I understand why it is possible for trigonometric ratios to have many angles. Despite this I am unsure of how to calculate the multiple values of angles when you have a trigonometric ratios. What do you have to do to calculate it?

Original post by VoiidDev

I understand why it is possible for trigonometric ratios to have many angles. Despite this I am unsure of how to calculate the multiple values of angles when you have a trigonometric ratios. What do you have to do to calculate it?

The two main ways (apart from just remembering formulae) are to understand it in terms of a unit circle with point (x,y)=(cos(theta),sin(theta)) or the trig graphs

https://www.bbc.co.uk/bitesize/guides/zsgjxfr/revision/6

https://www.mathcentre.ac.uk/resources/uploaded/mc-ty-trigeqn-2009-1.pdf

So for the circle, if you have cos (trig ratio), then this reprsents a point on the xaxis and the two solutions correspond to where a vertical line (passing through that point on the xaxis) intersect with the circle. The two values of theta correspond to unit right triangles defined by the intersection points and the origin. So the solutions will be symmetric (reflections) about the xaxis (0 and 180 degrees). So if theta is the principle solution say (above the xaxis so between 0 and 180), then the other solution will be its reflection in the xaxis or 360-theta.

Similarly for sin (trig rato) being a point on the y axis and the two values of theta correspond to where a horizontal line intersects with the circle, so symmetric (reflections) about the yaxis (90 and 270 degrees). So if theta is the principle solution to the right of the yaxis , the other will be its reflection in the yaxis so either180-theta or 540-theta, depending on whether the principle is above (0 to 90) or below (270 to 360) the xaxis.

This is equivalent to thinking about the trig graphs (previous links) and cos is symmetric about 0 and 180 (for instance). Similarly sin is symmetric about 90 and 270. Drawing a horizontal line from the y axis (trig ratio/value) and noting where it intersects with the trig curve gives the mutiple solutions as illustrated in both links. You can remember the previous formulae, though its as easy/easier to derive the appropriate one using the circle or trig graph sketch.

(edited 4 months ago)

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