Question

For \[A={{133}^{\circ }},2\cos \left( \dfrac{A}{2} \right)\] is equal to
A. \[-\sqrt{1+\sin (A)}-\sqrt{1-\sin (A)}\]
B. \[-\sqrt{1+\sin (A)}+\sqrt{1-\sin (A)}\]
C. \[\sqrt{1+\sin (A)}-\sqrt{1-\sin (A)}\]
D. \[\sqrt{1+\sin (A)}+\sqrt{1-\sin (A)}\]

Answer 1

Hint: According to given question to find the value of \[2\cos \left( \dfrac{A}{2} \right)\] we have to solve each option (hit and try method) by applying Trigonometry property to get the answer. \[\sqrt{1+\sin (A)}-\sqrt{1-\sin (A)}\]By solving this we get the answer as\[2\cos \left( \dfrac{A}{2} \right)\].

Complete step by step answer:
Given data is:
\[A={{133}^{\circ }}----(1)\]
If we divide this equation\[(1)\] we get:
\[\dfrac{A}{2}=\dfrac{{{133}^{\circ }}}{2}=\,{{66.5}^{\circ }}\] (Acute angle)

Now, we have to find the term of \[\sqrt{1+\sin (A)}\]
By applying trigonometry identity formula, \[{{\sin }^{2}}\left( \dfrac{A}{2} \right)+{{\cos }^{2}}\left( \dfrac{A}{2} \right)=1\]
By solving substituting the formula we get:
\[\sqrt{1+\sin (A)}=\sqrt{{{\sin }^{2}}\left( \dfrac{A}{2} \right)+{{\cos }^{2}}\left( \dfrac{A}{2} \right)+\sin (A)}--(2)\]
By using the half angle formula,
\[\sin (A)=2\sin \left( \dfrac{A}{2} \right)\cos \left( \dfrac{A}{2} \right)\]
By substituting the half angle formula in equation\[(2)\]
\[\sqrt{1+\sin (A)}=\sqrt{{{\sin }^{2}}\left( \dfrac{A}{2} \right)+{{\cos }^{2}}\left( \dfrac{A}{2} \right)+2\sin \left( \dfrac{A}{2} \right)\cos \left( \dfrac{A}{2} \right)}\]
By using the property of \[{{(a+b)}^{2}}={{a}^{2}}+2ab+{{b}^{2}}\]
We get,
\[\sqrt{1+\sin (A)}=\sqrt{{{\left( \sin \left( \dfrac{A}{2} \right)+\cos \left( \dfrac{A}{2} \right) \right)}^{2}}}\]
By simplifying the further equation
We get:
\[\sqrt{1+\sin (A)}=\sin \left( \dfrac{A}{2} \right)+\cos \left( \dfrac{A}{2} \right)----(3)\]
Similarly we can solve for \[\sqrt{1-\sin (A)}\]
We get:
By using trigonometry identity property,
\[{{\sin }^{2}}\left( \dfrac{A}{2} \right)+{{\cos }^{2}}\left( \dfrac{A}{2} \right)=1\]
By substituting the formula we get:
\[\sqrt{1-\sin (A)}=\sqrt{{{\sin }^{2}}\left( \dfrac{A}{2} \right)+{{\cos }^{2}}\left( \dfrac{A}{2} \right)-\sin (A)}--(4)\]
By using the half angle formula,
\[\sin (A)=2\sin \left( \dfrac{A}{2} \right)\cos \left( \dfrac{A}{2} \right)\]
By substituting the half angle formula in equation \[(4)\]
\[\sqrt{1-\sin (A)}=\sqrt{{{\sin }^{2}}\left( \dfrac{A}{2} \right)+{{\cos }^{2}}\left( \dfrac{A}{2} \right)-2\sin \left( \dfrac{A}{2} \right)\cos \left( \dfrac{A}{2} \right)}\]
By using the property of
\[{{(a-b)}^{2}}={{a}^{2}}-2ab+{{b}^{2}}\]
\[\sqrt{1-\sin (A)}=\sqrt{{{\left( \sin \left( \dfrac{A}{2} \right)-\cos \left( \dfrac{A}{2} \right) \right)}^{2}}}\]
By simplifying the further equation we get:
\[\sqrt{1-\sin (A)}=\sin \left( \dfrac{A}{2} \right)-\cos \left( \dfrac{A}{2} \right)----(5)\]
By subtracting the two equation \[(3)\]and \[(5)\]
\[\sqrt{1+\sin (A)}-\sqrt{1-\sin (A)}=\sin \left( \dfrac{A}{2} \right)+\cos \left( \dfrac{A}{2} \right)-\left( \sin \left( \dfrac{A}{2} \right)-\cos \left( \dfrac{A}{2} \right) \right)\]
By simplifying the equation further we get:
\[\sqrt{1+\sin (A)}-\sqrt{1-\sin (A)}=\sin \left( \dfrac{A}{2} \right)+\cos \left( \dfrac{A}{2} \right)-\sin \left( \dfrac{A}{2} \right)+\cos \left( \dfrac{A}{2} \right)\]
So here, \[\sin \left( \dfrac{A}{2} \right)\] get cancelled out
\[\sqrt{1+\sin (A)}-\sqrt{1-\sin (A)}=\cos \left( \dfrac{A}{2} \right)+\cos \left( \dfrac{A}{2} \right)\]
\[\sqrt{1+\sin (A)}-\sqrt{1-\sin (A)}=2\cos \left( \dfrac{A}{2} \right)\]

So, the correct answer is “Option C”.

Note: In this question asked to find the value of \[2\cos \left( \dfrac{A}{2} \right)\] then we can subtract the two equation in the above solution if the question is asked about \[2\sin \left( \dfrac{A}{2} \right)\] then we have to add the two equation to get the value. So in this way sometimes we can use the hit and try method and also apply the trigonometry property and use the half angle formula in the option to get the correct options.