# Thread: Basic Tan Sin Cos Questions

1. ## Basic Tan Sin Cos Questions

Hello again! I got some basic trig questions...

Complete the following list:

sin A = $\displaystyle$\dfrac{a}{c}$$(this one was done as an example) sin B = \displaystyle \dfrac{b}{c}$$

m∠A + m∠B = 90°

cos A = $\displaystyle$\dfrac{b}{c}$$cos B = \displaystyle \dfrac{a}{c}$$

a2 + b2 = c2

tan A = $\displaystyle$\dfrac{a}{b}$$tan B = \displaystyle \dfrac{b}{a}$$

Circle the following statement we can conclude from the information above.
(A) sin A = cos B
(B) cos A = sin (90 -A)
(C) sin2 B + cos2 B = 1
(D) All of these

[I think it's A or D]

Complete the following:
sin 45° = $\displaystyle$\dfrac{1}{ \sqrt[]{2}}$$=\displaystyle \dfrac{ \sqrt[]{2}}{2}$$ (this one was done as an example)

sin 30° =

cos 45° =

cos 30° =

tan 45° =

tan 30° =
[I actually do not understand what they are asking me to do here. If someone could explain what's being asked I'd really appreciate it.]

13. Circle the item below that is equal to $\displaystyle \sqrt{3}$
(A) sin 60° (B) tan 60° (C) cos 60° (D) tan-1 $\displaystyle$\dfrac{ \sqrt[]{3}}{3}$$[I think it's B] 2. ## Re: Basic Tan Sin Cos Questions For the first question you might as well get a calculator draw an a triangle with valid angles and experiment with each option given, either way it is option D). For the second part we are given the triangle with all three side lengths. We know that sin Q = opposite/hypotenuse cos Q= adj./hypotenuse, tan Q= opp./adj. So if Q = 30 degrees just pick the required side lengths and place them in the ratio required. So, sin 30 = 1/2 , can you do the rest. Finally it is option B) because if we consult the triangle diagram, when the angle is 60 degrees, the trigonometric ratio tan 60 = root 3 / 1 = root 3 3. ## Re: Basic Tan Sin Cos Questions Hey thanks for the quick reply! I thought that sin 30 = 1/2 but I was just confused because the example had two statements :/ So then sin 30° = 1/2 cos 45° = 1 / \displaystyle \sqrt{2} cos 30° = \displaystyle \sqrt{3} / 2 tan 45° = \displaystyle \dfrac{1}{1}$$

tan 30° = $\displaystyle$\dfrac{1}{ \sqrt{3}}$$Right? I still don't understand the last part of the example sin 45° = \displaystyle \dfrac{1}{ \sqrt[]{2}}$$ = $\displaystyle$\dfrac{ \sqrt[]{2}}{2} I understand what the first "fraction" refers to but not the second...

4. ## Re: Basic Tan Sin Cos Questions

We are just multiplying $\frac{1}{\sqrt2}$ by $\frac{\sqrt2}{\sqrt2}$ which is the same as multiplying by 1. However the purpose here is to turn our denominator into a whole number instead of $\sqrt2$. Its mathematical convention.

So, $\frac{1}{\sqrt2} \times \frac{\sqrt2}{\sqrt2}$
$= \frac{\sqrt2}{\sqrt2 \times \sqrt2}$
$= \frac{\sqrt2}{2}$

5. ## Re: Basic Tan Sin Cos Questions

Oh duh, I feel like an idiot! I though they were trying to make a similarity statement with another triangle.

Thanks for explaining things clearly to the oblivious!