I want the solution for sin9A=sinA.

Open in App

Solution

If sin(x)=sin(y) then it's general solution is

x=nπ+{(-1)^n}y ,n€z.

Hence, for sin(9A) =sin(A)

x=9A ; y= A

So, it's general solution is

9A= nπ+{(-1)^n}×A

=> 9A - {(-1)^n}A = nπ

=> A[9-(-1)^n] = nπ

Hence, A = (nπ)/[9-(-1)^n]

Suggest Corrections

Similar questions

\frac{s i n A + s i n 5 A + s i n 9 A}{c o s A + c o s 5 A + c o s 9 A} = t a n 5 A

\frac{sin A - sin 5 A + sin 9 A - sin 13 A}{cos A - cos 5 A - cos 9 A + cos 13 A} = cot 4 A

\frac{\sin A + \sin B}{\sin A - \sin B} = \tan (\frac{A + B}{2}) \cot (\frac{A - B}{2})

\frac{\sin A + \sin 3 A}{\cos A - \cos 3 A} = \cot A

\frac{\sin 9 A - \sin 7 A}{\cos 7 A - \cos 9 A} = \cot 8 A

\frac{\cos A + \cos B}{\cos B - \cos A} = \cot (\frac{A + B}{2}) \cot (\frac{A - B}{2})

\frac{\sin A - \sin B}{\cos A + \cos B} = \tan \frac{A - B}{2}

I want worksheet on indices for practice. Can solution also be provided

Join BYJU'S Learning Program