Trigonometric ratios of angles above 90 degrees

[rudransh verma]

Homework Statement

$\sin 235$

Relevant Equations

$\sin 135$

I have been doing the resolutions of vectors on x and y-axis with making triangles and reference angles in all quadrants. But I want to calculate now how to find something like $\sin 235$ without the help of reference angles. I know we don’t need to. Calculator and Taylor theorem is handy here but I want to calculate using unit circles.
Thanks for the help.

 

[gleem]

If you are doing vector resolution you should have studied trigonometry. The trigonometric functions have what are called identities (of which there are many) which show how these functions are related to one another. For sine then the sin(x+y) = sin(x)cos(y) +cos(x)sin(y). 135° = 90° + 45° and 235° = 180° +45°. Of course you know sin(90°) and you need to know the sin(180°) to which I refer you to this https://byjus.com/maths/trigonometric-identities/ for more relations and identities.

You mentioned Taylors Theorem, so you know Calculus? If so, you had a precalculus course? If not and you are studying even non calculus physics, it is mandatory to know trigonometry.

 

[Steve4Physics]

But I want to calculate now how to find something like $\sin 235$ without the help of reference angles. ... I want to calculate using unit circles

The unit circle is used in conjunction with reference angles. You can't have your cake and eat it!

If you don’t want to use reference angles, you can use identities:
sin(A) = -sin(-A)
sin(A) = -sin(A – 180º) (assuming A is in degrees)
sin(A) = sin(A – 360º) (assuming A is in degrees)

E.g. Using the above identities gives:
sin(235º) = -sin(235º – 180º) = -sin(55º) (then a calculator is needed)
sin(135º) = -sin(135º – 180º) = -sin(-45º) = sin(45º) (standard value)
sin(300º) = sin(300º – 360º) = sin(-60º) = -sin(60º) (standard value)

See if you find the equivalent identities for cos and tan.

Also, note that values of angles given with no units are usually taken to be in radians
"sin 235" would generally be taken to mean sin (235 radians). If the angle is meant to be in degrees, write "sin(235º)".

 

[FactChecker]

Calculator and Taylor theorem is handy here but I want to calculate using unit circles.

What form of "calculation" are you talking about? Many (most?) calls to the trig functions in the software will allow large angles already and (I assume) the standard trig functions reduce large angles to the reference angles as the first step. So you don't have to worry about it if you are calling standard utility trig functions.

 

[robphy]

Is this what you are after?

If \theta is the counterclockwise angle from the positive-x axis,
then a vector \vec V has components
V_x=\left|\vec V\right|\cos\theta
V_y=\left|\vec V\right|\sin\theta,
without having to ask which quadrant is the tip of \vec V is in,
and then building a right-triangle in that quadrant to work with acute angles.

 

[rudransh verma]

The unit circle is used in conjunction with reference angles. You can't have your cake and eat it!

If you don’t want to use reference angles, you can use identities:
sin(A) = -sin(-A)
sin(A) = -sin(A – 180º) (assuming A is in degrees)
sin(A) = sin(A – 360º) (assuming A is in degrees)

E.g. Using the above identities gives:
sin(235º) = -sin(235º – 180º) = -sin(55º) (then a calculator is needed)
sin(135º) = -sin(135º – 180º) = -sin(-45º) = sin(45º) (standard value)
sin(300º) = sin(300º – 360º) = sin(-60º) = -sin(60º) (standard value)

Yaa you are right . I do need a calculator eventually in many cases.
Also I didn’t knew what I was doing when finding components using reference angles. So I was using unit circles too in that. Right! By putting + and - signs to which quadrant they are in. Thanks