what if a problem just asks you to find the sin, cos, and tan of 30 degrees , 60 ...

[abel muroi]

what if a problem just asks you to find the sin, cos, and tan of 30 degrees , 60 degrees and 45 degrees WITHOUT giving you a triangle.

how do i start finding the functions here?

btw is it even possible to find sin 45 degrees , ect, without a triangle?

 

[Ishuda]

what if a problem just asks you to find the sin, cos, and tan of 30 degrees , 60 degrees and 45 degrees WITHOUT giving you a triangle.

how do i start finding the functions here?

btw is it even possible to find sin 45 degrees , ect, without a triangle?

Well you can start with one of them and use trig identities to find the others or you can just remember them.

Yes, it is possible to find sin(x) without the triangle. You can define the sine (and other trig quantities) without relying on geometry (triangles) although it is sometimes much easier to use triangles to explain the sine, cosine, and tangent (and other trig functions).

 

[abel muroi]

Well you can start with one of them and use trig identities to find the others or you can just remember them.

Yes, it is possible to find sin(x) without the triangle. You can define the sine (and other trig quantities) without relying on geometry (triangles) although it is sometimes much easier to use triangles to explain the sine, cosine, and tangent (and other trig functions).

By trig identities i assume you mean something like

sin(of theta) = 2/4

and i would use the 2 and the 4 to figure out the other side of the right triangle by using that famous theorem (a² + b² = c²)...

but what if i never was assigned those "identities"?

how would i solve, lets say, cos 45 degrees?

 

[Quaid]

A trigonometric identity is an equation that's true for all values of the variable, where each side is defined.

EGs

sin(x)^2 + cos(x)^2 = 1

tan(x) = sin(x)/cos(x)

sin(x) = 1/csc(x)

Each of these equations is true for all values of x in their respective domains, but, regarding the last two identities, remember that a domain cannot include any value of x which lead to zero in a denominator. This is what's meant by the condition that each side must be defined, in the definition above.

sin(x) = 2/4 is not an identity because this equation is not true for all values of x.

Why are you asking about determining trig values without using triangles, for the most-common angles? By the way, if you're not given a triangle, you can always make one up.

Check out the derivations on this page.

Cheers

 

[HallsofIvy]

Think of an equilateral triangle. Since it has all three sides the same, of length s, say, it must have all three angles the same. And since the angles in a triangle add to 180 degrees, each must be 180/3= 60. Now draw a line from one vertex to the midpoint of the opposite side. That divides the equilateral triangle into two congruent triangles and it is easy to show that this new line is also perpendicular to opposite side and bisects the angle at the vertex. That means we now have two right triangles with angle 60 degrees and 30 degrees. The hypotenuse of either is a side of the original triangle, of length s. The leg opposite the 30 degree angle is half a side of the original triangle, of length s/2. Calling the length of other leg "x", by the Pythagorean theorem we must have \(\displaystyle s^2= x^2+ s^2/4\) so \(\displaystyle x^2= (3/4)s^2\) and \(\displaystyle x= \frac{\sqrt{3}}{2}s\).

Now we can write down all of the trig functions for both 30 and 60 degree. We have a right triangle with hypotenuse of length s, leg opposite the 30 degree angle of length s/2, and leg adjacent to the 30 degree angle of length \(\displaystyle \frac{\sqrt{3}}{2}s\). sin(30)= "opposite side over hypotenuse"= (s/2)/s= 1/2. cos(30)= "adjacent side over hypotenuse"\(\displaystyle = (\sqrt{3}/2)s/s= \frac{\sqrt{3}}{2}\). tan(30), cot(30), sec(30), and csc(30) can now be calculated from their definitions in terms of cosine and sine.

sin(60)= "opposite side over hypotenuse" but now the side opposite the 60 degree angle has length \(\displaystyle \frac{\sqrt{3}}{2}s\) so \(\displaystyle sin(60)= (\frac{\sqrt{3}}{2})s/s= \frac{\sqrt{3}}{2}\).

You can do the same thing for 45 degrees. An isosceles right triangle has both legs the same length, s, say, so both angles the same, 90/2= 45 degrees. The hypotenuse has length, by the Pythagorean theorem, has length \(\displaystyle \sqrt{2}s\) so sin(45) and cos(45) are both \(\displaystyle \frac{s}{\sqrt{2}s}= \frac{1}{\sqrt{2}}= \frac{\sqrt{2}}{2}\). From that, tan(45)= cot(45)= 1 and \(\displaystyle sec(45)= csc(45)= \sqrt{2}\).

 

[Steven G]

what if a problem just asks you to find the sin, cos, and tan of 30 degrees , 60 degrees and 45 degrees WITHOUT giving you a triangle.

how do i start finding the functions here?

btw is it even possible to find sin 45 degrees , ect, without a triangle?

You need to picture the triangle. For example, if a right triangle has a 45 degree angle then in fact it has two 45 degrees angles. Call the sides opposite the 45 degrees 1 and then Pythagoras would come to the rescue and tell you that the hypotenuse is sqrt(2). Armed with all sides and angles of the triangle you can compute all trig functions for 45 degrees.