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ss7ja [257]
3 years ago
9

[Sin A + Sin 3A + Sin 5A + Sin 7A]÷[Cos A + Cos 3A + Cos 5A + Cos 7A]=Tan 4A​

Mathematics
1 answer:
Ainat [17]3 years ago
6 0

Do the following rewrites:

sin(<em>x</em>) = sin(4<em>x</em> - 3<em>x</em>) = sin(4<em>x</em>) cos(3<em>x</em>) - cos(4<em>x</em>) sin(3<em>x</em>)

sin(7<em>x</em>) = sin(4<em>x</em> + 3<em>x</em>) = sin(4<em>x</em>) cos(3<em>x</em>) + cos(4<em>x</em>) sin(3<em>x</em>)

sin(3<em>x</em>) = sin(4<em>x</em> - <em>x</em>) = sin(4<em>x</em>) cos(<em>x</em>) - cos(4<em>x</em>) sin(<em>x</em>)

sin(5<em>x</em>) = sin(4<em>x</em> + <em>x</em>) = sin(4<em>x</em>) cos(<em>x</em>) + cos(4<em>x</em>) sin(<em>x</em>)

cos(<em>x</em>) = cos(4<em>x</em> - 3<em>x</em>) = cos(4<em>x</em>) cos(3<em>x</em>) + sin(4<em>x</em>) sin(3<em>x</em>)

cos(7<em>x</em>) = cos(4<em>x</em> + 3<em>x</em>) = cos(4<em>x</em>) cos(3<em>x</em>) - sin(4<em>x</em>) sin(3<em>x</em>)

cos(3<em>x</em>) = cos(4<em>x</em> - <em>x</em>) = cos(4<em>x</em>) cos(<em>x</em>) + sin(4<em>x</em>) sin(<em>x</em>)

cos(5<em>x</em>) = cos(4<em>x</em> + <em>x</em>) = cos(4<em>x</em>) cos(<em>x</em>) - sin(4<em>x</em>) sin(<em>x</em>)

So in the numerator, we have

sin(<em>x</em>) + sin(3<em>x</em>) + sin(5<em>x</em>) + sin(7<em>x</em>)

= 2 sin(4<em>x</em>) cos(3<em>x</em>) + 2 sin(4<em>x</em>) cos(<em>x</em>)

= 2 sin(4<em>x</em>) (cos(3<em>x</em>) + cos(<em>x</em>))

In the denominator,

cos(<em>x</em>) + cos(3<em>x</em>) + cos(5<em>x</em>) + cos(7<em>x</em>)

= 2 cos(4<em>x</em>) cos(3<em>x</em>) + 2 cos(4<em>x</em>) cos(<em>x</em>)

= 2 cos(4<em>x</em>) (cos(3<em>x</em>) + cos(<em>x</em>))

So we have

(sin(<em>x</em>) + sin(3<em>x</em>) + sin(5<em>x</em>) + sin(7<em>x</em>)) / (cos(<em>x</em>) + cos(3<em>x</em>) + cos(5<em>x</em>) + cos(7<em>x</em>))

= (2 sin(4<em>x</em>) (cos(3<em>x</em>) + cos(<em>x</em>))) / (2 cos(4<em>x</em>) (cos(3<em>x</em>) + cos(<em>x</em>)))

= sin(4<em>x</em>) / cos(4<em>x</em>)

= tan(4<em>x</em>)

QED

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