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Prove that (cosx+cos3x+cos5x)/(sinx+sin3x+sin5x) = cot3x​

User Sajas
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Answer:

We can use the identity cos(A+B) = cosAcosB - sinAsinB to simplify the expression in the numerator:

cos(x) + cos(3x) + cos(5x)

= cos(x) + (cos(2x)cos(x) - sin(2x)sin(x)) + (cos(4x)cos(x) - sin(4x)sin(x))

= cos(x) + cos(x)(2cos^2(x) - 1) - sin(x)(2cos(x)sin(x)) + cos(x)(2cos^2(2x) - 1) - sin(x)(2cos(2x)sin(2x))

= cos(x)(4cos^2(2x) + 2cos^2(x) - 2) - sin(x)(4cos(2x)sin(2x) + 2cos(x)sin(x))

Similarly, we can use the identity sin(A+B) = sinAcosB + cosAsinB to simplify the expression in the denominator:

sin(x) + sin(3x) + sin(5x)

= sin(x) + (sin(2x)cos(x) + cos(2x)sin(x)) + (sin(4x)cos(x) + cos(4x)sin(x))

= sin(x) + cos(x)(2sin(x)cos(2x)) + sin(x)(2cos(2x)sin(2x)) + cos(x)(2cos(4x)sin(x)) - sin(x)(2cos(4x)cos(x))

= sin(x)(4cos^2(2x) + 2cos^2(x) - 2) + cos(x)(4cos(2x)sin(2x) + 2cos(x)sin(x))

We can simplify the expression further by dividing both the numerator and denominator by sin(3x):

(cos(x) + cos(3x) + cos(5x))/(sin(x) + sin(3x) + sin(5x))

= [cos(x)/sin(3x) + cos(3x)/sin(3x) + cos(5x)/sin(3x)] / [sin(x)/sin(3x) + sin(3x)/sin(3x) + sin(5x)/sin(3x)]

= [cot(3x) + 1 + cot(3x)*cot(2x)] / [csc(3x) + 1 + csc(3x)*csc(2x)]

= [cot(3x) + cot(3x)*cot(2x) + 1] / [csc(3x) + csc(3x)*csc(2x) + 1]

= cot(3x)

User Jar
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