if tan 5θ + tan 4θ =1 find 3θ


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Question Number 61157 by Askash last updated on 29/May/19

$i f$ $t a n 5 θ + t a n 4 θ = 1$ $f i n d 3 θ$
Commented by mr W last updated on 30/May/19

$t h e r e a r e t o t a l l y 9 v a l u e s f o r \tan 3 θ .$ $o n e o f t h e m i s θ = \frac{π}{4}, a n d \tan 3 θ = - 1 .$ $w i t h θ = \frac{π}{4}$ $\tan 3 θ = \tan (\frac{3 π}{4}) = - 1$ $\tan 5 θ + \tan 4 θ = \tan (\frac{5 π}{4}) + \tan (π) = 1$
Commented by Askash last updated on 29/May/19

$t a n 3 θ$
Commented by mr W last updated on 30/May/19

	Answered by tanmay last updated on 30/May/19

	$t a n (α + β + γ + . . .) = \frac{S_{1} - S_{3} + S_{5} . . .}{1 - S_{2} + S_{4} - S_{6} + . .}$ $t a n θ = a$ $t a n 5 θ = \frac{5 a - 5 C_{3} a^{3} + 5 C_{5} a^{5}}{1 - 5 C_{2} a^{2} + 5 C_{4} a^{4}} = \frac{5 a - 10 a^{3} + a^{5}}{1 - 10 a^{2} + 5 a^{4}}$ $t a n 4 θ = \frac{4 a - 4 C_{3} a^{3}}{1 - 4 C_{2} a^{2} + 4 C_{4} a^{4}} = \frac{4 a - 4 a^{3}}{1 - 6 a^{2} + a^{4}}$ $t a n 5 θ + t a 4 θ = 1$ $\frac{5 a - 10 a^{2} + a^{5}}{1 - 10 a^{2} + 5 a^{4}} + \frac{4 a - 4 a^{3}}{1 - 6 a^{2} + a^{4}} = 1$ $w a i t . . .$ $a n i t h e r a p p r o a c h . . .$ $h e r e t a n 5 θ + t a n 4 θ = 1$ $s o θ = s m a l l$ $t a n 5 θ + t a n 4 θ \approx 5 θ + 4 θ$ $9 θ = 1$ $θ = \frac{1}{9}$ $t a n 3 θ \approx 3 θ$ $= 3 \times \frac{1}{9} = 0.33$ $s o t a n 3 θ = 0.33 (a p p r o a x)$
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