Question

If $\cos {20^ \circ } - \sin {20^ \circ } = p$, then $\cos {40^ \circ }$ is equal to
A. \[{p^2}\sqrt {\left( {2 - {p^2}} \right)} \]
B. \[p\sqrt {\left( {2 - {p^2}} \right)} \]
C. \[p + \sqrt {\left( {2 - {p^2}} \right)} \]
D. \[p - \sqrt {\left( {2 - {p^2}} \right)} \]

Answer 1

Hint: In order to find the value of $\cos {40^ \circ }$, square both the sides of the equation given $\cos {20^ \circ } - \sin {20^ \circ } = p$, then using the basic formulas of trigonometry and, multiples and sub- multiple angles, substitute the values, simplify it and get the results.

Formula used:
${\left( {a - b} \right)^2} = {a^2} + {b^2} - 2ab$
$\sin 2\theta = 2\sin \theta \cos \theta $
${\cos ^2}\theta + {\sin ^2}\theta = 1$
$\cos \theta = \sqrt {1 - {{\left( {\sin \theta } \right)}^2}} $
$\sqrt {{x^2}} = x$
Complete step-by-step solution:
We are given the equation $\cos {20^ \circ } - \sin {20^ \circ } = p$.
We can start with modifying the equations to find the value of $\cos {40^ \circ }$.
Starting with squaring both the sides of the equation $\cos {20^ \circ } - \sin {20^ \circ } = p$, we get:
${\left( {\cos {{20}^ \circ } - \sin {{20}^ \circ }} \right)^2} = {p^2}$
From the basic formulas, we know that:
${\left( {a - b} \right)^2} = {a^2} + {b^2} - 2ab$
Comparing and substituting the expanded value in ${\left( {\cos {{20}^ \circ } - \sin {{20}^ \circ }} \right)^2}$, we get:
${\left( {\cos {{20}^ \circ }} \right)^2} + {\left( {\sin {{20}^ \circ }} \right)^2} - 2\cos {20^ \circ }\sin {20^ \circ } = {p^2}$
Pairing the first two operands in one parenthesis, we get:
$\left( {{{\left( {\cos {{20}^ \circ }} \right)}^2} + {{\left( {\sin {{20}^ \circ }} \right)}^2}} \right) - 2\cos {20^ \circ }\sin {20^ \circ } = {p^2}$ ……(1)
From trigonometric formulas, we know that ${\cos ^2}\theta + {\sin ^2}\theta = 1$. So, comparing this equation with ${\left( {\cos {{20}^ \circ }} \right)^2} + {\left( {\sin {{20}^ \circ }} \right)^2}$ we can write ${\left( {\cos {{20}^ \circ }} \right)^2} + {\left( {\sin {{20}^ \circ }} \right)^2} = 1$.
Substituting this value in $\left( {{{\left( {\cos {{20}^ \circ }} \right)}^2} + {{\left( {\sin {{20}^ \circ }} \right)}^2}} \right) - 2\cos {20^ \circ }\sin {20^ \circ } = {p^2}$, we get:
$\left( {{{\left( {\cos {{20}^ \circ }} \right)}^2} + {{\left( {\sin {{20}^ \circ }} \right)}^2}} \right) - 2\cos {20^ \circ }\sin {20^ \circ } = {p^2}$
$ \Rightarrow 1 - 2\cos {20^ \circ }\sin {20^ \circ } = {p^2}$ …….(2)
From multiple and sub multiple angles, we know that:
$ \Rightarrow \sin 2\theta = 2\sin \theta \cos \theta $
Comparing this value with the equation 2, we get:
$\theta = {20^ \circ }$
So, the value becomes:
$ \Rightarrow 2\sin {20^ \circ }\cos {20^ \circ } = \sin \left( {2 \times {{20}^ \circ }} \right)$
$ \Rightarrow 2\sin {20^ \circ }\cos {20^ \circ } = \sin {40^ \circ }$
Substituting this value with the equation 2, we get:
$ \Rightarrow 1 - 2\cos {20^ \circ }\sin {20^ \circ } = {p^2}$
$ \Rightarrow 1 - \sin {40^ \circ } = {p^2}$
Subtracting both sides by $1$:
$ \Rightarrow 1 - \sin {40^ \circ } - 1 = {p^2} - 1$
$ \Rightarrow - \sin {40^ \circ } = {p^2} - 1$
Dividing both sides by $ - 1$:
$ \Rightarrow \dfrac{{ - \sin {{40}^ \circ }}}{{ - 1}} = \dfrac{{{p^2} - 1}}{{ - 1}}$
$ \Rightarrow \sin {40^ \circ } = 1 - {p^2}$ ……(3)

From Trigonometric Formulas, we know that:
$ \Rightarrow \cos \theta = \sqrt {1 - {{\left( {\sin \theta } \right)}^2}} $

So, substituting the value of $\theta = {40^ \circ }$:
$ \Rightarrow \cos {40^ \circ } = \sqrt {1 - {{\left( {\sin {{40}^ \circ }} \right)}^2}} $
Substituting the equation 3 in the above value, we get:
$ \Rightarrow \cos {40^ \circ } = \sqrt {1 - {{\left( {1 - {p^2}} \right)}^2}} $
Expanding the brackets ${\left( {1 - {p^2}} \right)^2}$ using the formula ${\left( {a - b} \right)^2} = {a^2} + {b^2} - 2ab$, we get:
$ \Rightarrow \cos {40^ \circ } = \sqrt {1 - \left( {{1^2} + {{\left( {{p^2}} \right)}^2} - 2{p^2}} \right)} $
$ \Rightarrow \cos {40^ \circ } = \sqrt {1 - 1 - {p^4} + 2{p^2}} $
$ \Rightarrow \cos {40^ \circ } = \sqrt {2{p^2} - {p^4}} $
Taking ${p^2}$ common inside the bracket, and we get:
$ \Rightarrow \cos {40^ \circ } = \sqrt {{p^2}\left( {2 - {p^2}} \right)} $
Since, we know that $\sqrt {{x^2}} = x$, using this in the above equation and taking ${p^2}$ outside the bracket, we get:
$ \Rightarrow \cos {40^ \circ } = p\sqrt {\left( {2 - {p^2}} \right)} $
Hence, the value of $\cos {40^ \circ } = p\sqrt {\left( {2 - {p^2}} \right)} $.
Therefore, Option 2 is correct.

Note: Since, we wrote that $\cos \theta = \sqrt {1 - {{\left( {\sin \theta } \right)}^2}} $ which can be proved using the formula ${\cos ^2}\theta + {\sin ^2}\theta = 1$. In which subtract both the sides by ${\sin ^2}\theta $, and take square root on both the sides and we first obtain the equation, ${\cos ^2}\theta = 1 - {\sin ^2}\theta $ and then $\cos \theta = \sqrt {1 - {{\sin }^2}\theta } $.