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Question: If \(\cos {20^ \circ } - \sin {20^ \circ } = p\), then \(\cos {40^ \circ }\) is equal to A. \[{p^2...

If cos⁡20∘−sin⁡20∘=p\cos {20^ \circ } - \sin {20^ \circ } = p, then cos⁡40∘\cos {40^ \circ } is equal to
A. p2(2−p2){p^2}\sqrt {\left( {2 - {p^2}} \right)}
B. p(2−p2)p\sqrt {\left( {2 - {p^2}} \right)}
C. p+(2−p2)p + \sqrt {\left( {2 - {p^2}} \right)}
D. p−(2−p2)p - \sqrt {\left( {2 - {p^2}} \right)}

Explanation

Solution

In order to find the value of cos⁡40∘\cos {40^ \circ }, square both the sides of the equation given cos⁡20∘−sin⁡20∘=p\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:
(a−b)2=a2+b2−2ab{\left( {a - b} \right)^2} = {a^2} + {b^2} - 2ab
sin⁡2θ=2sin⁡θcos⁡θ\sin 2\theta = 2\sin \theta \cos \theta
cos⁡2θ+sin⁡2θ=1{\cos ^2}\theta + {\sin ^2}\theta = 1
cos⁡θ=1−(sin⁡θ)2\cos \theta = \sqrt {1 - {{\left( {\sin \theta } \right)}^2}}
x2=x\sqrt {{x^2}} = x
Complete step-by-step solution:
We are given the equation cos⁡20∘−sin⁡20∘=p\cos {20^ \circ } - \sin {20^ \circ } = p.
We can start with modifying the equations to find the value of cos⁡40∘\cos {40^ \circ }.
Starting with squaring both the sides of the equation cos⁡20∘−sin⁡20∘=p\cos {20^ \circ } - \sin {20^ \circ } = p, we get:
(cos⁡20∘−sin⁡20∘)2=p2{\left( {\cos {{20}^ \circ } - \sin {{20}^ \circ }} \right)^2} = {p^2}
From the basic formulas, we know that:
(a−b)2=a2+b2−2ab{\left( {a - b} \right)^2} = {a^2} + {b^2} - 2ab
Comparing and substituting the expanded value in (cos⁡20∘−sin⁡20∘)2{\left( {\cos {{20}^ \circ } - \sin {{20}^ \circ }} \right)^2}, we get:
(cos⁡20∘)2+(sin⁡20∘)2−2cos⁡20∘sin⁡20∘=p2{\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:
((cos⁡20∘)2+(sin⁡20∘)2)−2cos⁡20∘sin⁡20∘=p2\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⁡2θ+sin⁡2θ=1{\cos ^2}\theta + {\sin ^2}\theta = 1. So, comparing this equation with (cos⁡20∘)2+(sin⁡20∘)2{\left( {\cos {{20}^ \circ }} \right)^2} + {\left( {\sin {{20}^ \circ }} \right)^2} we can write (cos⁡20∘)2+(sin⁡20∘)2=1{\left( {\cos {{20}^ \circ }} \right)^2} + {\left( {\sin {{20}^ \circ }} \right)^2} = 1.
Substituting this value in ((cos⁡20∘)2+(sin⁡20∘)2)−2cos⁡20∘sin⁡20∘=p2\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:
((cos⁡20∘)2+(sin⁡20∘)2)−2cos⁡20∘sin⁡20∘=p2\left( {{{\left( {\cos {{20}^ \circ }} \right)}^2} + {{\left( {\sin {{20}^ \circ }} \right)}^2}} \right) - 2\cos {20^ \circ }\sin {20^ \circ } = {p^2}
⇒1−2cos⁡20∘sin⁡20∘=p2\Rightarrow 1 - 2\cos {20^ \circ }\sin {20^ \circ } = {p^2} …….(2)
From multiple and sub multiple angles, we know that:
⇒sin⁡2θ=2sin⁡θcos⁡θ\Rightarrow \sin 2\theta = 2\sin \theta \cos \theta
Comparing this value with the equation 2, we get:
θ=20∘\theta = {20^ \circ }
So, the value becomes:
⇒2sin⁡20∘cos⁡20∘=sin⁡(2×20∘)\Rightarrow 2\sin {20^ \circ }\cos {20^ \circ } = \sin \left( {2 \times {{20}^ \circ }} \right)
⇒2sin⁡20∘cos⁡20∘=sin⁡40∘\Rightarrow 2\sin {20^ \circ }\cos {20^ \circ } = \sin {40^ \circ }
Substituting this value with the equation 2, we get:
⇒1−2cos⁡20∘sin⁡20∘=p2\Rightarrow 1 - 2\cos {20^ \circ }\sin {20^ \circ } = {p^2}
⇒1−sin⁡40∘=p2\Rightarrow 1 - \sin {40^ \circ } = {p^2}
Subtracting both sides by 11:
⇒1−sin⁡40∘−1=p2−1\Rightarrow 1 - \sin {40^ \circ } - 1 = {p^2} - 1
⇒−sin⁡40∘=p2−1\Rightarrow - \sin {40^ \circ } = {p^2} - 1
Dividing both sides by −1 - 1:
⇒−sin⁡40∘−1=p2−1−1\Rightarrow \dfrac{{ - \sin {{40}^ \circ }}}{{ - 1}} = \dfrac{{{p^2} - 1}}{{ - 1}}
⇒sin⁡40∘=1−p2\Rightarrow \sin {40^ \circ } = 1 - {p^2} ……(3)

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

So, substituting the value of θ=40∘\theta = {40^ \circ }:
⇒cos⁡40∘=1−(sin⁡40∘)2\Rightarrow \cos {40^ \circ } = \sqrt {1 - {{\left( {\sin {{40}^ \circ }} \right)}^2}}
Substituting the equation 3 in the above value, we get:
⇒cos⁡40∘=1−(1−p2)2\Rightarrow \cos {40^ \circ } = \sqrt {1 - {{\left( {1 - {p^2}} \right)}^2}}
Expanding the brackets (1−p2)2{\left( {1 - {p^2}} \right)^2} using the formula (a−b)2=a2+b2−2ab{\left( {a - b} \right)^2} = {a^2} + {b^2} - 2ab, we get:
⇒cos⁡40∘=1−(12+(p2)2−2p2)\Rightarrow \cos {40^ \circ } = \sqrt {1 - \left( {{1^2} + {{\left( {{p^2}} \right)}^2} - 2{p^2}} \right)}
⇒cos⁡40∘=1−1−p4+2p2\Rightarrow \cos {40^ \circ } = \sqrt {1 - 1 - {p^4} + 2{p^2}}
⇒cos⁡40∘=2p2−p4\Rightarrow \cos {40^ \circ } = \sqrt {2{p^2} - {p^4}}
Taking p2{p^2} common inside the bracket, and we get:
⇒cos⁡40∘=p2(2−p2)\Rightarrow \cos {40^ \circ } = \sqrt {{p^2}\left( {2 - {p^2}} \right)}
Since, we know that x2=x\sqrt {{x^2}} = x, using this in the above equation and taking p2{p^2} outside the bracket, we get:
⇒cos⁡40∘=p(2−p2)\Rightarrow \cos {40^ \circ } = p\sqrt {\left( {2 - {p^2}} \right)}
Hence, the value of cos⁡40∘=p(2−p2)\cos {40^ \circ } = p\sqrt {\left( {2 - {p^2}} \right)} .
Therefore, Option 2 is correct.

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