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If Siny = x Sin (a+y). Prove that
dy/dx=[Sin²(a+y)]/Sina​

User Patryk Obara
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1 Answer

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22 votes

Answer:

Use Trigonometric formulae to simplify expressions

Explanation:

Re-order to group the x and y,

Sin(y) / Sin(a+y) = x

to differentiate apply d/dx to both sides


(d)/(dx)Sin(y) / Sin(a+y) =
(d)/(dx)x

use division rule on left side which will give


(dy)/(dx)Cos(y)Sin(a+y) -
(dy)/(dx)Cos(a+y)Sin(y) / Sin²(a+y) = 1

Group them so dy/dx is on the left hand side:


(dy)/(dx) = Sin²(a+y) / Cos(y)Sin(a+y) - Cos(a+y)Sin(y)

Apply trigonometric sum formula to expand Sin(a+y) and Cos(a+y),


(dy)/(dx) = Sin²(a+y) / Cos(y)[Sin(a)Cos(y) + Cos(a)Sin(y)] - Sin(y)[Cos(a)Cos(y) - Sin(a)Sin(y)]

Simplify it


(dy)/(dx) = Sin²(a+y) / (Cos²(y) + Sin²(y))(Sin(a))


(dy)/(dx) = Sin²(a+y) / (1)(Sin(a))

User Verthosa
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