First slide

Compound angles

Question

If tan $α$ is equal to the integral solution of the inequality $4 x^{2} - 16 x + 15 < 0$ and $cosβ$ it equal to the slope of the bisector of the first quadrant, then $\sin (α + β) \sin (α - β)$ is

Moderate

A

3/5
B

5/3
C

$\frac{2}{\sqrt{5}}$
D

4/5

Solution

We have $4 x^{2} - 16 x + 15 < 0$
$\Rightarrow \frac{3}{2} < x < \frac{5}{2}$
Therefore, the integral solution of
$4 x^{2} - 16 x + 15 < 0 is x = 2$
$Thus, \tan α = 2 . It is given that \cos β = \tan 45^{\circ} = 1 .$
$\begin{matrix} ∴ \sin (α + β) \sin (α - β) = \sin^{2} α - \sin^{2} β \\ = \frac{1}{1 + \cot^{2} α} - (1 - \cos^{2} β) \\ = \frac{1}{1 + \frac{1}{4}} - 0 = \frac{4}{5} \end{matrix}$

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