^ \circ }$
All students must know signs of different trigonometric functions in all four quadrants so that conversion of angle becomes easy with that knowledge. Below all signs quadrant wise are mentioned: -
First quadrant = All trigonometric functions are positive (sine, cosine, tan, sec, cosec, cot)
Second quadrant=Positive (sine, cosec) Negative (cosine, tan, sec, cot)
Third quadrant= Positive (tan, cot) Negative (sine, cosine, sec, cosec)
Fourth quadrant= Positive (cosine, sec) Negative (sine, tan, cot, cosec)