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Use the Special Right Triangle to evaluate sin 45°, cos 45° and tan 45°. Your answers should be exact (not a decimal).

Asin 45 = √2/2, cos 45 = √2/2, tan 45 = 1
Bsin 45 = 1, cos 45 = 1, tan 45 = √2/2
Csin 45 = 1/2, cos 45 = √3/2, tan 45 = √3
Dsin 45 = 0, cos 45 = 1, tan 45 = 0

Verified Answer :

syed514

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

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) \(\sin 45^{\circ} =\frac{\sqrt{2}}{2},\ \cos 45^{\circ} =\frac{\sqrt{2}}{2} \ \text{and} \ \tan 45^{\circ} =1\)

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