Tabansi

2020-10-18

If $\mathrm{sin}x+\mathrm{sin}y=a\phantom{\rule{1em}{0ex}}\text{and}\phantom{\rule{1em}{0ex}}\mathrm{cos}x+\mathrm{cos}y=b$ then find $\mathrm{tan}\left(x-\frac{y}{2}\right)$

Benedict

$\mathrm{sin}\left(A+B\right)=\mathrm{sin}A\mathrm{cos}B+\mathrm{cos}A\mathrm{sin}B\phantom{\rule{1em}{0ex}}\text{and}\phantom{\rule{1em}{0ex}}\mathrm{sin}\left(A-B\right)=\mathrm{sin}A\mathrm{cos}B-\mathrm{cos}A\mathrm{sin}B.$
$\mathrm{cos}\left(A+B\right)=\mathrm{cos}A\mathrm{cos}B-\mathrm{sin}A\mathrm{sin}B\phantom{\rule{1em}{0ex}}\text{and}\phantom{\rule{1em}{0ex}}\mathrm{cos}\left(A-B\right)=\mathrm{cos}A\mathrm{cos}B+\mathrm{sin}A\mathrm{sin}B.$
These are true for all A and B. So, $\mathrm{sin}\left(A+B\right)+\mathrm{sin}\left(A-B\right)=2\mathrm{sin}A\mathrm{cos}B\phantom{\rule{1em}{0ex}}\text{and}\phantom{\rule{1em}{0ex}}\mathrm{cos}\left(A+B\right)+\mathrm{cos}\left(A-B\right)=2\mathrm{cos}A\mathrm{cos}B.$
Therefore relating A and B to x and y we get$A+B=x\phantom{\rule{1em}{0ex}}\text{and}\phantom{\rule{1em}{0ex}}A-B=y.FromtheseA=\frac{x+y}{2}\phantom{\rule{1em}{0ex}}\text{and}\phantom{\rule{1em}{0ex}}B=\frac{x-y}{2}.$
We can now write $\mathrm{sin}x+\mathrm{sin}y=2\frac{\mathrm{sin}1}{2}\left(x+y\right)\frac{\mathrm{cos}1}{2}\left(x-y\right)=a\phantom{\rule{1em}{0ex}}\text{and}\phantom{\rule{1em}{0ex}}\mathrm{cos}x+\mathrm{cos}y=2\frac{\mathrm{cos}1}{2}\left(x+y\right)\frac{\mathrm{cos}1}{2}\left(x-y\right)=b.$
Therefore, dividing these two we get $\frac{\mathrm{tan}1}{2}\left(x+y\right)=\frac{a}{b}.$
If we square each of the original equations we get:
${\mathrm{sin}}^{2}x+{\mathrm{sin}}^{2}y+2\mathrm{sin}x.\mathrm{sin}y={a}^{2}\phantom{\rule{1em}{0ex}}\text{and}\phantom{\rule{1em}{0ex}}{\mathrm{cos}}^{2}x+{\mathrm{cos}}^{2}y+2\mathrm{cos}x.\mathrm{cos}y={b}^{2}$
Adding these two equations we get
We can expand $\mathrm{cos}\left(x-y\right)$ into $2co{s}^{2}\left(\left(x-y\right)/2\right)-1$, so we can write

Jeffrey Jordon

The value of $\mathrm{tan}\left(\frac{x-y}{2}\right)$

Step-by-step explanation:

Given: $\mathrm{sin}x+\mathrm{sin}y=a,\mathrm{cos}x+\mathrm{cos}y=b$

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

Solution:

$\mathrm{sin}x+\mathrm{sin}y=a$

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

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

$\mathrm{cos}x+\mathrm{cos}y=b$

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

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

Divide (1) and (2),

$\frac{2\mathrm{sin}\left(\frac{x+y}{2}\right)\mathrm{cos}\left(\frac{x-y}{2}\right)}{2\mathrm{cos}\left(\frac{x+y}{2}\right)\mathrm{sin}\left(\frac{x-y}{2}\right)}$

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

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