Question

Could someone please help me:) I am stick and I am not sure what to do ​

Answer 1

Answer:

Part 5.1.1:

$\displaystyle \cos 2A = \frac{7}{8}$

Part 5.1.2:

$\displaystyle \cos A = \frac{\sqrt{15}}{4}$

Step-by-step explanation:

We are given that:

$\displaystyle \sin 2A = \frac{\sqrt{15}}{8}$

Part 5.1.1

Recall that:

$\displaystyle \sin^2 \theta + \cos^2 \theta = 1$

Let θ = 2A. Hence:

$\displaystyle \sin ^2 2A + \cos ^2 2A = 1$

Square the original equation:

$\displaystyle \sin^2 2A = \frac{15}{64}$

Hence:

$\displaystyle \left(\frac{15}{64}\right) + \cos ^2 2A = 1$

Subtract:

$\displaystyle \cos ^2 2A = \frac{49}{64}$

Take the square root of both sides:

$\displaystyle \cos 2A = \pm\sqrt{\frac{49}{64}}$

Since 0° ≤ 2A ≤ 90°, cos(2A) must be positive. Hence:

$\displaystyle \cos 2A = \frac{7}{8}$

Part 5.1.2

Recall that:

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

We can use the third form. Substitute:

$\displaystyle \left(\frac{7}{8}\right) = 2\cos^2 A - 1$

Solve for cosine:

$\displaystyle \begin{aligned} \frac{15}{8} &= 2\cos^2 A\\ \\ \cos^2 A &= \frac{15}{16} \\ \\ \cos A& = \pm\sqrt{\frac{15}{16}} \\ \\ \Rightarrow \cos A &= \frac{\sqrt{15}}{4}\end{aligned}$

In conclusion:

$\displaystyle \cos A = \frac{\sqrt{15}}{4}$

(Note that since 0° ≤ 2A ≤ 90°, 0° ≤ A ≤ 45°. Hence, cos(A) must be positive.)