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

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Rico Grant eNotes educator | Certified Educator

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We are asked to show that (sin3a+sina+sin5a)/(cos3a+cosa+cos5a)=tan3a.

Note that sin3a=sina(-1+4cos^2a) and sin5a=sina(1-12cos^2a+16cos^4a)
Similarly cos3a=cosa(1-4sin^2a) and cos5a=cosa(1-12sin^2a+16sin^4a)

Consider the numerator. Substituting and factoring out the common sina yields:

sina((-1+4cos^2a)+1+(1-12cos^2a+16cos^4a)) or
sina(1-8cos^2a+16cos^4a)=sina(4cos^2a-1)^2

The denominator, after substituting and factoring out a common cosa, yields:

cosa((1-4sin^2a)+1+(1-12sin^2a+16sin^4a)) or
cosa(3-16sin^2a+16sin^4a)=cosa(-4sin^2a+3)(-4sin^2a+1)

So the rational expression can be expressed as the product of three rational expressions:

sina/cosa * (4cos^2a-1)/(-4sin^2a+3) * (4cos^2a-1)/(-4sin^2a+1)

4cos^2a-1 = -4sin^2+3, so the middle factor is 1. The first factor is tana.

The last factor:

(4cos^2a-1)/(-4sin^2+1)=(2cos^2a+cos2a)/(cos2a-2sin^2a)
=(2cos^2a+cos^2a-sin^2a)/(cos^2a-sin^2a-2sin^2a)
=(3cos^2a-sin^2a)/(cos^2a-3sin^2a)
=(3-sin^2a/cos^2a)/(1-3sin^2a/cos^2a)
=(3-tan^2a)/(1-3tan^2a)

Thus we have tana * 1 * (3-tan^2a)/(1-3tan^2a)
=(3tana-tan^3a)/(1-3tan^2a)
=tan(3a) as required.

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justaguide eNotes educator | Certified 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

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lochana2500 | Student

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