trigonometry

Level pending

Compute: ( t a n ( 22.5 ) ) 4 + 12 2 (tan(22.5))^{4} + 12\sqrt{2}


The answer is 17.

Aditya Todkar by Aditya Todkar , 24, India

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1 solution

Pi Han Goh
Nov 21, 2014

With tan ⁡ ( 4 5 ∘ ) = 1 \tan (45^\circ) = 1 , apply Double Angle Formula, tan ⁡ ( 2 x ) = 2 tan ⁡ ( x ) 1 − tan ⁡ 2 ( x ) \tan(2x) = \frac { 2 \tan (x) }{1 - \tan^2 (x) } , we have tan ⁡ ( 22. 5 ∘ ) = 2 − 1 \tan(22.5^\circ) = \sqrt2 - 1

⇒ tan ⁡ 4 ( 22. 5 ∘ ) = ( 2 − 1 ) 4 = ( ( 2 − 1 ) 2 ) 2 = ( 3 − 2 2 ) 2 = 17 − 12 2 \Rightarrow \tan^4(22.5^\circ) = (\sqrt2 - 1)^4 = ( (\sqrt2 - 1)^2 )^2 = (3- 2\sqrt2)^2 = 17 - 12\sqrt2

⇒ tan ⁡ 4 ( 22. 5 ∘ ) + 12 2 = 17 \Rightarrow \tan^4(22.5^\circ) + 12\sqrt2 = \boxed{17}

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how did u got tan(22.5)=\sqrt{2} - 1

Aditya Todkar - 6 years, 6 months ago

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Substitute x = 22. 5 ∘ x = 22.5^\circ into the trigonometric equation given, then tan ⁡ ( 2 x ) = 1 \tan(2x) = 1 , Let y = tan ⁡ ( 22. 5 ∘ ) y = \tan(22.5^\circ) , rearrange, solve for y y using the quadratic formula.

Pi Han Goh - 6 years, 6 months ago

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