Respuesta :
Using trigonometric identities, it is found that the measures are given as follows:
a) [tex]\cos{2\theta} = -\frac{6\sqrt{14}}{25}[/tex]
b) [tex]\sin{\left(\frac{\theta}{2}\right)} = \sqrt{\frac{5 + \sqrt{7}}{10}}[/tex]
How to find the value of cosine of 2θ?
cos(2θ) is determine according to the following identity:
cos(2θ) = 2sin(θ)cos(θ)
Given the sine, we find the cosine according to the following identity:
[tex]\sin^{2}{\theta} + \cos^2{\theta} = 1[/tex]
For this problem, the sine is given by:
[tex]\sin{\theta} = \frac{3\sqrt{2}}{5}[/tex]
Hence the cosine is found as follows:
[tex]\left(\frac{3\sqrt{2}}{5}\right)^2 + \cos^2{\theta} = 1[/tex]
[tex]\frac{18}{25} + \cos^2{\theta} = 1[/tex]
[tex]\cos^2{\theta} = \frac{7}{25}[/tex]
[tex]\cos{\theta} = \pm \sqrt{\frac{7}{25}}[/tex]
Since the angle is on the second quadrant, the cosine is negative, hence:
[tex]\cos{\theta} = -\frac{\sqrt{7}}{5}[/tex]
Hence the cosine of the doubled angle is:
[tex]\cos{2\theta} = 2\sin{\theta}\cos{\theta} = 2 \times \frac{3\sqrt{2}}{5} \times -\frac{\sqrt{7}}{5} = -\frac{6\sqrt{14}}{25}[/tex]
How to find the sine of the halved angle?
The sine of the halved angle is given by the following identity:
[tex]\sin{\left(\frac{\theta}{2}\right)} = \sqrt{\frac{1 - \cos{\theta}}{2}}[/tex]
Hence:
[tex]\sin{\left(\frac{\theta}{2}\right)} = \sqrt{\frac{1 + \frac{\sqrt{7}}{5}}{2}} = \sqrt{\frac{5 + \sqrt{7}}{10}}[/tex]
More can be learned about trigonometric identities at https://brainly.com/question/26676095
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Answer:
[tex]\textsf{A)} \quad \cos (2\theta)=-\dfrac{11}{25}[/tex]
[tex]\textsf{B)} \quad \sin \left(\dfrac{\theta}{2}\right) = \sqrt{\dfrac{5+\sqrt{7}}{10}[/tex]
Step-by-step explanation:
Given:
[tex]\sin \theta=\dfrac{3\sqrt{2}}{5}, \quad \quad \dfrac{\pi}{2} < \theta < \pi[/tex]
Part A
[tex]\boxed{\begin{minipage}{4 cm}\underline{Trigonometric Identities}\\\\$\cos (2 \theta)=\cos^2 \theta - \sin^2 \theta\\\\\sin^2 \theta + \cos^2 \theta=1$\\\end{minipage}}[/tex]
Using the trigonometric identities, rewrite cos(2θ) using sinθ only:
[tex]\begin{aligned}\implies \cos (2 \theta) & =\cos^2 \theta - \sin^2 \theta\\&=1-\sin^2 \theta - \sin^2 \theta\\&=1 - 2 \sin^2 \theta\end{aligned}[/tex]
Substitute the given value of sinθ into the rewritten cos double angle identity:
[tex]\begin{aligned}\implies \cos (2 \theta) & =1 - 2 \sin^2 \theta \\ & =1 - 2 \left(\dfrac{3\sqrt{2}}{5}\right)^2 \\& = 1-2\left(\dfrac{18}{25}\right)\\& = 1-\dfrac{36}{25}\\& = -\dfrac{11}{25}\end{aligned}[/tex]
Part B
[tex]\boxed{\begin{minipage}{5.5cm}\underline{Trigonometric Identities}\\\\$ \sin \left(\dfrac{\theta}{2}\right)=\pm \sqrt{\dfrac{1-\cos \theta}{2}} $\\\\\sin^2 \theta + \cos^2 \theta=1\\\end{minipage}}[/tex]
Rewrite the second identity to isolate cosθ:
[tex]\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}[/tex]
As θ is obtuse, cosθ < 0. Therefore:
[tex]\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}[/tex]
Substitute this into the Half Angle Identity, using the positive root as sinθ is positive:
[tex]\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}[/tex]
