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Question: Evaluate \( 4\cos {20^ \circ } - \sqrt 3 cot{20^ \circ } \)...

Evaluate 4cos⁡20∘−3cot20∘4\cos {20^ \circ } - \sqrt 3 cot{20^ \circ }

Explanation

Solution

Hint : Cot can also be written as the ratio of cosine to sine. Replace the cot with the ratio of cosine and sine. Use the formulas of sin⁡Acos⁡B\sin A\cos B and cos⁡Asin⁡B\cos A\sin B wherever it is necessary. And also remember that sine of negative angle is equal to negative sine angle.
Formulas used:
1. 2cos⁡Asin⁡B=sin⁡(A+B)−sin⁡(A−B)2\cos A\sin B = \sin \left( {A + B} \right) - \sin \left( {A - B} \right)
2. 2sin⁡Acos⁡B=sin⁡(A+B)+sin⁡(A−B)2\sin A\cos B = \sin \left( {A + B} \right) + \sin \left( {A - B} \right)
3. sin⁡(−θ)=−sin⁡θ\sin \left( { - \theta } \right) = - \sin \theta
4. sin⁡A−sin⁡B=2cos⁡(A+B2)sin⁡(A−B2)\sin A - \sin B = 2\cos \left( {\dfrac{{A + B}}{2}} \right)\sin \left( {\dfrac{{A - B}}{2}} \right)

