# Thread: Two questions regarding sin cos and tan relations.

1. ## Two questions regarding sin cos and tan relations.

Hi, this is a bit embarrassing but I'm a bit rusty with this stuff.
My first question:
a) If sin(a°) = 0.951, then what is tan a equal to?
The answer is 3.076 according to the book, however i'm not sure how this conclusion is reached? If someone could explain that'd be nice.

Question 2:
If 0°<a<90° and cos a<0.5, then which of the following is correct?
A. a<30° B. a>30° C. a<60°
D. a<45° E. a>60°

Once again, no idea on how to go about this and need clearing up. Do any of these questions relate to :

tan(x) = sin(x) / cos(x)
or
sinē(x) + cosē(x) = 1
Thanks.

2. Originally Posted by 99.95
Hi, this is a bit embarrassing but I'm a bit rusty with this stuff.
My first question:
a) If sin(a°) = 0.951, then what is tan a equal to?
The answer is 3.076 according to the book, however i'm not sure how this conclusion is reached? If someone could explain that'd be nice.

...
Once again, no idea on how to go about this and need clearing up. Do any of these questions relate to :

tan(x) = sin(x) / cos(x)
or
sinē(x) + cosē(x) = 1
Thanks.
$\sin^2(x) + \cos^2(x)=1~\implies~\cos(x)=\sqrt{1-\sin^2(x)}$

Plug in the value of $\sin(a)$ to calculate the value of $\cos(a)$

$\tan(a)=\dfrac{\sin(a)}{\cos(a)}$

Plug in the values of $\sin(a)$ and $\cos(a)$ to calculate the value of $\tan(a)$.

3. Thanks, sloppy of me not to get that -_-.
Sorry for the typo in Question 2, that's fixed now, any ideas?

4. Originally Posted by 99.95
Hi, this is a bit embarrassing but I'm a bit rusty with this stuff.
...
Question 2:
If 0°<a<90° and cos a<0.5, then which of the following is correct?
A. a<30° B. a>30° C. a<60°
D. a<45° E. a>60°

Once again, no idea on how to go about this and need clearing up. Do any of these questions relate to :

...
Maybe the following list helps:

$\begin{array}{c|c|c}\alpha & \sin(\alpha) & \cos(\alpha)\\ \hline\\ 0^\circ & \frac12 \sqrt{0} & \frac12 \sqrt{4} \\ 30^\circ & \frac12 \sqrt{1} & \frac12 \sqrt{3} \\ 45^\circ & \frac12 \sqrt{2} & \frac12 \sqrt{2} \\ 60^\circ & \frac12 \sqrt{3} & \frac12 \sqrt{1} \\ 90^\circ & \frac12 \sqrt{4} & \frac12 \sqrt{0} \end{array}$

5. Ah i see now, thank you very much.

6. Another way to find tan(a) given that sin(a) = 0.951:

Imagine a triangle with one leg of length 0.951 and hypotenuse 1. The angle opposite that leg is a and, by the Pythagorean theorem, the "near side" to that angle has length $\sqrt{1- .951^2}= \sqrt{0.095599}= 0.3092$. tan(a) is "opposite side divided by near side", $\frac{.951}{.3092}= 3.076$. It's exactly the same calculation except that I have phrased it in terms of the Pythagorean theorem, $a^2+ b^2= c^2$ rather than the trigonometric identity, $sin^2(a)+ cos^2(a)= 1$.

A way of "remembering" the values Earboth gives without having to "memorize" them:

Since 45+ 45= 90, a right triangle having one angle 45 degrees also has the other 45 degrees and so is an isosceles right triangle- its legs are the same length. If we take that length to be "1", by the Pythagorean theorem, the hypotenuse has length $\sqrt{2}$. From that, $sin(45)= cos(45)= \frac{1}{\sqrt{2}}= \frac{\sqrt{2}}{2}$ when we "rationalize the denominator".

For the 40 and 60 degree angles, remember that an equilateral triangle has all three angle 60 degrees. If draw a line from one vertex perpendicular to the opposite side, that also bisects the side and bisects the angle: we now have two right triangles with angle 30 degrees opposite the "bisected" side and angle 60 degree adjacent to that side. If we take the "bisected" side to have length 1, then a whole side of the original equilateral triangle, the hypotenuse of the right triangle, has length 2. By the Pythagorean theorem again, the other leg, the perpendicular line, has length $\sqrt{4- 1}= \sqrt{3}$. The leg of length 1 is "opposite" the 30 degree angle and "adjacent" to the 60 degree angle. The leg of length $\sqrt{3}$ is "opposite" the 60 degree angle and "adjacent" to the 30 degree angle.

$sin(30)= cos(60)= \frac{1}{2}$ and $sin(60)= cos(30)= \frac{\sqrt{3}}{2}$.