
Evaluate \[\sin {57^0} - \cos {33^0}\]
Answer
560.7k+ views
Hint: Firstly, it is a simple problem and we use the angle conversion in this method. We convert it into \[\sin ({90^0} - \theta )\] and simplify, we get the required answer. Further we note that \[\sin ({90^0} - \theta ) = \cos \theta \], 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,\[\theta = {90^0}\]
Given,\[\sin {57^0} - \cos {33^0}\]
It can be written as,
\[ = \sin ({90^0} - {33^0}) - \cos {33^0}\]
Using, \[\sin ({90^0} - \theta ) = \cos \theta \], and substituting we get,
\[ = \cos {33^0} - \cos {33^0}\]
\[ = 0\]
Thus we get, \[\sin {57^0} - \cos {33^0} = 0\]
We can also convert \[\cos \theta \] into \[\sin \theta \] 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 \[\cos {33^0}\] and see the solution. you will get the same answer. That is we are going to convert \[\cos {33^0}\]as\[\cos {33^0} = \cos ({90^0} - {57^0}) = \sin {57^0}\]. Either way it’s going to cancel out again.Hence the solution is zero.
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,\[\theta = {90^0}\]
Given,\[\sin {57^0} - \cos {33^0}\]
It can be written as,
\[ = \sin ({90^0} - {33^0}) - \cos {33^0}\]
Using, \[\sin ({90^0} - \theta ) = \cos \theta \], and substituting we get,
\[ = \cos {33^0} - \cos {33^0}\]
\[ = 0\]
Thus we get, \[\sin {57^0} - \cos {33^0} = 0\]
We can also convert \[\cos \theta \] into \[\sin \theta \] 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 \[\cos {33^0}\] and see the solution. you will get the same answer. That is we are going to convert \[\cos {33^0}\]as\[\cos {33^0} = \cos ({90^0} - {57^0}) = \sin {57^0}\]. Either way it’s going to cancel out again.Hence the solution is zero.
Recently Updated Pages
Why are manures considered better than fertilizers class 11 biology CBSE

Find the coordinates of the midpoint of the line segment class 11 maths CBSE

Distinguish between static friction limiting friction class 11 physics CBSE

The Chairman of the constituent Assembly was A Jawaharlal class 11 social science CBSE

The first National Commission on Labour NCL submitted class 11 social science CBSE

Number of all subshell of n + l 7 is A 4 B 5 C 6 D class 11 chemistry CBSE

Trending doubts
Differentiate between an exothermic and an endothermic class 11 chemistry CBSE

10 examples of friction in our daily life

One Metric ton is equal to kg A 10000 B 1000 C 100 class 11 physics CBSE

Difference Between Prokaryotic Cells and Eukaryotic Cells

1 Quintal is equal to a 110 kg b 10 kg c 100kg d 1000 class 11 physics CBSE

State the laws of reflection of light

