My knowledge of trigonometry is still insufficient to resolve this

Bradlee Hooper

Bradlee Hooper

Answered question

2022-02-26

My knowledge of trigonometry is still insufficient to resolve this problem. Any help would be greatly appreciated.
tanα+2tan2α+4tan4α=cotα8cot8α

Answer & Explanation

faraidz3i

faraidz3i

Beginner2022-02-27Added 10 answers

Hoping this is not homework:
cotx8cot8x=cotx8cot8x=cotx8cot24x12cot4x
=cotx4cot24x1cot4x=cotx4cot4x+4tan4x
=cotx4cot22x12cot2x+4tan4x=cotx2cor2x+2tan2x+4tan4x
=tanx+2tan2x+4tan4x
Liyana Iles

Liyana Iles

Beginner2022-02-28Added 6 answers

This follows easily from the identity:
cotx2cot2x=tanx
Which can easily be seen by using sin2x=2sinxcosx and cos2x=cos2xsin2x as follows:
cotxtanx=cosxsinxsinxcosx=cos2xsin2xsinxcosx=2cot2x
We now have
cotα2cot2α=tanα
2cot2α4cot4α=2tan2α
4cot2α8cot4α=4tan4α
Adding gives us the result

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