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I am trying to simplify this trigonometric equation to a single trigonometric ratio. I had this question on my mathematical test today, and I was not able to get the correct answer, and I decided to ask the mathematical community, this way I can seek the answer, and see a solution of how this can be simplified.

The question gave me this expression, and it asked to reduce it to a single trigonometric ratio.

The expression is : $$\frac{3tan(x) - tan^3(x)}{1- 3tan^2(x)}$$

Thank You!

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  • $\begingroup$ Tangent of thrice angle $\endgroup$ – Narasimham Dec 16 '15 at 22:44
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The expression gave you a few hints with the number $3$ all over the place but, if you're not feeling inspired to guess the solution, you sometimes have to brute force your way. This can be much harder than proving that your guess is correct. I would suggest starting your exploration by expanding all your coefficients and your exponents:

$$ \frac{\tan x + \tan x + \tan x - \tan x \tan x \tan x}{1 - \tan x \tan x - \tan x \tan x - \tan x \tan x} \\ $$

For the numerator, I would try a few combinations:

$$ \tan x + \tan x (2 - \tan^2 x) \\ 2 \tan x + \tan x (1 - \tan^2 x) \\ $$

For the denominator, I would try the same thing:

$$ 1 - \tan x \tan x - \tan x (2 \tan x) \\ 1 - \tan^2 x - 2 \tan^2 x \\ $$

Aha! $1 - \tan^2 x$ is common in the numerator and the denominator, so I exploit it: $$ \frac{\frac{2\tan x}{1 - \tan^2 x} + \tan x}{1 - \frac{2 \tan^2 x}{1 - \tan^2 x}} \\ $$

Aha! $\frac{2 \tan x}{1 - \tan^2 x}$ is an identity I recognize for $\tan 2x$, so I replace it:

$$ \frac{\tan 2x + \tan x}{1 - \tan 2x \tan x} \\ $$

Aha! $\frac{\tan x + \tan y}{1 - \tan x \tan y}$ is another identity I recognize for $\tan (x + y)$, so I apply it:

$$ \tan (2x + x) \\ \tan 3x \\ $$

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Tan 3x, tan of thrice of angle. Tan is sin / cos.

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$$ \tan(x+y) = \frac{\tan x + \tan y }{ 1- \tan x \tan y }$$

$$ \implies \tan(2x) = \frac{2\tan x }{ 1- \tan^2 x }$$

$$ \implies \tan(3x) = \frac{\tan x + \frac{2\tan x }{ 1- \tan^2 x }}{ 1- \tan x \frac{2\tan x }{ 1- \tan^2 x } }$$

$$=\frac{( 1- \tan^2 x)\tan x + 2\tan x }{ ( 1- \tan^2 x)- 2\tan^2 x }$$

$$=\frac{3 \tan x - \tan^3 x }{ 1- 3\tan^2 x }$$

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  • $\begingroup$ Thank You, for the solution ! $\endgroup$ – Viktor Raspberry Dec 16 '15 at 23:13
  • $\begingroup$ somebody down-voted this answer without any explanation $\endgroup$ – WW1 Dec 17 '15 at 18:03

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