MHB Helping my daughter study for a Trig test....

[ok782]

My daughter is a Sophmore in high school and is taking Calculus with a current focus on Trig. I am trying to help her study but this stuff is way over my head...but I am trying! Can anyone help me with this? Here are a few of her study problems that we cannot figure out. Can you help me to solve but also explain how so we can learn? Thank you so much!Solve the following triangle and find it's area. Round all sides and all angles to the nearest tenth.
B =72 degrees, a=10, c=7Find the exact values for the following without using a calculator:

1 - Sin(Arcsin2)
2 - Arc cos(cos 7pi/4)
3 - csc(arc cos 2/9)
4 - tan(arc sin 3/7)Solve the following trigonometric equation for the domain 0 <= x <2pi

1- 2sin^2 x=0Thank you so much to anyone who has the chance to look this over. The test is tomorrow, so hopefully someone will see this today!

 

[MarkFL]

For the first problem, to get the unknown side $b$, we can use the Law of Cosines as follows:

$$b=\sqrt{a^2+c^2-2ac\cos\left(72^{\circ}\right)}$$

Then, to get the two unknown angles $A$ and $C$, we can use the Law of Sines as follows:

$$\frac{\sin(A)}{a}=\frac{\sin(B)}{b}\implies A=\arcsin\left(\frac{a\sin(B)}{b}\right)$$

$$\frac{\sin(C)}{c}=\frac{\sin(B)}{b}\implies C=\arcsin\left(\frac{c\sin(B)}{b}\right)$$

Actually, once you get one of the unknown angles, you can use the fact that the sum of all 3 interior angles must be $180^{\circ}$ to get the remaining unknown angle.

And finally to get the area $Z$, we can use the following formula:

$$Z=\frac{1}{2}ac\sin(B)$$

 

[Prove It]

Find the exact values for the following without using a calculator:

1 - Sin(Arcsin2)
2 - Arc cos(cos 7pi/4)
3 - csc(arc cos 2/9)
4 - tan(arc sin 3/7)Solve the following trigonometric equation for the domain 0 <= x <2pi

1- 2sin^2 x=0Thank you so much to anyone who has the chance to look this over. The test is tomorrow, so hopefully someone will see this today!

Some hints for the first lot here:

$\displaystyle \begin{align*} -1 \leq \sin{(x)} \leq 1 , \, -1 \leq \cos{(x)} \leq 1 \end{align*}$

The inverse trigonometric functions are defined over the following regions:

$\displaystyle \begin{align*} -\frac{\pi}{2} \leq \arcsin{(x)} \leq \frac{\pi}{2}, \, 0 \leq \arccos{(x)} \leq \pi , \, -\frac{\pi}{2} \leq \tan{(x)} \leq \frac{\pi}{2} \end{align*}$

I would be inclined to think that the first does not exist, as $\displaystyle \begin{align*} \arcsin{(2)} \end{align*}$ is saying there is an angle such that $\displaystyle \begin{align*} \sin{\left( \theta \right) } = 2 \end{align*}$, which is impossible.

2.
$\displaystyle \begin{align*} \cos{ \left( \frac{7\,\pi}{4} \right) } &= \cos{ \left( 2\,\pi - \frac{\pi}{4} \right) } \\ &= \cos{ \left( \frac{\pi}{4} \right) } \end{align*}$

and since $\displaystyle \begin{align*} 0 \leq \arccos{ (x) } \leq \pi \end{align*}$, that would have to mean

$\displaystyle \begin{align*} \arccos{ \left[ \cos{ \left( \frac{7\,\pi}{4} \right) } \right] } &= \arccos{ \left[ \cos{ \left( \frac{\pi}{4} \right) } \right] } \\ &= \frac{\pi}{4} \end{align*}$

3.
$\displaystyle \begin{align*} \arccos{ \left( \frac{2}{9} \right) } \end{align*}$ is in the first quadrant (can you see why?)

Drawing up a right angle triangle with an angle $\displaystyle \begin{align*} \theta \end{align*}$ having adjacent side 2 units and hypotenuse 9 units, by Pythagoras the opposite side must be $\displaystyle \begin{align*} \sqrt{77} \end{align*}$ units, therefore $\displaystyle \begin{align*} \arccos{ \left( \frac{2}{9} \right) } = \arcsin{ \left( \frac{\sqrt{77}}{9} \right) } \end{align*}$. So

$\displaystyle \begin{align*} \csc{ \left[ \arccos{ \left( \frac{2}{9} \right) } \right] } &= \csc{ \left[ \arcsin{ \left( \frac{\sqrt{77}}{9} \right) } \right] } \\ &= \frac{1}{\sin{ \left[ \arcsin{ \left( \frac{\sqrt{77}}{9} \right) } \right] }} \\ &= \frac{1}{\frac{\sqrt{77}}{9}} \\ &= \frac{9}{\sqrt{77}} \\ &= \frac{9\,\sqrt{77}}{77} \end{align*}$

Do you think your daughter could follow a similar process for question 4?

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1- 2sin^2 x=0

$\displaystyle \begin{align*} 1 - 2\sin^2{(x)} &= 0 \\ 2\sin^2{(x)} &= 1 \\ \sin^2{(x)} &= \frac{1}{2} \\ \sin{(x)} &= \pm \frac{1}{\sqrt{2}} \\ x &= \left\{ \frac{\pi}{4} , \, \frac{3\,\pi}{4} , \, \frac{5\,\pi}{4} , \, \frac{7\,\pi}{4} \right\} + 2\,\pi\,n \textrm{ where }n \in \mathbf{Z} \\ x &= \left\{ \frac{\pi}{4}, \, \frac{3\,\pi}{4}, \, \frac{5\,\pi}{4} , \, \frac{7\,\pi}{4} \right\} \textrm{ as } 0 \leq x \leq 2\,\pi \end{align*}$

 

[suluclac]

For the first problem, to get the unknown side $b$, we can use the Law of Cosines as follows:

$$b=\sqrt{a^2+c^2-2ac\cos\left(72^{\circ}\right)}$$

And finally to get the area $Z$, we can use the following formula:

$$Z=\frac{1}{2}ac\sin(B)$$

This is correct, but also note that $$Z=\frac{1}{4}\sqrt{4a^2b^2-(a^2+b^2-c^2)^2}$$.