
Given $\cos \left( x \right)=-\dfrac{3}{-7}$ and $\dfrac{\pi }{2} < x < \pi $ , how do you find $\cos \left( \dfrac{x}{2} \right)$ ?
(a) Using trigonometric identities
(b) Using linear formulas
(c) Using trigonometric table
(d) All of the above
Answer
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Hint: We are to find the value of $\cos \left( \dfrac{x}{2} \right)$ in a specific domain when the value of cos x is given to us. We will deal this problem using the trigonometric identity $\cos 2x=2{{\cos }^{2}}x-1$
Here, we will use the $\dfrac{x}{2}$ in place of x and now we will check in which quadrant the value of cosine takes place and thus we can choose the sign of the value.
Complete step by step solution:
According to the question, we have $\cos \left( x \right)=-\dfrac{3}{-7}$
This can also be written as, $\cos \left( x \right)=\dfrac{3}{7}$
Now, again, x can be written as, $2\times \dfrac{x}{2}$ ,
So, we have,
$\cos \left( x \right)=\cos \left( 2\times \dfrac{x}{2} \right)$
Again from the formula given as, $\cos 2x=2{{\cos }^{2}}x-1$ we have,
$\Rightarrow {{\cos }^{2}}\left( \dfrac{x}{2} \right)-{{\sin }^{2}}\left( \dfrac{x}{2} \right)$
Again, ${{\sin }^{2}}x=1-{{\cos }^{2}}x$ , so we can write,
$\Rightarrow {{\cos }^{2}}\left( \dfrac{x}{2} \right)-\left( 1-{{\cos }^{2}}\left( \dfrac{x}{2} \right) \right)$
Simplifying,
$\Rightarrow 2{{\cos }^{2}}\left( \dfrac{x}{2} \right)-1$
So, from $\cos \left( x \right)=\dfrac{3}{7}$ ,
We are having,
$\Rightarrow 2{{\cos }^{2}}\left( \dfrac{x}{2} \right)-1=\dfrac{3}{7}$
Adding 1 on both sides,
$\Rightarrow 2{{\cos }^{2}}\left( \dfrac{x}{2} \right)=1+\dfrac{3}{7}$
$\Rightarrow 2{{\cos }^{2}}\left( \dfrac{x}{2} \right)=\dfrac{10}{7}$
So, we have the value of ${{\cos }^{2}}\left( \dfrac{x}{2} \right)$ as $\dfrac{5}{7}$ .
Then, $\cos \left( \dfrac{x}{2} \right)=\pm \sqrt{\dfrac{5}{7}}$
The domain was originally $\dfrac{\pi }{2}So, we have, $\cos \left( \dfrac{x}{2} \right)=\sqrt{\dfrac{5}{7}}$, as cosine is positive in the first and fourth quadrant and negative in the second and the third quadrant.
Hence the solution is, $\cos \left( \dfrac{x}{2} \right)=\sqrt{\dfrac{5}{7}}$.
So, the correct answer is “Option a”.
Note: In mathematics, trigonometric identities are equalities that involve trigonometric functions and are true for every value of the occurring variables for which both sides of the equality are defined. Geometrically, these are identities involving certain functions of one or more angles. They are distinct from triangle identities, which are identities potentially involving angles but also involving side lengths or other lengths of a triangle.
Here, we will use the $\dfrac{x}{2}$ in place of x and now we will check in which quadrant the value of cosine takes place and thus we can choose the sign of the value.
Complete step by step solution:
According to the question, we have $\cos \left( x \right)=-\dfrac{3}{-7}$
This can also be written as, $\cos \left( x \right)=\dfrac{3}{7}$
Now, again, x can be written as, $2\times \dfrac{x}{2}$ ,
So, we have,
$\cos \left( x \right)=\cos \left( 2\times \dfrac{x}{2} \right)$
Again from the formula given as, $\cos 2x=2{{\cos }^{2}}x-1$ we have,
$\Rightarrow {{\cos }^{2}}\left( \dfrac{x}{2} \right)-{{\sin }^{2}}\left( \dfrac{x}{2} \right)$
Again, ${{\sin }^{2}}x=1-{{\cos }^{2}}x$ , so we can write,
$\Rightarrow {{\cos }^{2}}\left( \dfrac{x}{2} \right)-\left( 1-{{\cos }^{2}}\left( \dfrac{x}{2} \right) \right)$
Simplifying,
$\Rightarrow 2{{\cos }^{2}}\left( \dfrac{x}{2} \right)-1$
So, from $\cos \left( x \right)=\dfrac{3}{7}$ ,
We are having,
$\Rightarrow 2{{\cos }^{2}}\left( \dfrac{x}{2} \right)-1=\dfrac{3}{7}$
Adding 1 on both sides,
$\Rightarrow 2{{\cos }^{2}}\left( \dfrac{x}{2} \right)=1+\dfrac{3}{7}$
$\Rightarrow 2{{\cos }^{2}}\left( \dfrac{x}{2} \right)=\dfrac{10}{7}$
So, we have the value of ${{\cos }^{2}}\left( \dfrac{x}{2} \right)$ as $\dfrac{5}{7}$ .
Then, $\cos \left( \dfrac{x}{2} \right)=\pm \sqrt{\dfrac{5}{7}}$
The domain was originally $\dfrac{\pi }{2}
Hence the solution is, $\cos \left( \dfrac{x}{2} \right)=\sqrt{\dfrac{5}{7}}$.
So, the correct answer is “Option a”.
Note: In mathematics, trigonometric identities are equalities that involve trigonometric functions and are true for every value of the occurring variables for which both sides of the equality are defined. Geometrically, these are identities involving certain functions of one or more angles. They are distinct from triangle identities, which are identities potentially involving angles but also involving side lengths or other lengths of a triangle.
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