Question

Please help with number 12

Answer 1

Answer:

Step-by-step explanation:

sin(195º)= -√6+√2/4

cos(195º)=-√6-√2/4

tan(195º)=2-√3

Answer 2

Recall the rules

$\sin(a-b)=\sin(a) \cos(b) - \cos(a) \sin(b)$

$\cos(a-b)=\sin(a) \sin(b) + \cos(a) \cos(b)$

Use the suggested difference:

$\sin(195)=\sin(225-30)=\sin(225) \cos(30) - \cos(225) \sin(30)$

$\cos(195)=\cos(225-30)=\sin(225) \sin(30) + \cos(225) \cos(30)$

Since 225 and 30 are known angles, we can plug the values:

$\sin(225)=\cos(225)=-\dfrac{1}{\sqrt{2}},\quad \sin(30)=\dfrac{1}{2},\quad \cos(30)=\dfrac{\sqrt{3}}{2}$

The expressions become

$\sin(195)=-\dfrac{1}{\sqrt{2}} \cdot \dfrac{\sqrt{3}}{2} - \left(-\dfrac{1}{\sqrt{2}}\right) \cdot \dfrac{1}{2} = \dfrac{1-\sqrt{3}}{2\sqrt{2}}$

$\cos(195)=-\dfrac{1}{\sqrt{2}} \cdot \dfrac{1}{2} + \left(-\dfrac{1}{\sqrt{2}}\right) \dfrac{\sqrt{3}}{2}=\dfrac{-1-\sqrt{3}}{2\sqrt{2}}$

As usual, we just use the definition of the tangent:

$\tan(195)=\dfrac{\sin(195)}{\cos(195)}=\dfrac{\frac{1-\sqrt{3}}{2\sqrt{2}}}{\frac{-1-\sqrt{3}}{2\sqrt{2}}}=\dfrac{1-\sqrt{3}}{-1-\sqrt{3}}$