Wrap+up+to+Investigation+1B

math

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Today in class we started with a quiz. Ms. Tyson then did one of the problems on the board to correctly demonstrate how to properly solve the equation. Ex:

Solve.

$$

6sin(x+1)-4=1

6sin(x+1)=5

sin(x+1) = \frac{5}{6}

x+1=\sin^{-1}(\frac{5}{6})

$$

Here it is important to note that the $ \sin^{-1}(\frac{5}{6}) $ has two answers.

Possibility 1: $

x+1= .985+2\pi K; K \epsilon\mathbb{Z}

x= -.015+2\pi K; K \epsilon \mathbb{Z}

$

Possibility 2:

$

\Pi - .985+ 2\pi K; K\epsilon \mathbb{Z} $

One could have also done this problem by leaving their answers in $ \sin^{-1}(x) $ form.

For these types of problems, it is important to remember that there are two answers between $ (0, 2 \Pi) $ for $ sin(x) $ and $ cos(x) $. For $ tan(x) $ there are also two answers, but the period is $ \pi $ and it is a one to one function.$

Therefore, for answers involving $ tan(x) $, the equation for all solutions has it reoccurring every $ \pi $ times, not $ 2 \pi $.

Relating to this topic, Ms. Tyson brought up the homework from $\S$ 1.9 #7d. For that problem it is important to watch what is inside the of $sin(x)$. If there is something multiplying or dividing the x, the period must also be multiplied or divided. If you want more problems like this to practice, Ms. Tyson pointed out $\S$ 1.9 CYC 5,6; OYO 7, 9-13 and $\S$ 1.10 CYC 3,4,6,7; OYO 10-13 are similar to the above problem.

Then Ms. Tyson asked for comments on Investigation 1B. In the Reflections 1B #5, some important relationships were pointed out: math math

$ With this image, Ms. Tyson reminded us of supplementary and complementary angles and how we used them in geometry. She then helped us prove the Pythagorean identifies. The radius of the unit circle is 1 so: $ \sin(x)^{2} + \cos(x)^{2}=1$ as $sin(\alpha)$ and $cos(\alpha)$ are the legs of a right triangle and the Pythagorean theorem states that $ a^{2}+b^{2}=c^{2}$. Another right triangle in the picture above proves $ tan(\alpha )^{2}+1=\sec (\alpha )^{2}$. Then, if you divide the original equation by $\cos(x)^{2}$ to get $\cot ^{2}(x)+1=\csc ^{2}(x)$.

math