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Question: Evaluate \[\sin {57^0} - \cos {33^0}\]...

Evaluate sin⁡570−cos⁡330\sin {57^0} - \cos {33^0}

Explanation

Solution

Hint : Firstly, it is a simple problem and we use the angle conversion in this method. We convert it into sin⁡(900−θ)\sin ({90^0} - \theta ) and simplify, we get the required answer. Further we note that sin⁡(900−θ)=cos⁡θ\sin ({90^0} - \theta ) = \cos \theta , it lies in the first quadrant. So that it takes positive values. If not, we use the trigonometric ratios to solve these questions that is by expressing trigonometric functions in terms of their complements. We can also express the trigonometric functions in terms of their supplements.

Complete step-by-step answer :
There are six trigonometric ratios, namely
Sine, cosine, tangent, cosecant, secant and cotangent.
Also remember, by convection, positive angles are measured in the anti-clockwise direction starting from the positive x axis. Domain ranges between -1 and 1 only. We use right angle triangles while defining the trigonometric relations. That is,θ=900\theta = {90^0}
Given,sin⁡570−cos⁡330\sin {57^0} - \cos {33^0}
It can be written as,
=sin⁡(900−330)−cos⁡330= \sin ({90^0} - {33^0}) - \cos {33^0}
Using, sin⁡(900−θ)=cos⁡θ\sin ({90^0} - \theta ) = \cos \theta , and substituting we get,
=cos⁡330−cos⁡330= \cos {33^0} - \cos {33^0}
=0= 0
Thus we get, sin⁡570−cos⁡330=0\sin {57^0} - \cos {33^0} = 0
We can also convert cos⁡θ\cos \theta into sin⁡θ\sin \theta by using trigonometric ratios and solving the same as we did above, we will get the same answer.
So, the correct answer is “0”.

Note : The given problem is simple and straightforward. Simply using the trigonometric ratios and substituting. Convert cos⁡330\cos {33^0} and see the solution. you will get the same answer. That is we are going to convert cos⁡330\cos {33^0}ascos⁡330=cos⁡(900−570)=sin⁡570\cos {33^0} = \cos ({90^0} - {57^0}) = \sin {57^0}. Either way it’s going to cancel out again.Hence the solution is zero.