# Prove that : sin 3a+sin a+sin 5a/cos 3a+cos a+cos 5a=tan 3a

justaguide | College Teacher | (Level 2) Distinguished Educator

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We have to prove [sin 3a + sin a + sin 5a]/ [cos 3a+cos a+cos 5a] = tan 3a

We start with [sin 3a + sin a + sin 5a]/ [cos 3a+cos a+cos 5a]

We use the formula: sin A + sin B = 2 sin [ (A + B) / 2 ] cos [ (A - B) / 2 ] and cos A + cos B = 2 cos [ (A + B) / 2 ] cos [ (A - B) / 2 ]

[sin 3a + sin a + sin 5a]/ [cos 3a+cos a+cos 5a]

=> [sin 3a + 2 sin [(a + 5a) / 2]  cos [ (5a - a) / 2 ]/ [cos 3a+ 2 cos [ (a + 5a) / 2] cos [ (5a - a) / 2 ]

=> [sin 3a + 2 sin 3a * cos 2a]/ [cos 3a+ 2 cos 3a cos 2a ]

=> [sin 3a ( 1 + 2 cos 2a)] / [cos 3a ( 1 + 2 cos 2a)]

=> sin 3a / cos 3a

=> tan 3a

This proves [sin 3a + sin a + sin 5a]/ [cos 3a+cos a+cos 5a] = tan 3a

lochana2500 | Student, Undergraduate | (Level 1) Valedictorian

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L:H:S ≡ (sin3A + sinA + sin5A)/(cos3A + cosA + cos5A)

⇒ use sin C + sin D = 2sin[(C+D)/2].cos[(C-D)/2]

& cos C + cos D = 2cos[(C+D)/2].cos[(C-D)/2]

= [sin3A + (sinA + sin5A)]/[cos3A + (cosA + cos5A)]

= (sin3A + 2sin3A.cos2A)/(cos3A + 2cos3A.cos2A)

= sin3A(1 + 2cos2A)/cos3A(1 + 2cos2A)

= sin3A/cos3A

= tan3A

Therefore L:H:S ≡ R:H:S