Math help: trigonometric functions of a given angle?

For Exact values, consider that 150 = 180 - 30 and is in the second quadrant. This will affect the signs (+ or -) You also need to know the ratios for a 30-60-90 right triangle. You will use it for several years, expecially if you go on to calculus or take physics.

sin 150 = sin 30 = 1/2.

csc 150 = 1/sin 150 = 1/(1/2) = 2

cos 150 = –cos 30 = – √3/2.

sec 150 = 1/cos 150 = 1/(– √3/2) = – 2/ √3 = – 2√3/3

tan 150 = –tan 30 = –1/√3 = –√3/3.

cot 150 = 1/tan 150 = 1/(–1/√3) = –√3

Well, 150 degrees is a common angle that you'll be seeing. So memorize all the values of cosine and sine for every 30 degrees. (you really only need to remember one quadrant and just change the signs.) Once you know all the cosine and sine values the rest are easy. tan (x) = sin(x)/cos(x) etc.

But if you are supposed to find them without already knowing the values of cosine and sine at 150 degrees... then you have to use some geometry.

On a unit circle the radius will always be one.

At 150 degrees, you can make a right triangle. The right triangle will be composed of a 30 degree angle, a 60 degree angle and a 90 degree angle. This is a special type of triangle. the side opposite 60 degree angle will be sqrt(3)/2 times the hypotenuse (which is one in this case) so the bottom of the triangle will be sqrt(3)/2

then the cosine would be CAH (cosine = adjacent side over hypotenuse) which would be sqrt(3)/2 divided by 1. So you have to use these special properties of 30-60-90 triangles.

(btw the side opposite the 30 degree angle is 1/2 times the hypotenuse.

I know its hard to understand it... It would be much easier if I showed you a picture... mmm

Edit: By the way, the cosine at 150 degrees is actually -sqrt(3)/2 because its in the negative x direction. the sine will still be positive because the vertical side of the triangle will still be on the positive part of the y-axis.

well okay, this is going to take some explaining: first i hope you know the Cartesian plane well enough and the unit circle. along with knowing where 90, 180, 270, and 360 degrees spots are (id show you but i am in the internet so i can only try to explain). well lets talk quadrants first, all four of em, you know that the first one is the upper right, then second is upper left, then 3rd is lower left and the 4th is the lower left. well now you are going to learn an acronym, A.S.T.C, it stand for "all students take calculus" and you use this to help know which whether to use positive or negative for the given trig function. for example, lets say you wants to find the sine of 150 degrees (one part of your problem), what you do is picture or draw which quadrant 150 will land on. it turns out it lands on the second one, so lets look at the second letter of our favorite acronym. the second letter is "S". this "S" means that the only thing positive in this quadrant is sine. so since we are using the sine function we know our answer will be positive. now we need to know how to find the sine of 150 degrees. you have choices, either use a chart, a unit circle, or just do what i do, try to get it to two digits instead of three. like this : 180 - 150 = 30. yes just subtract 180 from it to make it 30. now you must memorize that the sine of 30 degrees is 1/2. just got to. you need to memorize at least 30, 45, and 60 degrees for the sin, cos, and tan. to find their reciprocals, just flip it. from their corresponding functions (ex. sin 30 is 1/2 so csc 30 is 2).

so lemme do one more before i lose you some more: the cos of 150

180 - 150 = 30

remember 150 lands on the second quadrant and the only thing positive there is the sine function, that means everything else is a negative answer, just like ours (since its cos)

cos 30 is sqrt(3) / 2

but remember, it is negative

-sqrt(3) / 2

try the other ones your self

hope this kind of helps

We use the fact that 150° = 90° + 60° along with the angle addition formulas. (Or use a reference angle of 30 degrees)

sin(A + B) = sinA cosB + sinB cos A = sin90°cos60° + sin60°cos90° = cos60° = ½

150° is in Q2, so cos150° is negative. cos150° = √[1² − (½)²] = −√(3)/2

sin150° = 1/2

cos150° = −√(3)/2

csc150° = 1/sin150° = 2

sec150° = 1/cos150° = −2√(3)/3

tan150° = sin150°/cos150° = (1/2)/(−√(3)/2) = −√(3)/3

cot150° = (−√(3)/2)/(1/2) = −√(3)

I'm guessing this is for a triangle. Do you have at least one side?

Do this:

O= opposite Sode

H= hypotenuse

A= Adjacent Side

Cos 150 = A/H

Sin 150 = O/H

Tan 150= O/A

Csc 150= flip the sin one

sec 150= flip the cosine one

cot 150= flip the tan one

hope it helped

Note that 150° = 5π/6, whose sine and cosine values are 1/2 and -√3/2 respectively. The other function values can be computed using their formulas (tan = sin/cos, etc.).

Remember that sine is positive in quadrants I and II and cosine is positive in quadrants I and IV.

WolframAlpha can be a big help when you can't remember specific values.

sin(150) = sin(30) = ½

cos(150) = -cos(30) = -√3/2

tan(150) = sin(150)/cos(150) = -1/√3

csc(150) = 1/sin(150) = 2

sec(150) = 1/cos(150) = -2/√3

cot(150) = 1/tan(150) = -√3