Question

If $\cos \left( {x - y} \right),\cos x,\cos \left( {x + y} \right)$ are in H.P, then $\cos x\sec \left( {\dfrac{y}{2}} \right)$

Answer 1

Hint:
Since we are given that $\cos \left( {x - y} \right), \cos x,\cos \left( {x + y} \right)$ are in HP we can use the relationship $b = \dfrac{{2ac}}{{a + c}}$ when a , b , c are in HP and then by using the identity $\cos x + \cos y = 2\left[ {\cos \left( {\dfrac{{x - y}}{2}} \right)\cos \left( {\dfrac{{x + y}}{2}} \right)} \right]$ we get $\cos x = \dfrac{{\cos 2x\cos 2y}}{{\cos \left( {x - y} \right) + \cos \left( {x + y} \right)}}$ and then by using the identity $2{\cos ^2}x = 1 + \cos 2x$ we get the required value.

Complete step by step solution:
We are given that $\cos \left( {x - y} \right), \cos x, \cos \left( {x + y} \right)$ are in HP.
We know that when a , b, c are in H.P then
$ \Rightarrow b = \dfrac{{2ac}}{{a + c}}$
Using this we get
$ \Rightarrow \cos x = \dfrac{{2\cos \left( {x - y} \right)\cos \left( {x + y} \right)}}{{\cos \left( {x - y} \right) + \cos \left( {x + y} \right)}}$
Now let's proceed to solve this using the identities to get the required result
By using the identity $\cos x + \cos y = 2\left[ {\cos \left( {\dfrac{{x - y}}{2}} \right)\cos \left( {\dfrac{{x + y}}{2}} \right)} \right]$
We get our right hand side to be
$ \Rightarrow \cos x = \dfrac{{\cos 2x\cos 2y}}{{\cos \left( {x - y} \right) + \cos \left( {x + y} \right)}}$…………….(1)
Now by using he identity $2{\cos ^2}x = 1 + \cos 2x$
We get $ \Rightarrow 2{\cos ^2}x - 1 = \cos 2x$
Using this in (1) we get
$
   \Rightarrow \cos x = \dfrac{{\left( {2{{\cos }^2}x - 1} \right) + \left( {2{{\cos }^2}y - 1} \right)}}{{2\cos x\cos y}} \\
   \Rightarrow \cos x = \dfrac{{2{{\cos }^2}x - 1 + 2{{\cos }^2}y - 1}}{{2\cos x\cos y}} \\
   \Rightarrow \cos x = \dfrac{{2{{\cos }^2}x + 2{{\cos }^2}y - 2}}{{2\cos x\cos y}} \\
 $
Cross multiplying we get
$
   \Rightarrow 2{\cos ^2}x\cos y = 2{\cos ^2}x + 2{\cos ^2}y - 2 \\
   \Rightarrow 2{\cos ^2}x\cos y - 2{\cos ^2}x = 2{\cos ^2}y - 2 \\
   \Rightarrow 2{\cos ^2}x\left( {\cos y - 1} \right) = 2\left( {{{\cos }^2}y - 1} \right) \\
   \Rightarrow 2{\cos ^2}x\left( {\cos y - 1} \right) = 2\left( {\cos y + 1} \right)\left( {\cos y - 1} \right) \\
   \Rightarrow {\cos ^2}x = \left( {\cos y + 1} \right) \\
 $
Using the identity $2{\cos ^2}x = 1 + \cos 2x$
We get $2{\cos ^2}\left( {\dfrac{y}{2}} \right) = 1 + \cos y$
Using this we get
$
   \Rightarrow {\cos ^2}x = 2{\cos ^2}\left( {\dfrac{y}{2}} \right) \\
   \Rightarrow \dfrac{{{{\cos }^2}x}}{{{{\cos }^2}\left( {\dfrac{y}{2}} \right)}} = 2 \\
   \Rightarrow {\cos ^2}x{\sec ^2}\left( {\dfrac{y}{2}} \right) = 2 \\
   \Rightarrow \cos x\sec \left( {\dfrac{y}{2}} \right) = \pm \sqrt 2 \\
 $
Hence we get the required value.

Note:
In mathematics, a harmonic progression (or harmonic sequence) is a progression formed by taking the reciprocals of an arithmetic progression. Equivalently, a sequence is a harmonic progression when each term is the harmonic mean of the neighboring terms.
Steps to keep in mind while solving trigonometric problems are
1) Always start from the more complex side
2) Express everything into sine and cosine
3) Combine terms into a single fraction
4) Use Pythagorean identities to transform between $\sin^{2} \theta$ and $\cos^{2} \theta$
5) Know when to apply double angle formula
6) Know when to apply addition formula
7) Good old expand/ factorize/ simplify/ cancelling