Grade 11Trigonometry

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

Abhinav Anand

9 Years agoGrade 11

4 Answers

Vikas TU

9 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.

Abhinav Anand

9 Years ago

Ans is 49/58 and I am not getting. Please be little more specific. Thanks in Advance And if possible show working .

govardhan

Approved Tutor Answer9 Years ago

siny=sinx/3

sin^2y=sin^2x/9

similarly,

cos^y=4cos^2x

then,sin^2y+cos^2y=sin^2x/9+4cos^2x

1=(sin^2x+36cos^2x)/9

9=sin^2x+36cos^2x

sin^2x+36(1-sin^2x)=9

by solving

sinx=27/35

now,sin^2x+cos^2x=36sin^2y+cos^2y

4=36(1-cos^2y)+cos^2y

by solving cosy=32/35

now, given that

sinx/siny=3 and cosx/cosy=1/2

sinx=3siny-------(1) and 2cosx=cosy---------(2)

multiplying (1) and (2)

(sinx)(2cosx)=3sinycosy

sin2x=(3/2)sin2y

sin2x/sin2y=3/2

and

sin^2x=27/35

cos^2x=1-27/35=8/35

cos^2y=32/35

sin^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

Srk

8 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