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

If sin⁡(x−y)\sin (x-y), sin⁡x\sin x and sin⁡(x+y)\sin (x+y) are in H.P., then sin⁡xsec⁡y2\sin x\sec \dfrac{y}{2} is equal to:

Explanation

Solution

Hint: Think of the basic definition of Harmonic progression, and use the harmonic mean of two numbers for sin⁡(x−y)\sin (x-y) and sin⁡(x+y)\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:1a,1a+d,1a+2d............\dfrac{1}{a},\dfrac{1}{a+d},\dfrac{1}{a+2d}............
Where a,a+d,a+2d...............a,a+d,a+2d............... will represent an arithmetic progression.
Also, the harmonic mean of two numbers a and b is 2aba+b\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.
∴sin⁡x=2sin⁡(x+y)sin⁡(x−y)sin⁡(x+y)+sin⁡(x−y)\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=2sin⁡(A+B2)cos⁡(A−B2)\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=2sin⁡(x+y)sin⁡(x−y)2sin⁡xcos⁡y\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)=cos2y−cos⁡2x\sin (x+y)\sin (x-y)=co{{s}^{2}}y-{{\cos }^{2}}x .
sin⁡x=cos⁡2y−cos⁡2xsin⁡xcos⁡y\sin x=\dfrac{{{\cos }^{2}}y-{{\cos }^{2}}x}{\sin x\cos y}
⇒sin⁡2xcos⁡y=cos⁡2y−cos⁡2x\Rightarrow {{\sin }^{2}}x\cos y={{\cos }^{2}}y-{{\cos }^{2}}x
Now we will use the identity that cos⁡2x=1−sin⁡2x{{\cos }^{2}}x=1-{{\sin }^{2}}x . On doing so, we get
sin⁡2xcos⁡y=cos⁡2y−1+sin⁡2x{{\sin }^{2}}x\cos y={{\cos }^{2}}y-1+{{\sin }^{2}}x
⇒sin⁡2xcos⁡y−sin⁡2x=cos⁡2y−1\Rightarrow {{\sin }^{2}}x\cos y-{{\sin }^{2}}x={{\cos }^{2}}y-1
⇒sin⁡2x(cos⁡y−1)=(cos⁡y−1)(cos⁡y+1)\Rightarrow {{\sin }^{2}}x\left( \cos y-1 \right)=\left( \cos y-1 \right)\left( \cos y+1 \right)
⇒sin⁡2x=cos⁡y+1\Rightarrow {{\sin }^{2}}x=\cos y+1
Now, we know that cos⁡2A=2cos⁡2A−1\cos 2A=2{{\cos }^{2}}A-1 . Therefore, our equation comes out to be:
sin⁡2x=2cos⁡2y2−1+1{{\sin }^{2}}x=2{{\cos }^{2}}\dfrac{y}{2}-1+1
⇒sin⁡2xcos⁡2y2=2\Rightarrow \dfrac{{{\sin }^{2}}x}{{{\cos }^{2}}\dfrac{y}{2}}=2
Now, we can represent 1cos⁡y2 as secy2\dfrac{1}{\cos \dfrac{y}{2}}\text{ as sec}\dfrac{y}{2} . On doing so, we get
⇒sin⁡2xsec⁡2y2=2\Rightarrow {{\sin }^{2}}x{{\sec }^{2}}\dfrac{y}{2}=2
So, the final equation we get is (sin⁡xsec⁡y2)2=2{{\left( \sin x\sec \dfrac{y}{2} \right)}^{2}}=2 which implies sin⁡xsec⁡y2=±2\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−11+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.