Question

whats the exact value of cos 75°?
a) $\frac{sqrt{6}-sqrt{2}}{4}$
b) $\frac{sqrt{6}+sqrt{2}}{2}$
c) $\frac{sqrt{6}+sqrt{2}}{4}$
d) $\frac{sqrt{6}-sqrt{2}}{2}$
question 3 (5 points)
find the period of the function $y = \frac{3}{2}\tan(\frac{1}{3}x)$.
a) $\frac{pi}{6}$
b) $pi$
c) $3pi$

Explanation:

Step1: Express 75° as a sum of known angles

We know that \(75^{\circ}=45^{\circ} + 30^{\circ}\). The cosine - addition formula is \(\cos(A + B)=\cos A\cos B-\sin A\sin B\). Here \(A = 45^{\circ}\) and \(B = 30^{\circ}\).
We know that \(\cos45^{\circ}=\frac{\sqrt{2}}{2}\), \(\sin45^{\circ}=\frac{\sqrt{2}}{2}\), \(\cos30^{\circ}=\frac{\sqrt{3}}{2}\), and \(\sin30^{\circ}=\frac{1}{2}\).

Step2: Substitute the values into the formula

\(\cos(45^{\circ}+30^{\circ})=\cos45^{\circ}\cos30^{\circ}-\sin45^{\circ}\sin30^{\circ}\)
\(=\frac{\sqrt{2}}{2}\times\frac{\sqrt{3}}{2}-\frac{\sqrt{2}}{2}\times\frac{1}{2}=\frac{\sqrt{6}-\sqrt{2}}{4}\)

Step3: Find the period of the tangent function

The general form of the tangent function is \(y = A\tan(Bx)\), and its period is given by \(T=\frac{\pi}{|B|}\).
For the function \(y=\frac{3}{2}\tan(\frac{1}{3}x)\), where \(A = \frac{3}{2}\) and \(B=\frac{1}{3}\).
The period \(T=\frac{\pi}{\frac{1}{3}} = 3\pi\)

Answer:

1. A. \(\frac{\sqrt{6}-\sqrt{2}}{4}\)
2. C. \(3\pi\)