Question

Найдите tg (а + п/4) , если cos2a=1/3 ; a э (0; п/2)

Answer 1

3 + 2√2 ≈ 5.8

Объяснение:

Cos 2α = 1/3

cos 2α = 1 - 2sin²α

1 - 2sin²α = 1/3

2sin²α = 2/3

sin²α = 1/3

cos²α = 1 - 1/3 = 2/3

tg²α = 1/3 : 2/3 = 1/2

tg α = +1/√2, так как α находится в 1-й четверти

tg (α + π/4) = (tgα + tg π/4)/(1 - tgα · tg π/4) = (1/√2 + 1)/(1 - 1/√2) =

= (1 + √2)/(√2 - 1) = (1 + √2)²/((√2 - 1)(√2 + 1)) = (1  + 2√2 + 2)/(2 - 1) = 3 + 2√2 ≈ 5.8

Answer 2

$\cos \big(2\alpha \big)=\dfrac 13\\\\\alpha \in \bigg(0;\dfrac{\pi}2\bigg)~~~\Rightarrow~~~2\alpha \in \big(0;\pi\big)~~~\Rightarrow~~~\sin\big(2\alpha\big)0\\\\\sin\big(2\alpha\big)=\sqrt{1-\cos^2 \big(2\alpha\big)}=\sqrt{1-\bigg(\dfrac13\bigg)^2}=\sqrt{\dfrac 89}=\dfrac{2\sqrt2}3$

$tg\bigg(\alpha +\dfrac{\pi}4\bigg)=\dfrac{tg \alpha +tg \big(\frac{\pi}4\big)}{1-tg \alpha \cdot tg \big(\frac{\pi}4\big)}=\dfrac{tg \alpha +1}{1-tg \alpha}=\\\\\\=\bigg(\dfrac{\sin \alpha }{\cos \alpha }+1\bigg):\bigg(1-\dfrac{\sin \alpha }{\cos \alpha }\bigg)=\\\\=\bigg(\dfrac{\sin \alpha +\cos \alpha }{\cos \alpha }\bigg)\cdot \bigg(\dfrac{\cos \alpha }{\cos \alpha -\sin \alpha }\bigg)=$

$=\dfrac{\sin \alpha +\cos \alpha }{\cos \alpha -\sin \alpha }=\dfrac{\sin \alpha +\cos \alpha }{\cos \alpha -\sin \alpha }\cdot \dfrac{\cos \alpha +\sin \alpha }{\cos \alpha +\sin \alpha }=\\\\=\dfrac{\big(\cos \alpha +\sin \alpha \big)^2}{\cos^2\alpha -\sin^2\alpha }=\dfrac{\cos^2\alpha +\sin^2\alpha+2\cos\alpha\sin\alpha }{\cos\big(2\alpha\big)}=\\\\=\dfrac{1+\sin\big(2\alpha\big) }{\cos\big(2\alpha\big)}=\bigg(1+\dfrac{2\sqrt2}3\bigg):\dfrac 13=\\\\=\bigg(1+\dfrac{2\sqrt2}3\bigg)\cdot3=3+2\sqrt2$

$\boxed{\boldsymbol{tg\bigg(\alpha +\dfrac{\pi}4\bigg)=3+2\sqrt2}}$

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Использованы формулы

$\sin^2\alpha +\cos^2\alpha =1\\\\\sin\big(2\alpha \big)=2\sin\alpha \cos\alpha \\\\\cos\big(2\alpha \big)=\cos^2\alpha -\sin^2\alpha \\\\tg\big(\alpha +\beta\big)=\dfrac{tg\alpha +tg\beta }{1-tg\alpha \cdot tg\beta }$

$(a-b)(a+b)=a^2-b^2$