Question

Use the Special Right Triangle to evaluate sin 45°, cos 45° and tan 45°. Your answers should be exact (not a decimal).
A. sin 45 = √2/2, cos 45 = √2/2, tan 45 = 1
B. sin 45 = 1, cos 45 = 1, tan 45 = √2/2
C. sin 45 = 1/2, cos 45 = √3/2, tan 45 = √3
D. sin 45 = 0, cos 45 = 1, tan 45 = 0

Answer 1

The best option is (A) sin 45 = √2/2, cos 45 = √2/2, tan 45 = 1.

Answer 2

Answer:

The correct option is A) $\sin 45^{\circ} =\frac{\sqrt{2}}{2},\ \cos 45^{\circ} =\frac{\sqrt{2}}{2} \ \text{and} \ \tan 45^{\circ} =1$

Step-by-step explanation:

we need to evaluate $\sin 45^{\circ}, \ \cos 45^{\circ} \ \text{and} \ \tan45^{\circ}$ with the use of special right triangle

In triangle ABC (figure -1 )

Since, $\sin \theta =\frac{opposite}{hypoteneous}$

$\cos  \theta =\frac{adjacent}{hypoteneous}$

$\tan  \theta =\frac{opposite}{adjacent}$

so,

$\sin 45^{\circ} =\frac{opposite}{hypoteneous}$

$\sin 45^{\circ} =\frac{x}{x\sqrt{2}}$

$\sin 45^{\circ} =\frac{1}{\sqrt{2}}$

$\cos 45^{\circ} =\frac{adjacent}{hypoteneous}$

$\cos 45^{\circ} =\frac{x}{x\sqrt{2}}$

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

and

$\tan  45^{\circ} =\frac{opposite}{adjacent}$

$\tan  45^{\circ} =\frac{x}{x}$

$\tan 45^{\circ} =1$

Therefore, the correct option is A) $\sin 45^{\circ} =\frac{\sqrt{2}}{2},\ \cos 45^{\circ} =\frac{\sqrt{2}}{2} \ \text{and} \ \tan 45^{\circ} =1$