Question

If $\cos \theta = \dfrac{3}{5}$and $\cos \phi = \dfrac{4}{5}$ where $\theta$and $\phi$are positive acute angles, then $\Rightarrow \cos \left( {\dfrac{{\theta - \phi }}{2}} \right) = \sqrt {\dfrac{{1 + \cos \theta \cos \phi + \sin \theta \sin \phi }}{2}}$
A. $\dfrac{7}{{\sqrt 2 }}$
B. $\dfrac{7}{{5\sqrt 2 }}$
C. $\dfrac{7}{{\sqrt 5 }}$
D. $\dfrac{7}{{2\sqrt 5 }}$

Answer 1

Here we have to find the value of$\cos \left( {\theta - \phi } \right)/2$. Since we don’t have any standard formula to find this value, we will derive this term from the existing standard formulas by changing the angles for our convenience. Then, by substituting the values given we will find the required value.
Formula: Some formulas that we need to know to solve this problem:
$\cos (2x) = 2{\cos ^2}(x) - 1$
$\cos (A - B) = \cos A\cos B + \sin A\sin B$
${\sin ^2}\theta + {\cos ^2}\theta = 1$

Complete step by step answer:
It is given that $\cos \theta = \dfrac{3}{5}$and$\cos \phi = \dfrac{4}{5}$ where $\theta$and $\phi$are positive acute angles. We aim to find the value of$\cos \left( {\theta - \phi } \right)/2$.
Since we don’t have any standard formula to find this term, we will now derive the required term using the standard formula that we have.
Consider the formula $\cos (2x) = 2{\cos ^2}(x) - 1$
Now we will modify this formula to get a cosine function with a half-angle.
$\Rightarrow \cos (x) = 2{\cos ^2}\left( {\dfrac{x}{2}} \right) - 1$
Now, let's rearrange this expression for our convenience.
$\Rightarrow 2{\cos ^2}\left( {\dfrac{x}{2}} \right) = 1 + \cos (x)$
$\Rightarrow {\cos ^2}\left( {\dfrac{x}{2}} \right) = \dfrac{{1 + \cos (x)}}{2}$
$\Rightarrow \cos \left( {\dfrac{x}{2}} \right) = \sqrt {\dfrac{{1 + \cos (x)}}{2}}$
Thus, we have attained the required cosine formula with a half-angle. Now let us substitute$x = \theta - \phi$.
$\Rightarrow \cos \left( {\dfrac{{\theta - \phi }}{2}} \right) = \sqrt {\dfrac{{1 + \cos (\theta - \phi )}}{2}}$
Here, we have a term$\cos (\theta - \phi )$. Let us expand this term using the formula$\cos (A - B) = \cos A\cos B + \sin A\sin B$.
Thus, we get$\cos (\theta - \phi ) = \cos \theta \cos \phi + \sin \theta \sin \phi$.
Let’s substitute it in the expression$\cos \left( {\dfrac{{\theta - \phi }}{2}} \right) = \sqrt {\dfrac{{1 + \cos (\theta - \phi )}}{2}}$,
We have the values of \cos \theta $$$$\& $$$$\cos \phi but we don’t have values of\sin \theta $$$$\& $$$$\sin \phi .
Consider the formula ${\sin ^2}\theta + {\cos ^2}\theta = 1$
Let us modify this formula to get the value of$\sin \theta$.
{\sin ^2}\theta + {\cos ^2}\theta = 1$$$$ \Rightarrow {\sin ^2}\theta = 1 - {\cos ^2}\theta
$\Rightarrow \sin \theta = \sqrt {1 - {{\cos }^2}\theta }$
We already have the value of$\cos \theta = \dfrac{3}{5}$. Let’s substitute it in the above expression and simplify it.
$\Rightarrow \sin \theta = \sqrt {1 - {{\left( {\dfrac{3}{5}} \right)}^2}}$
$\Rightarrow \sin \theta = \sqrt {1 - \left( {\dfrac{9}{{25}}} \right)}$
$\Rightarrow \sin \theta = \sqrt {\dfrac{{25 - 9}}{{25}}}$
$\Rightarrow \sin \theta = \sqrt {\dfrac{{16}}{{25}}}$
$\Rightarrow \sin \theta = \dfrac{4}{5}$
Now let us find the value of the term $\sin \phi$by changing the angle in the expression$\sin \theta = \sqrt {1 - {{\cos }^2}\theta }$.
Thus, we get$\sin \phi = \sqrt {1 - {{\cos }^2}\phi }$. Now substitute the value of$\cos \phi$ in this expression and simplify it to get the value of$\sin \phi$.
$\sin \phi = \sqrt {1 - {{\cos }^2}\phi } \Rightarrow \sin \phi = \sqrt {1 - {{\left( {\dfrac{4}{5}} \right)}^2}}$
$\Rightarrow \sin \phi = \sqrt {1 - \left( {\dfrac{{16}}{{25}}} \right)}$
$\Rightarrow \sin \phi = \sqrt {\dfrac{{25 - 16}}{{25}}}$
$\Rightarrow \sin \phi = \sqrt {\dfrac{9}{{25}}}$
$\Rightarrow \sin \phi = \dfrac{3}{5}$
Now we got the values of$\cos \theta ,\cos \phi ,\sin \theta \& \sin \phi$. Now let’s substitute it in$\cos (\theta - \phi ) = \cos \theta \cos \phi + \sin \theta \sin \phi$.
$\Rightarrow \cos (\theta - \phi ) = \left( {\dfrac{3}{5}} \right)\left( {\dfrac{4}{5}} \right) + \left( {\dfrac{4}{5}} \right)\left( {\dfrac{3}{5}} \right)$
$\Rightarrow \cos (\theta - \phi ) = \dfrac{{12}}{{25}} + \dfrac{{12}}{{25}}$
$\Rightarrow \cos (\theta - \phi ) = \dfrac{{24}}{{25}}$
Thus, we got the value of$\cos (\theta - \phi )$. Let us substitute it in$\cos \left( {\dfrac{{\theta - \phi }}{2}} \right) = \sqrt {\dfrac{{1 + \cos (\theta - \phi )}}{2}}$ and simplify it.
$\cos \left( {\dfrac{{\theta - \phi }}{2}} \right) = \sqrt {\dfrac{{1 + \left( {\dfrac{{24}}{{25}}} \right)}}{2}}$
$\Rightarrow \cos \left( {\dfrac{{\theta - \phi }}{2}} \right) = \sqrt {\dfrac{{\left( {\dfrac{{25 + 24}}{{25}}} \right)}}{2}}$
$\Rightarrow \cos \left( {\dfrac{{\theta - \phi }}{2}} \right) = \sqrt {\dfrac{{\left( {\dfrac{{49}}{{25}}} \right)}}{2}}$
$\Rightarrow \cos \left( {\dfrac{{\theta - \phi }}{2}} \right) = \sqrt {\dfrac{{49}}{{2 \times 25}}}$
$\cos \left( {\dfrac{{\theta - \phi }}{2}} \right) = \dfrac{7}{{5\sqrt 2 }}$
Thus, we got the value of$\cos \left( {\dfrac{{\theta - \phi }}{2}} \right)$.
So, the correct answer is “Option B”.

Note: Here we don’t have a standard for the required expression so we derived it from the existing formula. Also, we have to choose the formula that contains the terms which are available to us and the terms that we require. So that we can find the required value by substituting the values we already have.