Question

If sinx+siny=a and cosx+cosy=b then find tan(x-y/2)

Trigonometric equation and identitie
ANSWERED
asked 2020-10-18
If \(\displaystyle{\sin{{x}}}+{\sin{{y}}}={a}{\quad\text{and}\quad}{\cos{{x}}}+{\cos{{y}}}={b}\) then find \(\displaystyle{\tan{{\left({x}-\frac{{y}}{{2}}\right)}}}\)

Answers (2)

2020-10-19

\(\displaystyle{\sin{{\left({A}+{B}\right)}}}={\sin{{A}}}{\cos{{B}}}+{\cos{{A}}}{\sin{{B}}}{\quad\text{and}\quad}{\sin{{\left({A}-{B}\right)}}}={\sin{{A}}}{\cos{{B}}}-{\cos{{A}}}{\sin{{B}}}.\)
\(\displaystyle{\cos{{\left({A}+{B}\right)}}}={\cos{{A}}}{\cos{{B}}}-{\sin{{A}}}{\sin{{B}}}{\quad\text{and}\quad}{\cos{{\left({A}-{B}\right)}}}={\cos{{A}}}{\cos{{B}}}+{\sin{{A}}}{\sin{{B}}}.\)
These are true for all A and B. So, \(\displaystyle{\sin{{\left({A}+{B}\right)}}}+{\sin{{\left({A}-{B}\right)}}}={2}{\sin{{A}}}{\cos{{B}}}{\quad\text{and}\quad}{\cos{{\left({A}+{B}\right)}}}+{\cos{{\left({A}-{B}\right)}}}={2}{\cos{{A}}}{\cos{{B}}}.\)
Therefore relating A and B to x and y we get\(\displaystyle{A}+{B}={x}{\quad\text{and}\quad}{A}-{B}={y}.{F}{r}{o}{m}{t}{h}{e}{s}{e}{A}=\frac{{{x}+{y}}}{{2}}{\quad\text{and}\quad}{B}=\frac{{{x}-{y}}}{{2}}.\)
We can now write \(\displaystyle{\sin{{x}}}+{\sin{{y}}}={2}\frac{{\sin{{1}}}}{{2}}{\left({x}+{y}\right)}\frac{{\cos{{1}}}}{{2}}{\left({x}-{y}\right)}={a}{\quad\text{and}\quad}{\cos{{x}}}+{\cos{{y}}}={2}\frac{{\cos{{1}}}}{{2}}{\left({x}+{y}\right)}\frac{{\cos{{1}}}}{{2}}{\left({x}-{y}\right)}={b}.\)
Therefore, dividing these two we get \(\displaystyle\frac{{\tan{{1}}}}{{2}}{\left({x}+{y}\right)}=\frac{{a}}{{b}}.\)
If we square each of the original equations we get:
\(\displaystyle{{\sin}^{{2}}{x}}+{{\sin}^{{2}}{y}}+{2}{\sin{{x}}}.{\sin{{y}}}={a}^{{2}}{\quad\text{and}\quad}{{\cos}^{{2}}{x}}+{{\cos}^{{2}}{y}}+{2}{\cos{{x}}}.{\cos{{y}}}={b}^{{2}}\)
Adding these two equations we get \(\displaystyle{2}+{2}{\cos{{\left({x}-{y}\right)}}}={a}^{{2}}+{b}^{{2}}{\left[{{\sin}^{{2}}+}{{\cos}^{{2}}=}{1}\ {f}{\quad\text{or}\quad}\ \bot{h}\ {x}{\quad\text{and}\quad}{y}\right]}\)
We can expand \(\displaystyle{\cos{{\left({x}-{y}\right)}}}\) into \(2cos^2((x-y)/2)-1\), so we can write \(\displaystyle{4}{\left({{\cos}^{{2}}{\left({\left({x}-\frac{{y}}{{2}}\right)}={a}^{{2}}+{b}^{{2}}\right.}}\right.}\), because the 2's cancel out.
\(\displaystyle{{\cos}^{{2}}{\left(\frac{{{x}-{y}}}{{2}}\right)}}=\frac{{{a}^{{2}}+{b}^{{2}}}}{{4}}\) from which \(\displaystyle{{\sin}^{{2}}{\left(\frac{{{x}-{y}}}{{2}}\right)}}=\frac{{{4}-{a}^{{2}}-{b}^{{2}}}}{{4}}.\)
Therefore \(\displaystyle{{\tan}^{{2}}{\left(\frac{{{x}-{y}}}{{2}}\right)}}=\frac{{{4}-{a}^{{2}}-{b}^{{2}}}}{{{a}^{{2}}+{b}^{{2}}}}\) from which \(\displaystyle{\tan{{\left(\frac{{{x}-{y}}}{{2}}\right)}}}=\sqrt{{\frac{{{4}-{a}^{{2}}-{b}^{{2}}}}{{{a}^{{2}}+{b}^{{2}}}}}}.\)

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Best answer
2021-09-16

The value of \(\tan(\frac{x-y}{2})\)

Step-by-step explanation:

Given: \(\sin x+\sin y=a, \cos x + \cos y=b\)

To find: The value of \(\tan(\frac{x-y}{2})\)

Solution:

\(\sin x +\sin y=a\)

We know, \(\sin x+\sin y=2 \sin (\frac{x+y}{2}) \cos(\frac{x-y}{2}) \)

\(2 \sin (\frac{x+y}{2}) \cos(\frac{x-y}{2}) =a....(1)\)

\(\cos x+ \cos y=b\)

We know, \(\cos x+\cos y=2\cos(\frac{x+y}{2}) \sin (\frac{x-y}{2}) \)

\(2\cos(\frac{x+y}{2}) \sin (\frac{x-y}{2}) = b...(2)\)

Divide (1) and (2),

\(\frac{2\sin(\frac{x+y}{2}) \cos (\frac{x-y}{2})}{2\cos(\frac{x+y}{2}) \sin (\frac{x-y}{2}) }\)

Therefore, the value of \(\tan(\frac{x+y}{2})=\frac{a}{b}\)

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