Complete step-by-step answer :
We are given to evaluate the expression 4cos⁡20∘−3cot20∘4\cos {20^ \circ } - \sqrt 3 cot{20^ \circ }
We have cot function in the given expression; express it in terms of cosine and sine as cos⁡20∘sin⁡20∘\dfrac{{\cos {{20}^ \circ }}}{{\sin {{20}^ \circ }}}
Therefore, the expression becomes 4cos⁡20∘−3(cos⁡20∘sin⁡20∘)4\cos {20^ \circ } - \sqrt 3 \left( {\dfrac{{\cos {{20}^ \circ }}}{{\sin {{20}^ \circ }}}} \right)
We are next taking LCM and multiplying it to cosine
⇒4cos⁡20∘sin⁡20∘−3cos⁡20∘sin⁡20∘\Rightarrow \dfrac{{4\cos {{20}^ \circ }\sin {{20}^ \circ } - \sqrt 3 \cos {{20}^ \circ }}}{{\sin {{20}^ \circ }}}
Taking sin⁡20∘\sin {20^ \circ } out, we get
⇒1sin⁡20∘(4cos⁡20∘sin⁡20∘−3cos⁡20∘)\Rightarrow \dfrac{1}{{\sin {{20}^ \circ }}}\left( {4\cos {{20}^ \circ }\sin {{20}^ \circ } - \sqrt 3 \cos {{20}^ \circ }} \right)
⇒1sin⁡20∘[2(2cos⁡20∘sin⁡20∘)−3cos⁡20∘]\Rightarrow \dfrac{1}{{\sin {{20}^ \circ }}}\left[ {2\left( {2\cos {{20}^ \circ }\sin {{20}^ \circ }} \right) - \sqrt 3 \cos {{20}^ \circ }} \right]
As we can see the first term inside the bracket is of the form 2cos⁡Asin⁡B2\cos A\sin B which is equal to sin⁡(A+B)−sin⁡(A−B)\sin \left( {A + B} \right) - \sin \left( {A - B} \right) , where A and B are equal to 20∘{20^ \circ }
Therefore, 2cos⁡20∘sin⁡20∘=sin⁡(20∘+20∘)−sin⁡(20∘−20∘)=sin⁡40∘−0=sin⁡40∘2\cos {20^ \circ }\sin {20^ \circ } = \sin \left( {{{20}^ \circ } + {{20}^ \circ }} \right) - \sin \left( {{{20}^ \circ } - {{20}^ \circ }} \right) = \sin {40^ \circ } - 0 = \sin {40^ \circ }
On substituting the obtained value of 2cos⁡20∘sin⁡20∘2\cos {20^ \circ }\sin {20^ \circ } , we get
⇒1sin⁡20∘[2sin⁡40∘−3cos⁡20∘]\Rightarrow \dfrac{1}{{\sin {{20}^ \circ }}}\left[ {2\sin {{40}^ \circ } - \sqrt 3 \cos {{20}^ \circ }} \right]
Taking 2 out common, we get
⇒2sin⁡20∘[sin⁡40∘−32cos⁡20∘]\Rightarrow \dfrac{2}{{\sin {{20}^ \circ }}}\left[ {\sin {{40}^ \circ } - \dfrac{{\sqrt 3 }}{2}\cos {{20}^ \circ }} \right]
32\dfrac{{\sqrt 3 }}{2} can also be written as sin⁡60∘\sin {60^ \circ }
Therefore, the expression becomes ⇒2sin⁡20∘[sin⁡40∘−sin⁡60∘cos⁡20∘]\Rightarrow \dfrac{2}{{\sin {{20}^ \circ }}}\left[ {\sin {{40}^ \circ } - \sin {{60}^ \circ }\cos {{20}^ \circ }} \right]
Now we are sending 2 back inside
⇒1sin⁡20∘[2sin⁡40∘−2sin⁡60∘cos⁡20∘]\Rightarrow \dfrac{1}{{\sin {{20}^ \circ }}}\left[ {2\sin {{40}^ \circ } - 2\sin {{60}^ \circ }\cos {{20}^ \circ }} \right]
As we can see the second term inside the bracket is in the form 2sin⁡Acos⁡B2\sin A\cos B which is equal to sin⁡(A+B)+sin⁡(A−B)\sin \left( {A + B} \right) + \sin \left( {A - B} \right) , where A is 60∘{60^ \circ } and B is 20∘{20^ \circ }
Therefore, 2sin⁡60∘cos⁡20∘=sin⁡(60∘+20∘)+sin⁡(60∘−20∘)=sin⁡80∘+sin⁡40∘2\sin {60^ \circ }\cos {20^ \circ } = \sin \left( {{{60}^ \circ } + {{20}^ \circ }} \right) + \sin \left( {{{60}^ \circ } - {{20}^ \circ }} \right) = \sin {80^ \circ } + \sin {40^ \circ }
On substituting the obtained value of 2sin⁡60∘cos⁡20∘2\sin {60^ \circ }\cos {20^ \circ } , we get
⇒1sin⁡20∘[2sin⁡40∘−(sin⁡80∘+sin⁡40∘)]\Rightarrow \dfrac{1}{{\sin {{20}^ \circ }}}\left[ {2\sin {{40}^ \circ } - \left( {\sin {{80}^ \circ } + \sin {{40}^ \circ }} \right)} \right]
⇒1sin⁡20∘[2sin⁡40∘−sin⁡80∘−sin⁡40∘]=1sin⁡20∘[sin⁡40∘−sin⁡80∘]\Rightarrow \dfrac{1}{{\sin {{20}^ \circ }}}\left[ {2\sin {{40}^ \circ } - \sin {{80}^ \circ } - \sin {{40}^ \circ }} \right] = \dfrac{1}{{\sin {{20}^ \circ }}}\left[ {\sin {{40}^ \circ } - \sin {{80}^ \circ }} \right]
The above expression in the bracket is in the form sin⁡A−sin⁡B\sin A - \sin B which is equal to 2cos⁡(A+B2)sin⁡(A−B2)2\cos \left( {\dfrac{{A + B}}{2}} \right)\sin \left( {\dfrac{{A - B}}{2}} \right) , where A is 40∘{40^ \circ } and B is 80∘{80^ \circ }
Therefore, sin⁡40∘−sin⁡80∘=2cos⁡(40∘+80∘2)sin⁡(40∘−80∘2)=2cos⁡60∘sin⁡(−20∘)\sin {40^ \circ } - \sin {80^ \circ } = 2\cos \left( {\dfrac{{{{40}^ \circ } + {{80}^ \circ }}}{2}} \right)\sin \left( {\dfrac{{{{40}^ \circ } - {{80}^ \circ }}}{2}} \right) = 2\cos {60^ \circ }\sin \left( { - {{20}^ \circ }} \right) .
We know that sin⁡(−θ)=−sin⁡θ\sin \left( { - \theta } \right) = - \sin \theta , this means sin⁡(−20∘)=−sin⁡20∘\sin \left( { - {{20}^ \circ }} \right) = - \sin {20^ \circ }
Therefore, sin⁡40∘−sin⁡80∘=−2cos⁡60∘sin⁡20∘\sin {40^ \circ } - \sin {80^ \circ } = - 2\cos {60^ \circ }\sin {20^ \circ }
The expression becomes
1sin⁡20∘(−2cos⁡60∘sin⁡20∘)=−sin⁡20∘sin⁡20∘(2cos⁡60∘)=−2×cos⁡60∘\dfrac{1}{{\sin {{20}^ \circ }}}\left( { - 2\cos {{60}^ \circ }\sin {{20}^ \circ }} \right) = \dfrac{{ - \sin {{20}^ \circ }}}{{\sin {{20}^ \circ }}}\left( {2\cos {{60}^ \circ }} \right) = - 2 \times \cos {60^ \circ }
The value of cos⁡60∘=12\cos {60^ \circ } = \dfrac{1}{2}
Therefore, −2×cos⁡60∘=−2×12=−1- 2 \times \cos {60^ \circ } = - 2 \times \dfrac{1}{2} = - 1
The value of 4cos⁡20∘−3cot20∘4\cos {20^ \circ } - \sqrt 3 cot{20^ \circ } is -1.
So, the correct answer is “-1”.

Note : Cot is the inverse of tan, tan is the ratio of sine and cosine. Do not confuse that cot is the ratio of sine and cosine; rather it is the ratio of cosine and sine. For these types of questions, one needs to know all the formulas involving sine and cosine. Be careful with the signs of the terms while writing the formulas.