Question

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

Answer 1

Part 5.1.1:

`\displaystyle \cos 2A = (7)/(8)`

Part 5.1.2:

`\displaystyle \cos A = (√(15))/(4)`

Explanation:

We are given that:

`\displaystyle \sin 2A = (√(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 = (15)/(64)`

Hence:

`\displaystyle \left((15)/(64)\right) + \cos ^2 2A = 1`

Subtract:

`\displaystyle \cos ^2 2A = (49)/(64)`

Take the square root of both sides:

`\displaystyle \cos 2A = \pm\sqrt{(49)/(64)}`

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

`\displaystyle \cos 2A = (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((7)/(8)\right) = 2\cos^2 A - 1`

Solve for cosine:

`\displaystyle \begin{aligned} (15)/(8) &= 2\cos^2 A\\ \\ \cos^2 A &= (15)/(16) \\ \\ \cos A& = \pm\sqrt{(15)/(16)} \\ \\ \Rightarrow \cos A &= (√(15))/(4)\end{aligned}`

In conclusion:

`\displaystyle \cos A = (√(15))/(4)`

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