4 multiple choice 1 point give the exact value. cos 45° 5 multiple choice 1 point give the exact value. sec…

Explanation:

Step1: Recall the cosine value of special angles

We know that for a (45^{\circ}-45^{\circ}-90^{\circ}) triangle, if the legs are of length (a) and the hypotenuse is (c), by the Pythagorean theorem (c^{2}=a^{2}+a^{2}=2a^{2}), so (c = a\sqrt{2}). Then (\cos45^{\circ}=\frac{\text{adjacent}}{\text{hypotenuse}}=\frac{a}{a\sqrt{2}}=\frac{1}{\sqrt{2}}). Rationalizing the denominator, we get (\frac{\sqrt{2}}{2}).

Step2: Recall the secant formula and reference angle for (150^{\circ})

We know that (\sec\theta=\frac{1}{\cos\theta}). The angle (150^{\circ}) has a reference angle of (180 - 150=30^{\circ}). And (\cos150^{\circ}=-\cos30^{\circ}) (since (150^{\circ}) is in the second - quadrant where cosine is negative). Since (\cos30^{\circ}=\frac{\sqrt{3}}{2}), then (\cos150^{\circ}=-\frac{\sqrt{3}}{2}). So (\sec150^{\circ}=\frac{1}{\cos150^{\circ}}=\frac{1}{-\frac{\sqrt{3}}{2}}=-\frac{2}{\sqrt{3}}). Rationalizing the denominator gives (-\frac{2\sqrt{3}}{3}).

Answer:

For (\cos45^{\circ}): (\frac{\sqrt{2}}{2}) (the fourth option). For (\sec150^{\circ}): (-\frac{2\sqrt{3}}{3}) (the second option).