Question

someone pls answer- will give brainliest

Answer 1

Hello there!

Formulas

$\displaystyle \large{ \sin (2 \theta) = 2 \sin \theta \cos \theta} \\ \displaystyle \large{ \cos (2 \theta) = \begin{cases} 2 { \cos}^{2} \theta - 1 \\ 1 - 2 { \sin}^{2} \theta \\ { \cos}^{2} \theta - { \sin}^{2} \theta \end{cases}}$

First, we have to find cos(theta). We know that the theta lies in Q3 which is negative for both sin and cos. To find cos, you can use the identity or Pythagoras.

Recall

$\displaystyle \large \tt{ \sin \theta = \frac{opposite}{hypotenuse} } \\ \displaystyle \large \tt{ \cos \theta = \frac{adjacent}{hypotenuse} }$

Finding cos, I will use Pythagoras.

Recall Pythagorean

• Define a or b = adjacent or opposite
• Define c = hypotenuse

$\displaystyle \large{ {a}^{2} + {b}^{2} = {c}^{2} }$

From sin = -1/6

• opposite = 1
• hypotenuse = 6
• adjacent = x

$\displaystyle \large{ {1}^{2} + {b}^{2} = {6}^{2} } \\ \displaystyle \large{1 + {b}^{2} = 36 } \\ \displaystyle \large{ {b}^{2} = 36 - 1 } \\ \displaystyle \large{ {b}^{2} = 35 } \\ \displaystyle \large{ {b} = \sqrt{35} }$

b is our adjacent. Hence,

$\displaystyle \large{ \cos \theta = \frac{ \sqrt{35} }{6} }$

Since theta is in Q3, cos < 0 or in negative.

$\displaystyle \large{ \cos \theta = - \frac{ \sqrt{35} }{6} }$

Finding sin(2theta)

$\displaystyle \large{ \sin \theta = - \frac{ 1 }{6} }$

Hence,

$\displaystyle \large{ \sin 2 \theta = 2( - \frac{ 1 }{6} )( - \frac{ \sqrt{35} }{6}) } \\ \displaystyle \large{ \sin 2 \theta = ( - \frac{ 1 }{3} )( - \frac{ \sqrt{35} }{6}) } \\ \displaystyle \large{ \sin 2 \theta = \frac{ \sqrt{35} }{18} }$

Recall

• Negative × Negative = Positive

Finding cos(2theta)

The cos of 2theta has multiple formulas. All of them are same and also identity to each others.

I will be using the first one as it only involves cos, not sin.

$\displaystyle \large{ \cos \theta = - \frac{ \sqrt{35} }{6} }$

Therefore,

$\displaystyle \large{ \cos2 \theta = 2 { ( - \frac{ \sqrt{35} }{6} )}^{2} - 1 } \\ \displaystyle \large{ \cos2 \theta = 2 (\frac{ 35}{36}) - 1 } \\ \displaystyle \large{ \cos2 \theta = \frac{ 35}{18} - 1 } \\ \displaystyle \large{ \cos2 \theta = \frac{ 35}{18} - \frac{18}{18} \longrightarrow \boxed{\frac{17}{18} }}$

Answer

• sin(2theta) = √35/18
• cos(2theta) = 17/18

The last choice.

Let me know if you have any questions!