Question

calcule o valor do cosseno de 75° ​

Answer 1

(√6-√2)/4

Explicação passo-a-passo:

Cos 75= cos (30+45)

Cos (a+b) = cos a. cos b - sen a. sen b

Cos (30+45) = cos 30. cos 45 - sen 30 - sen 45

Cos (30+45) = √3/2 . √2/2 - 1/2 . √2/2

Cos (30+45) = √6/4 - √2/4

Cos (30+45) = (√6-√2)/4

Answer 2

Explicação passo-a-passo:

4)

$\sf cos~(a+b)=cos~a\cdot cos~b-sen~a\cdot sen~b$

$\sf cos~(30^{\circ}+45^{\circ})=cos~30^{\circ}\cdot cos~45^{\circ}-sen~30^{\circ}\cdot sen~45^{\circ}$

$\sf cos~75^{\circ}=\dfrac{\sqrt{3}}{2}\cdot\dfrac{\sqrt{2}}{2}-\dfrac{1}{2}\cdot\dfrac{\sqrt{2}}{2}$

$\sf cos~75^{\circ}=\dfrac{\sqrt{6}}{4}-\dfrac{\sqrt{2}}{4}$

$\sf cos~75^{\circ}=\dfrac{\sqrt{6}-\sqrt{2}}{4}$

5)

$\sf tg~(a-b)=\dfrac{tg~a-tg~b}{1+tg~a\cdot tg~b}$

$\sf tg~(45^{\circ}-30^{\circ})=\dfrac{tg~45^{\circ}-tg~30^{\circ}}{1+tg~45^{\circ}\cdot tg~30^{\circ}}$

$\sf tg~15^{\circ}=\dfrac{1-\frac{\sqrt{3}}{3}}{1+1\cdot\frac{\sqrt{3}}{3}}$

$\sf tg~15^{\circ}=\dfrac{\frac{3-\sqrt{3}}{3}}{\frac{3+\sqrt{3}}{3}}$

$\sf tg~15^{\circ}=\dfrac{3-\sqrt{3}}{3+\sqrt{3}}\cdot\dfrac{3-\sqrt{3}}{3-\sqrt{3}}$

$\sf tg~15^{\circ}=\dfrac{9-3\sqrt{3}-3\sqrt{3}+3}{3^2-(\sqrt{3})^2}$

$\sf tg~15^{\circ}=\dfrac{12-6\sqrt{3}}{9-3}$

$\sf tg~15^{\circ}=\dfrac{6\cdot(2-\sqrt{3})}{6}$

$\sf tg~15^{\circ}=2-\sqrt{3}$