The expression of x in terms of y for the case when cos(xº) = sin(yº) is x° = cos^(-1)sin(y°)
What are the six trigonometric ratios?
Trigonometric ratios for a right angled triangle are from the perspective of a particular non-right angle.
In a right angled triangle, two such angles are there which are not right angled(not of 90 degrees).
The slant side is called hypotenuse.
From the considered angle, the side opposite to it is called perpendicular, and the remaining side will be called base.
From that angle (suppose its measure is θ),
[tex]\sin(\theta) = \dfrac{\text{Length of perpendicular}}{\text{Length of Hypotenuse}}\\\cos(\theta) = \dfrac{\text{Length of Base }}{\text{Length of Hypotenuse}}\\\\\tan(\theta) = \dfrac{\text{Length of perpendicular}}{\text{Length of base}}\\\\\cot(\theta) = \dfrac{\text{Length of base}}{\text{Length of perpendicular}}\\\\\sec(\theta) = \dfrac{\text{Length of Hypotenuse}}{\text{Length of base}}\\\\\csc(\theta) = \dfrac{\text{Length of Hypotenuse}}{\text{Length of perpendicular}}\\[/tex]
For this case, we have to express x in terms of y for [tex]\cos(x^\circ) = \sin(y^\circ)[/tex]
Since 0 < x < 90, and 0 < x < 90 so taking inverse cosine gives:
[tex]x^\circ= \cos^{-1}(\sin(y^\circ))[/tex]
Thus, the expression of x in terms of y for the case when cos(xº) = sin(yº) is [tex]x^\circ= \cos^{-1}(\sin(y^\circ))[/tex]
Learn more about inverse trigonometric functions here:
https://brainly.com/question/12551115
