# My theory is that: as cos 143 should initially be negative as it belongs to the 2nd quadrant, then if in the question it appears as positive, we should at the end change the sign of our answer, is this right? Even though it is simple, I got rather confused at the end. When dealing with extended igcse trigonometry, we face a question which could say: express cos 143 in terms of the cosine of another angle. If for example we were to find the equivalent as a cosine acute angle, we would first: 180-143= 37, so cos 36 should equal cos 143 as cos 36 belongs to the first quadrant of angles in a circle so it should be positve. However, in the calclator we are shown: -cos36=cos143. You should reduce the cosine of `143^o`  to the cosine of the equivalent angle, located in the quadrant 1 such that:

`cos 143^o = cos (180^o - 143^o) `

You need to expand the right side using the following formula:

`cos(a - b) = cos a*cos b + sin a*sin b`

Reasoning by analogy yields:

`cos (180^o - 143^o) = cos 180^o* cos 143^o - sin 143^o* sin 180^o`

Since `cos 180^o = -1`  and `sin 180^o = 0`  yields:

`cos (180^o - 143^o) = - cos 143^o`

Substituting `37^o`  for `180^o - 143^` o yields:

`cos 37^o = - cos 143^o`

Hence, reducing to the first quadrant the argument 143^o yields  `cos 143^o = -cos 37^o` .

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