+0  
 
0
288
5
avatar+845 

please show workings thank you

 May 12, 2019
 #1
avatar+111431 
+3

r^2 [sin2 θ]  = 6

r^2 (2sin θ cos θ)  =  6     divide through by 2

r^2 (sin θ cos  θ)  =  3        [ sin θ = y/r ,  cos θ= x / r ]

 

So....

 

r^2 ( y/r)(x/r)  = 3

r^(2) ( xy) / r^2   = 3

xy  =  3

y =  3 / x       

 

 

cool cool cool    

 May 12, 2019
 #2
avatar+845 
0

how does sin20 become 2sin0 cos0?

sorry im terrible at this topic

YEEEEEET  May 12, 2019
 #4
avatar+111431 
+2

This is a trig identity

 

sin (2 θ)  =  2sin θ cos  θ

 

 

cool cool cool

CPhill  May 12, 2019
 #5
avatar+845 
+1

ok thank you very much

YEEEEEET  May 12, 2019
 #3
avatar+111431 
+2

x^4 = x^2 + y^2       [x^2 + y^2  = r^2,   x = r cos θ, y = r sin θ  ]

 

(r cos θ)^4   =  r^2

 

r^4 (cos  θ)^4  = r^2

 

cool cool cool

 May 12, 2019

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