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```
Sin x / sin y = 3 and cos x / cos y = 1/2 find the value of sin 2x/sin 2y + cos 2x/cos 2y

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4 years ago

```							Write siny = sinx/3  and squaring both sides,(siny)^2 = (sinx)^2/9..............(1)(cosy)^2 = 4(cosx)^2........(2)add both of them,and equate it to 1.and solve in sinx or cosx in any of them by writing cosx = 1 – sin^2x or vice versa.get x values and then simply y values. and then u can find the final value.
```
4 years ago
```							Ans is 49/58 and I am not getting. Please be little more specific. Thanks in Advance And if possible show working .
```
4 years ago
```							siny=sinx/3sin^2y=sin^2x/9similarly,cos^y=4cos^2xthen,sin^2y+cos^2y=sin^2x/9+4cos^2x1=(sin^2x+36cos^2x)/99=sin^2x+36cos^2xsin^2x+36(1-sin^2x)=9by solvingsinx=27/35   now,sin^2x+cos^2x=36sin^2y+cos^2y             4=36(1-cos^2y)+cos^2yby solving cosy=32/35 now, given thatsinx/siny=3 and cosx/cosy=1/2sinx=3siny-------(1) and 2cosx=cosy---------(2)multiplying (1) and (2)(sinx)(2cosx)=3sinycosysin2x=(3/2)sin2ysin2x/sin2y=3/2      andsin^2x=27/35cos^2x=1-27/35=8/35cos^2y=32/35sin^2y=1-32/35=3/35 and finally,(sin2x/sin2y)+(cos2x/cos2y)=(3/2)+[(8/35)-(27/35)]/[(32/35)-(3/25)]                                              =3/2-19/29                                              =49/58
```
4 years ago
```							siny=sinx/3sin^2y=sin^2x/9similarly,cos^y=4cos^2xthen,sin^2y+cos^2y=sin^2x/9+4cos^2x1=(sin^2x+36cos^2x)/99=sin^2x+36cos^2xsin^2x+36(1-sin^2x)=9by solvingsinx=27/35   now,sin^2x+cos^2x=36sin^2y+cos^2y             4=36(1-cos^2y)+cos^2yby solving cosy=32/35 now, given thatsinx/siny=3 and cosx/cosy=1/2sinx=3siny-------(1) and 2cosx=cosy---------(2)multiplying (1) and (2)(sinx)(2cosx)=3sinycosysin2x=(3/2)sin2ysin2x/sin2y=3/2      andsin^2x=27/35cos^2x=1-27/35=8/35cos^2y=32/35sin^2y=1-32/35=3/35 and finally,(sin2x/sin2y)+(cos2x/cos2y)=(3/2)+[(8/35)-(27/35)]/[(32/35)-(3/25)]                                              =3/2-19/29                                              =49/58
```
3 years ago
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