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My daughter (in 5th grade) is interested in learning advanced maths, so we looked at sine and cosine. Question: is there an easy, geometrical way to "show" tha
by isolli 3y ago
My daughter (in 5th grade) is interested in learning advanced maths, so we looked at sine and cosine.
Question: is there an easy, geometrical way to "show" that `cos(60º)=1/2` using the intuitive definition based on the trigonometric circle?
- orlandpm 3y agoYes! Draw an equilateral triangle with side length one. Then cut it in half and reason from there.
- sublinear 3y agoPretty sure we used to have posters of this and similar figures in middle school classrooms.
- gjm11 3y agoYes. (I assume that by "the intuitive definition based on the trigonometric circle" you mean that (cos t, sin t) is the point at angle t on the unit circle. If you meant something else, then what follows may not be helpful.) APPROACH 1 If you take the point (1,0) and rotate it anticlockwise through 60º about the origin, you get (cos 60º, sin 60º), by definition. Look at the triangle formed by (0,0), (1,0), and (cos 60º, sin 60º). The angle at the (0,0) vertex is 60º. The other two angles are equal because the triangle is isosceles (rotating a line segment doesn't change its length) so they must also be 60º since the angles in a triangle add up to 180º. So now drop a perpendicular from (cos 60º, sin 60º) down to the horizontal side of the triangle; it lands at (cos 60º, 0), so cos 60º is the distance from (0,0) to that point, and 1-cos 60º is the distance from (1,0) to that point. But those distances are the same, by symmetry, and now we're done. APPROACH 2 Draw the unit circle and inscribe a regular hexagon in it with (1,0) as one vertex. Divide it up into six triangles by drawing radii. As above, these are equilateral triangles. So now look at the horizontal separation between the opposite points (-1,0) and (1,0). Obviously the distance is 2 units. But we can split it up a different way: we have a horizontal distance 1-cos 60º from either of those to the next point around the circle, then 1 from that to the next point, then 1-cos 60º back to the opposite point. So 2 = (1-cos 60º)+1+(1-cos 60º) and again we're done. (But I think any proof of this particular thing is going to be basically about equilateral triangles more than it's about circles. It might be easier first to show that for angles between 0 and 90º the trig functions have their "traditional" definition in terms of right-angled triangles, and then bisect an equilateral triangle.)
- adrian_b 3y agoIf you look at a regular hexagon inscribed in a circle (i.e. whose vertices are determined by 6 rotations of 60 degrees whose end point returns to the starting position), the length of one edge is equal to the radius (this could be demonstrated based on symmetry, because in the triangles formed by 1 edge and 2 radii all angles are equal, therefore also all edges). The cosine of 60 degrees is one half of the edge divided by the radius, i.e. divided by the edge, therefore it is 1/2.
- pvg 3y agoAs a side comment to sibling comments with various approaches, there's a handy bit of intuition/technique to this that's useful for 5th to 105th graders - when you notice 'useful' angles in a plane geometry problem, make the thing they suggest out of them. So, 60 deg angle is a piece of an equilateral triangle - finish the whole triangle and see if you get any ideas. The unit circle sin/cos thing teaches you all sorts of stuff about right-angled triangles. See a problem without a right angled triangle? Draw a line to add one, etc.