Question

If $\sin (x-y)$, $\sin x$ and $\sin (x+y)$ are in H.P., then $\sin x\sec \dfrac{y}{2}$ is equal to:

Answer 1

Hint: Think of the basic definition of Harmonic progression, and use the harmonic mean of two numbers for $\sin (x-y)$ and $\sin(x+y)$ and equate it with sin x. Solve the equation to reach the required result.

Complete step-by-step solution -
Before starting with the solution to the above question, let us talk about harmonic progression.
In mathematics, harmonic progression is defined as a sequence of numbers, for which the sequence of the reciprocals of its terms gives an arithmetic progression.
We can represent a general harmonic progression as:$\dfrac{1}{a},\dfrac{1}{a+d},\dfrac{1}{a+2d}............$
Where $a,a+d,a+2d...............$ will represent an arithmetic progression.
Also, the harmonic mean of two numbers a and b is $\dfrac{2ab}{a+b}$.
Now, let us start with the solution to the above question. According to the question sin(x-y), sinx and sin(x+y) are in H.P., which implies that H.M. of numbers for sin(x-y) and sin(x+y) is sinx.
$\therefore \sin x=\dfrac{2\sin (x+y)\sin (x-y)}{\sin (x+y)+\sin (x-y)}$
Now we know that $\sin A+\sin B=2\sin \left( \dfrac{A+B}{2} \right)\cos \left( \dfrac{A-B}{2} \right)$ . So, our equation becomes:
$\sin x=\dfrac{2\sin (x+y)\sin (x-y)}{2\sin x\cos y}$
We also know that $\sin (x+y)\sin (x-y)=co{{s}^{2}}y-{{\cos }^{2}}x$ .
\[\sin x=\dfrac{{{\cos }^{2}}y-{{\cos }^{2}}x}{\sin x\cos y}\]
\[\Rightarrow {{\sin }^{2}}x\cos y={{\cos }^{2}}y-{{\cos }^{2}}x\]
Now we will use the identity that ${{\cos }^{2}}x=1-{{\sin }^{2}}x$ . On doing so, we get
\[{{\sin }^{2}}x\cos y={{\cos }^{2}}y-1+{{\sin }^{2}}x\]
\[\Rightarrow {{\sin }^{2}}x\cos y-{{\sin }^{2}}x={{\cos }^{2}}y-1\]
\[\Rightarrow {{\sin }^{2}}x\left( \cos y-1 \right)=\left( \cos y-1 \right)\left( \cos y+1 \right)\]
\[\Rightarrow {{\sin }^{2}}x=\cos y+1\]
Now, we know that $\cos 2A=2{{\cos }^{2}}A-1$ . Therefore, our equation comes out to be:
\[{{\sin }^{2}}x=2{{\cos }^{2}}\dfrac{y}{2}-1+1\]
\[\Rightarrow \dfrac{{{\sin }^{2}}x}{{{\cos }^{2}}\dfrac{y}{2}}=2\]
Now, we can represent $\dfrac{1}{\cos \dfrac{y}{2}}\text{ as sec}\dfrac{y}{2}$ . On doing so, we get
\[\Rightarrow {{\sin }^{2}}x{{\sec }^{2}}\dfrac{y}{2}=2\]
So, the final equation we get is \[{{\left( \sin x\sec \dfrac{y}{2} \right)}^{2}}=2\] which implies \[\sin x\sec \dfrac{y}{2}=\pm \sqrt{2}\] .

Note: Be careful about the calculation and the signs while opening the brackets. The general mistake that a student can make is $1+x-(x-1)=1+x-x-1$. Also, you need to remember the properties related to complementary angles and trigonometric ratios, along with all the identities related to them.