# How to convince the students of grade 8 that $\sin 90^\circ =1$? ( calculator not allowed )

In general the secondary student may not ask why is $\sin 90^\circ = 1$ because they can see the answer in the graph of the sine wave. However the students in grade 8 are not familiar with the graph of sine, so how the teacher could find a simple way to convince the student?? All what they have is the sine definition as a ratio between the lenght of the opposite side to the angle and the length of the side opposite to the right angle.

• I've never understood why educators insist on starting everyone off with the restrictive, unnatural, damaging triangle definition. All it does is sit around instilling bad mental habits and making things like polar coordinates much harder to understand. Is a unit circle really so advanced that middle schoolers wouldn't be able to handle it? Nov 5, 2014 at 23:06
• Agree with @JackM. If you teach the unit circle, it's obvious that $90^\circ$ corresponds to the point $(0,1)$. We are not training people to survey fields anymore. The unit circle is the modern viewpoint. And then you don't have to "memorize" the values in the different quadrants, you just read them off the circle. I wish this point would receive some pedagogical attention. The modern importance of sin and cos is their functional relation to the circle and (eventually) to complex numbers and the exponential function. Nobody's ever going to have to solve a triangle or know what a cosecant is. Nov 15, 2014 at 4:17
• @JackM, why do you feel it's unnatural and damaging to start with the triangle definition? I think that definition is simpler, and in Precalc, I do two units on trig, the first with triangles. Is there any research that this is damaging? Nov 27, 2015 at 18:10
• @SueVanHattum The biggest, but not the only reason, is that you're quickly expected to deal with things like $\sin(110°)$ or $\sin(-33°)$, which makes no sense with the triangle definition. Nov 27, 2015 at 18:23

Maybe draw this picture? Make it clear that the green hypothenuse is fixed in length, but the red altitude is growing and approaching that hypothenuse in length as the angle approaches $90^\circ$. • This kind of plan can be very helpful in later courses as well, for determining something like $\lim_{x \to \infty} \mathrm{arctan }(x)$ Nov 4, 2014 at 18:02
• I've posted additional diagrams here along with a narrative that might help build intuition with the unit circle.
– WBT
Jul 15, 2016 at 22:08

By their definition (using a right triangle), $\sin 90^\circ$ is undefined, since a triangle cannot have two right angles. You need the definition based on a circle ($\sin \theta = y/r$) to have a value for $\sin 90^\circ$.

• Exactly. Joseph O'Rouke's picture above shows why this is a reasonable extension of their definition. But it is an extension! Nov 4, 2014 at 18:23
• A triangle where one side has length zero is doubly-right, isosceles, and illustrates the desired point quite well. Of course, mileage will vary with respect to kids' acceptance of degenerate shapes as mental models. Nov 4, 2014 at 19:07

I would simply define $\sin(x)$ from the unit circle. I would use a right-angled triangle exactly because (as others have point out already) this definition doesn't work for $90^\circ$. Note that is also seem to work well for negative values.

The sine is the ratio of the "opposite" axis (relative to theta, the angle in question) compared to the hypotenuse. The hypotenuse is always longer because it is opposite the right angle.

But as theta approaches 90 degrees, the length of the opposite axis approaches the hypotenuse as limit (in fact, they both approach infinity).

Conversely, as theta approaches 90 degrees, the third angle approaches zero (90 degrees minus theta), and the axis opposite this angle shrinks as theta angle does. T value of the axis approaches zero as the "non-theta" angle approaches zero.

Can't we think to find $\sin 90^\circ$ directly as the ratio between the side opposite to the angle and the side opposite to the right angle ( the basic defenition ) ?? in this privet case the side opposite to the right angle is the same side ! so $\sin 90^\circ = 1$. Notice the diagram. • Using this model, can you explain $cos (90)$? Nov 6, 2014 at 14:47
• In fact yes but it needs for a bit imagination, I mean, acoording to basic definition for the cosine as the ratio between the side ajacent to the angle and the side opposite to the right angle, which in this case equal to zero:AC = zero ! Further more we can conculde that $\tan 90 ^ \circ$ = AC: zero which infinity ! Nov 6, 2014 at 15:06
• I'm still lost - how is it that the side adjacent to B above has length zero? (And further to the point, aren't there two sides adjacent to B? How would we pick which one to use?) Nov 6, 2014 at 15:22
• THe ajacent side is the side that connects the angle we are seaching for its cosine and the right angle, but here in the special case what will be the length of the side that connects the right angle with itself ??? The limit is zero ! Nov 6, 2014 at 15:31