Question

prove: sin(5a-a)/sinacosa=4cos2a​

Answer 1

Answer:

To prove: \dfrac{\sin 5A}{\sin A}- \dfrac{cos 5A}{\cos A} = 4 \cos 2A

sinA

sin5A

−

cosA

cos5A

=4cos2A

Step-by-step explanation:

Consider L.H.S.

We have

\begin{gathered}\dfrac{\sin 5A}{\sin A}- \dfrac{\cos 5A}{\cos A} \\\\= \dfrac{\sin 5A \cos A-\cos 5A \sin A}{\sin A \cos A}\end{gathered}

sinA

sin5A

−

cosA

cos5A

=

sinAcosA

sin5AcosA−cos5AsinA

Now as we know \sin(x-y)= \sin x \cos y -\cos x \sin ysin(x−y)=sinxcosy−cosxsiny

Therefore we get

\begin{gathered}=\dfrac{\sin (5A-A)}{\cos A \sin A} \\\\=\dfrac{\sin 4A}{\cos A \sin A} \\\\ \text {Now } \because \sin 2A = 2 \sin A \cos A \text { we get }\\\\ =\dfrac{2\sin 2A \cos 2A}{\cos A \sin A} \\\\= \dfrac{2 \times 2 \sin A \cos A\cos 2A}{\cos A \sin A} =4 \cos A\end{gathered}

=

cosAsinA

sin(5A−A)

=

cosAsinA

sin4A

Now ∵sin2A=2sinAcosA we get

=

cosAsinA

2sin2Acos2A

=

cosAsinA

2×2sinAcosAcos2A

=4cosA

which is equal to R.H.S.

Hence, proved the required result that \dfrac{\sin 5A}{\sin A}- \dfrac{cos 5A}{\cos A} = 4 \cos 2A

sinA

sin5A

−

cosA

cos5A

=4cos2A