
Find the value of trigonometric identity for the given expression \[\cos (\dfrac{x}{2}) - \cos x = 0\,if\,0^\circ < x < 360^\circ \]?
Answer
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Hint: To solve this question we have to simplify the fraction and then compare the value of the trigonometric identity on both side of the equation, on comparison we can solve for the variable “x”, here on the second hand of equation we are provided with a number this we have to change for the value of “cos” by doing some arrangement we can find the exact value for the given numerical value.
Complete step by step solution:
Given question is \[\cos (\dfrac{x}{2}) - \cos x = 0\,\,if\,\,0^\circ < x <
360^\circ \]
Now we have to simplify the given expression, first we are going to solve for the left hand side of the equation,
Here we have to use the half angle property for angle in “cos”
The property is:
\[ \Rightarrow \cos 2A = 2{\cos ^2}\left( {\dfrac{A}{2}} \right) - 1\]
Using this property in our question we get:
\[
\Rightarrow \cos (\dfrac{x}{2}) - \cos x = 0 \Rightarrow \cos (\dfrac{x}{2}) - 2{\cos ^2}\left(
{\dfrac{x}{2}} \right) + 1 = 0 \\
\Rightarrow taking\,common\,\cos \left( {\dfrac{x}{2}}
\right)\,and\,transferring\,1\,on\,other\,side\,we\,get: \\
\Rightarrow \cos \left( {\dfrac{x}{2}} \right)\left[ {1 - 2\cos \left( {\dfrac{x}{2}} \right)} \right] = - 1
\\
\Rightarrow now,\,\cos \left( {\dfrac{x}{2}} \right) = - 1\,has\,1\,solutions\,that\,is\, \\
\Rightarrow \dfrac{x}{2} = \pi \\
\Rightarrow x = \,2\pi \\
and \\
\Rightarrow 2\cos \left( {\dfrac{x}{2}} \right) = 2\,has\,1\,solutions \\
\]
\[
\Rightarrow \cos \left( {\dfrac{x}{2}} \right) = 1 \\
\Rightarrow \dfrac{x}{2} = 0 \\
\Rightarrow x = 0 \\
answer\, = \pi \,and\,0\,is\,not\,accepted\,because\,not\,in\,the\,range\,of\,the\,question \\
\]
Hence we find the final value of angles for the variable.
Additional Information: While dealing with expressions for trigonometric identity, here we have used half angle property, and then after simplification we obtain the result after simplifying from equals to sign. For confirmation of your solution we can put the final obtained value in the main equation and can check for the answer.
Note: The trigonometric function used here can be replaced by expanding the single “cos” term by using the “cos(A)” for each term and then we can solve further, but it will become very lengthy, and steps used here are the same as we will do with the other process.
Complete step by step solution:
Given question is \[\cos (\dfrac{x}{2}) - \cos x = 0\,\,if\,\,0^\circ < x <
360^\circ \]
Now we have to simplify the given expression, first we are going to solve for the left hand side of the equation,
Here we have to use the half angle property for angle in “cos”
The property is:
\[ \Rightarrow \cos 2A = 2{\cos ^2}\left( {\dfrac{A}{2}} \right) - 1\]
Using this property in our question we get:
\[
\Rightarrow \cos (\dfrac{x}{2}) - \cos x = 0 \Rightarrow \cos (\dfrac{x}{2}) - 2{\cos ^2}\left(
{\dfrac{x}{2}} \right) + 1 = 0 \\
\Rightarrow taking\,common\,\cos \left( {\dfrac{x}{2}}
\right)\,and\,transferring\,1\,on\,other\,side\,we\,get: \\
\Rightarrow \cos \left( {\dfrac{x}{2}} \right)\left[ {1 - 2\cos \left( {\dfrac{x}{2}} \right)} \right] = - 1
\\
\Rightarrow now,\,\cos \left( {\dfrac{x}{2}} \right) = - 1\,has\,1\,solutions\,that\,is\, \\
\Rightarrow \dfrac{x}{2} = \pi \\
\Rightarrow x = \,2\pi \\
and \\
\Rightarrow 2\cos \left( {\dfrac{x}{2}} \right) = 2\,has\,1\,solutions \\
\]
\[
\Rightarrow \cos \left( {\dfrac{x}{2}} \right) = 1 \\
\Rightarrow \dfrac{x}{2} = 0 \\
\Rightarrow x = 0 \\
answer\, = \pi \,and\,0\,is\,not\,accepted\,because\,not\,in\,the\,range\,of\,the\,question \\
\]
Hence we find the final value of angles for the variable.
Additional Information: While dealing with expressions for trigonometric identity, here we have used half angle property, and then after simplification we obtain the result after simplifying from equals to sign. For confirmation of your solution we can put the final obtained value in the main equation and can check for the answer.
Note: The trigonometric function used here can be replaced by expanding the single “cos” term by using the “cos(A)” for each term and then we can solve further, but it will become very lengthy, and steps used here are the same as we will do with the other process.
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