Question

What is the exact value of cos(225)cos(75)

Answer 1

the answer is
-√3-1÷4

Answer 2

Answer:

$\cos \left ( 225^{\circ} \right )\cos \left ( 75^{\circ} \right )=\frac{1-\sqrt{3}}{4}$

Step-by-step explanation:

To find: $\cos \left ( 225^{\circ} \right )$

$\cos \left ( 225^{\circ} \right )=\cos \left ( 180^{\circ}+45^{\circ} \right )$

Now, angle $180^{\circ}+45^{\circ}$ lies in third quadrant in which $\cos$ is negative

So, $\cos \left ( 225^{\circ} \right )=\cos \left ( 180^{\circ}+45^{\circ} \right )=-\cos 45^{\circ}$

We know that $\cos 45^{\circ}=\frac{1}{\sqrt{2}}$

So, $\cos \left ( 225^{\circ} \right )=\frac{-1}{\sqrt{2}}$

To find: $\cos \left ( 75^{\circ} \right )$

$\cos \left ( 75^{\circ} \right )=\cos \left ( 90^{\circ}-15^{\circ} \right )$

We know that angle $90^{\circ}-15^{\circ}$ lies in first quadrant in which $\cos$ is positive. Also, $\cos \left ( 90-\alpha  \right )=\sin \alpha$

So, $\cos \left ( 75^{\circ} \right )=\cos \left ( 90^{\circ}-15^{\circ} \right )=\sin \left ( 15^{\circ} \right )$

We will use formula: $\sin \left ( A-B \right )=\sin A\cos B-\cos A\sin B$

Here,

$\sin \left ( 15^{\circ} \right )=\sin \left ( 45^{\circ}-30^{\circ} \right )=\sin \left ( 45^{\circ} \right )\cos \left ( 30^{\circ} \right )-\cos \left ( 45^{\circ} \right )\sin \left ( 30^{\circ} \right )=\left ( \frac{1}{\sqrt{2}} \right )\left ( \frac{\sqrt{3}}{2} \right )-\left ( \frac{1}{\sqrt{2}} \right )\left ( \frac{1}{2} \right )=\frac{\sqrt{3}}{2\sqrt{2}}-\frac{1}{2\sqrt{2}}=\frac{\sqrt{3}-1}{2\sqrt{2}}$

Therefore,

$\cos \left ( 225^{\circ} \right )\cos \left ( 75^{\circ} \right )=\left ( \frac{-1}{\sqrt{2}} \right )\frac{\sqrt{3}-1}{2\sqrt{2}}=\frac{1-\sqrt{3}}{4}$