Question

URGENT WILL GIVE BRAINLIEST

Answer 1

Answer:

$\textsf{A)} \quad \cos (2\theta)=-\dfrac{11}{25}$

$\textsf{B)} \quad \sin \left(\dfrac{\theta}{2}\right) = \sqrt{\dfrac{5+\sqrt{7}}{10}$

Step-by-step explanation:

Given:

$\sin \theta=\dfrac{3\sqrt{2}}{5}, \quad \quad \dfrac{\pi}{2} < \theta < \pi$

Part A

$\boxed{\begin{minipage}{4 cm}\underline{Trigonometric Identities}\\\\ \cos (2 \theta)=\cos^2 \theta - \sin^2 \theta\\\\\sin^2 \theta + \cos^2 \theta=1 \\\end{minipage}}$

Using the trigonometric identities, rewrite cos(2θ) using sinθ only:

$\begin{aligned}\implies \cos (2 \theta) & =\cos^2 \theta - \sin^2 \theta\\&=1-\sin^2 \theta - \sin^2 \theta\\&=1 - 2 \sin^2 \theta\end{aligned}$

Substitute the given value of sinθ into the rewritten cos double angle identity:

$\begin{aligned}\implies \cos (2 \theta) & =1 - 2 \sin^2 \theta \\ & =1 - 2 \left(\dfrac{3\sqrt{2}}{5}\right)^2 \\& = 1-2\left(\dfrac{18}{25}\right)\\& = 1-\dfrac{36}{25}\\& = -\dfrac{11}{25}\end{aligned}$

Part B

$\boxed{\begin{minipage}{5.5cm}\underline{Trigonometric Identities}\\\\ \sin \left(\dfrac{\theta}{2}\right)=\pm \sqrt{\dfrac{1-\cos \theta}{2}} \\\\\sin^2 \theta + \cos^2 \theta=1\\\end{minipage}}$

Rewrite the second identity to isolate cosθ:

$\begin{aligned}\sin^2 \theta+\cos ^2 \theta & =1\\\cos ^2 \theta & =1-\sin^2 \theta\\\cos \theta & = \pm \sqrt{1-\sin^2 \theta}\end{aligned}$

As θ is obtuse, cosθ < 0.  Therefore:

$\begin{aligned}cos \theta & =- \sqrt{1-\sin^2 \theta}\\& =- \sqrt{1-\left(\dfrac{3\sqrt{2}}{5}\right)^2}\\& =- \sqrt{1-\left(\dfrac{18}{2}\right)}\\& =- \sqrt{\dfrac{7}{25}}\\& =- \dfrac{\sqrt{7}}{5}\end{aligned}$

Substitute this into the Half Angle Identity, using the positive root as sinθ is positive:

$\begin{aligned}\sin \left(\dfrac{\theta}{2}\right) & = \sqrt{\dfrac{1-\cos \theta}{2}}\\& = \sqrt{\dfrac{\dfrac{5+\sqrt{7}}{5}}{2}}\\& = \sqrt{\dfrac{5+\sqrt{7}}{10}}\\\end{aligned}$