Prove: $\frac{\sin 4 A+\sin 5 A+\sin 6 A}{\cos 4 A+\cos 5 A+\cos 6 A}=\tan 5 A$

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written 5.6 years ago by

teamques10 ★ 69k

modified 3.0 years ago by

pedsangini276 • 4.8k

engineering mathematics

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written 5.6 years ago by

teamques10 ★ 69k

• modified 5.6 years ago

Solution:

$\frac{\sin 4 A+\sin 5 A+\sin 6 A}{\cos 4 A+\cos 5 A+\cos 6 A}$

$=\frac{(\sin 4 A+\sin 6 A)+\sin 5 A}{(\cos 4 A+\cos 6 A)+\cos 5 A}$

$=\frac{2 \sin \left(\frac{4 A+6 A}{2}\right) \cos \left(\frac{4 A-6 A}{2}\right)+\sin 5 A}{2 \cos \left(\frac{4 A+6 A}{2}\right) \cos \left(\frac{4 A-6 A}{2}\right)+\cos 5 A}$

$=\frac{ \sin 5 A \ [2 \cos (-A)+1]}{\cos 5 A [2 \cos (-A)+1]}$

$=\tan 5 A$

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5views

written 5.6 years ago by

teamques10 ★ 69k

Solution:

$\frac{\sin 4 A+\sin 5 A+\sin 6 A}{\cos 4 A+\cos 5 A+\cos 6 A}$

$=\frac{(\sin 4 A+\sin 6 A)+\sin 5 A}{(\cos 4 A+\cos 6 A)+\cos 5 A}$

$=\frac{2 \sin \left(\frac{4 A+6 A}{2}\right) \cos \left(\frac{4 A-6 A}{2}\right)+\cos 5 A}{2 \cos \left(\frac{4 A+6 A}{2}\right) \cos \left(\frac{4 A-6 A}{2}\right)+\cos 5 A}$

$=\frac{2 \sin 5 A \cos (-A)+\cos 5 A}{\cos 5 A}[2 \cos (-A)+1]$

$=\tan 5 A$

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