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# Prove $\cos {48^ \circ } \cdot \cos {12^ \circ } = \dfrac{{3 - \sqrt 5 }}{8}$ is true or false A.TrueB.False

Last updated date: 09th Sep 2024
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Hint: In this we use multiple theorem where $2\cos A\cos B = \cos \left( {A + B} \right) + \cos (A - B)$ and for simplifying we will use the value of $\cos {60^ \circ } = \dfrac{1}{2}$and $\cos {36^ \circ } = \dfrac{{\sqrt 5 + 1}}{4}$

Given $\cos {48^ \circ } \cdot \cos {12^ \circ }$
We know that $2\cos A\cos B = \cos \left( {A + B} \right) + \cos (A - B)$ substituting $A = {48^ \circ },B = {12^ \circ }$in formula we get
$2\cos {48^ \circ } \cdot \cos {12^ \circ } = \cos \left( {48 + 12} \right) + \cos \left( {48 - 12} \right)$
So, $\cos {48^ \circ } \cdot \cos {12^ \circ } = \dfrac{{\cos {{60}^ \circ } + \cos {{36}^ \circ }}}{2}$
Substituting $\cos {60^ \circ } = \dfrac{1}{2}$and $\cos {36^ \circ } = \dfrac{{\sqrt 5 + 1}}{4}$ ,we get,
$\Rightarrow \dfrac{{\dfrac{1}{2} + \dfrac{{\sqrt 5 + 1}}{4}}}{2}$
On simplifying we get
$\cos {48^ \circ } \cdot \cos {12^ \circ } = \dfrac{{\sqrt 5 + 3}}{8}$
So, we can say that the given statement is false
$2\cos A\cos B = \cos \left( {A + B} \right) + \cos (A - B) \\ 2\sin A\sin B = cos\left( {A - B} \right) - \cos \left( {A + B} \right) \\ 2\sin A\cos B = \sin \left( {A + B} \right) + \sin \left( {A - B} \right) \\ 2\sin B\cos A = \sin \left( {A + B} \right) - \sin \left( {A - B} \right) \\$
and must not confused in value of $\cos {36^ \circ } = \dfrac{{\sqrt 5 + 1}}{4}$