Question

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

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

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 $$\displaystyle{\tan{{\left(\frac{{{x}-{y}}}{{2}}\right)}}}=\sqrt{{\frac{{{4}-{a}^{{2}}-{b}^{{2}}}}{{{a}^{{2}}+{b}^{{2}}}}}}.$$

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 $$\tan(\frac{x+y}{2})=\frac{a}{b}$$