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Prove: $\frac{\sin 4 A+\sin 5 A+\sin 6 A}{\cos 4 A+\cos 5 A+\cos 6 A}=\tan 5 A$
written 4.7 years ago by
teamques10
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modified 2.1 years ago
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pedsangini276
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written 4.7 years ago by
teamques10
★ 65k
|
• modified 4.7 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 …
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written 4.7 years ago by
teamques10
★ 65k
|
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 …
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