Question

sin 5A = 16 $sin^{5}$A -20$sin^{3}$A+5sinA


a good answer will be marked as brainliest

Answer 1

Answer:

refer to this------

Step-by-step explanation:

We know De Moivre's theorem, applying which gives:-

(cos A+i sin A)5=cos 5A+i sin 5Acos A+i sin A5=cos 5A+i sin 5A

Applying Binomial Theorem on the LHS, we get:-

cos5A+5i cos4A sinA−10cos3A(i2sin2A)+10cos2A(i3sin3A)+5cosA(i4sin4A)+i5sin5A=cos5A+isin5Acos5A+5i cos4A sinA-10cos3Ai2sin2A+10cos2Ai3sin3A+5cosAi4sin4A+i5sin5A=cos5A+isin5A

⇒cos5A+5icos4A sinA−10cos3A sin2A−10i cos2A sin3A+5cosA sin4A+i sin5A=cos5A+i sin5A⇒cos5A+5icos4A sinA-10cos3A sin2A-10i cos2A sin3A+5cosA sin4A+i sin5A=cos5A+i sin5A

⇒(cos5A−10cos3A sin2A+5cosA sin4A)+i(5cos4A sinA−10cos2A sin3A+sin5A)=cos5A+i sin5A⇒cos5A-10cos3A sin2A+5cosA sin4A+i5cos4A sinA-10cos2A sin3A+sin5A=cos5A+i sin5A

Equating the real and imaginary terms of the equation, we have:-

cos 5A=cos5A−10cos3A sin2A+5cosA sin4A, andsin 5A=5cos4A sinA−10cos2A sin3A+sin5A