Question

Prove that
$\dfrac{\sin 5x+\sin 3x}{\cos 5x+\cos 3x}=\tan 4x$

Answer 1

Hint: Use the fact that $\sin \left( A \right)+\sin \left( B \right)=2\sin \left( \dfrac{A+B}{2} \right)\cos \left( \dfrac{A-B}{2} \right)$ and $\cos \left( A \right)+\cos \left( B \right)=2\cos \left( \dfrac{A+B}{2} \right)\cos \left( \dfrac{A-B}{2} \right)$. Hence prove that $\sin 5x+\sin 3x=2\sin 4x\cos x$ and $\cos 5x+\cos 3x=2\cos 4x\cos x$ and hence prove the above identity.
Complete step-by-step answer:
 Simplifying the Numerator:
We have
Numerator = sin5x+sin3x
We know that $\sin \left( A \right)+\sin \left( B \right)=2\sin \left( \dfrac{A+B}{2} \right)\cos \left( \dfrac{A-B}{2} \right)$
Put A = 5x and B = 3x, we get
$\sin 5x+\sin 3x=2\sin \left( \dfrac{5x+3x}{2} \right)\cos \left( \dfrac{5x-3x}{2} \right)=2\sin 4x\cos x$
Hence, we have Numerator = 2sin4xcosx
Simplifying the Denominator:
We have
Denominator = cos5x+cos3x
We know that $\cos \left( A \right)+\cos \left( B \right)=2\cos \dfrac{A+B}{2}\cos \dfrac{A-B}{2}$
Put A = 5x and B = 3x, we get
$\cos 5x+\cos 3x=2\cos \left( \dfrac{5x+3x}{2} \right)\cos \left( \dfrac{5x-3x}{2} \right)=2\cos 4x\cos x$
Hence, we have Denominator = 2cos4xcosx
Hence, we have
$\dfrac{\sin 5x+\sin 3x}{\cos 5x+\cos 3x}=\dfrac{2\sin 4x\cos x}{2\cos 4x\cos x}=\dfrac{\sin 4x}{\cos 4x}$
We know that $\dfrac{\sin x}{\cos x}=\tan x$
Hence, we have
$\dfrac{\sin 4x}{\cos 4x}=\tan 4x$
Hence, we have
$\dfrac{\sin 5x+\sin 3x}{\cos 5x+\cos 3x}=\tan 4x$
Hence, we have L.H.S. = R.H.S.
Q.E.D
Note: Aid to memory:
[i] S+S = 2SC
[ii] S-S = 2CS
[iii] C+C = 2CC
[iv] C-C = -2SS
Each one of the above parts helps to memorise two formula
Like from [ii], we have S-S = 2CS.
Hence, we have $\sin \left( A \right)-\sin \left( B \right)=2\cos \left( \dfrac{A+B}{2} \right)\sin \left( \dfrac{A-B}{2} \right)$ and $2\cos x\sin y=\sin \left( x+y \right)-\sin \left( x-y \right)$
Hence by memorising the above mnemonic, we can memorise 8 different formulae.