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In this video, weβll learn how to graph trigonometric functions such as sine, cosine, and tangent and deduce their properties.
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Weβll learn how to apply simple transformations to graph our functions in this form and recognize the relationship between each graph and the unit circle.
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Here is a unit circle.
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Itβs a circle with a radius of one unit, which weβve plotted with its center at the origin of a pair of π₯π¦-axes.
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Letβs add a point π, which can move around circumference of the circle.
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We could call its coordinates π₯, π¦.
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But what if we then add a right-angle triangle to our diagram and define an angle here.
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This is called the included angle, and weβre going to call it π.
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At the moment, we see that the length of the base of the triangle must be π₯ units, and its height must be π¦ units.
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We could, however, use trigonometry to find expressions for π₯ and π¦ in terms of π.
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We use the standard convention for labeling right-angled triangles.
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The side opposite the included angle is the opposite side.
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The longest side, thatβs the one that sits opposite the right angle, is the hypotenuse.
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And the other side, the side between the right angle and the included angle, is the adjacent.
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We recall the acronym SOHCAHTOA.
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And we see that we can use the cosine ratio to find a link between π₯ and π.
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This is cos of π is adjacent over hypotenuse.
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Well, the adjacent side in our triangle is π₯ units and the hypotenuse is one unit, so cos π is π₯ over one or simply cos π is equal to π₯.
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Similarly, weβll use the sine ratio to link π¦ and π.
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This time, the opposite is π¦ and the hypotenuse is one.
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So, we get sin π is π¦ divided by one, or simply sin π is equal to π¦.
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We can now say that the point π must have coordinates cos π, sin π.
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Now, as we move π around the circumference of the circle in an anticlockwise direction, the size of π₯ and π¦, and therefore the size of cos π and sin π, will change.
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In fact, they do something really interesting.
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Letβs see what that is.
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Letβs imagine our point lies on the positive π₯-axis here.
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At this point, π is zero.
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And the point has coordinates one, zero.
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When π is zero, then, we see that cos of π, which represents π₯, is equal to one.
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sin of π, which represents π¦, is equal to zero.
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Weβll now repeat this process here.
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This time, π is equal to 45 degrees.
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And we can add a right-angled triangle and we see the hypotenuse is equal to one unit.
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However, since this is a right-angled triangle with one angle of 45 degrees, we know in fact itβs an isosceles triangle.
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And so, we can say that its other two sides are equal to π units.
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The Pythagorean theorem tells us that the sum of the squares of the two shorter sides is equal to the square of the longer.
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That is, π squared plus π squared equals one squared or two π squared is equal to one.
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We divide through by two, and we find π squared is equal to one-half.
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And then we find the square root of both sides.
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Now, π is a length, so weβre going to say that itβs the positive square root of one-half, which we can write as root two over two.
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And so, we see when π is equal to 45 degrees, our point has coordinates root two over two, root two over two.
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This tells us that sin of π when π is 45 degrees is root two over two, as is cos π.
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And what about up here?
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Well, this time π is 90 degrees.
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And of course, our circle has a radius of one.
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So, this point has coordinates zero, one.
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And we see then when π is 90 degrees, cos of π, which is the π₯-value, is zero and sin of π, which is the π¦-value, is one.
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We can repeat this process for when π is equal to 180 degrees.
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At this point, we have a point with coordinates negative one, zero.
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So, when π is 180 degrees, cos of π is negative one and sin of π is zero.
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Then, when π is 270 degrees, our point has coordinates zero, negative one.
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So, sin of π is negative one, remember thatβs the π¦-coordinate, and cos of π is zero.
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And then, we continue around the circle and we see we get back to the start.
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So, when π is equal to 360 degrees, we have the same values for sin π and cos π as we did when π was equal to zero.
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sin π is zero, and cos π is one.
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Letβs fill in these gaps.
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At this point, π is 135 degrees.
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If we add in a right-angled triangle and use the fact that angles on a straight line sum to 180 degrees, we see we have a replica of the earlier triangle we looked at.
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It has a hypotenuse of one unit and an included angle of 45 degrees.
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This means the lengths of its other two sides must be root two over two units.
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In coordinate form then, this point must have coordinates negative root two over two, root two over two.
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And so, when π is 135 degrees, cos of π, which is our π₯-coordinate, is negative root two over two, and sin π is root two over two.
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When π is 225 degrees, we have a similar situation.
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We have another isosceles triangle with a hypotenuse of one unit and two other sides of root two over two units.
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This means when π is 225 degrees, our point must have coordinates negative root two over two, negative root two over two, meaning sin of π and cos of π are both negative root two over two.
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Finally, when π is 315 degrees, we have another isosceles triangle.
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This time, the coordinates of our point are root two over two and negative root two over two, meaning that when π is 315 degrees, sin π is negative root two over two and cos π is root two over two.
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Now, it should be quite clear that if we were to continue moving around this circle in an anticlockwise direction, we would meet all of these points once again.
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So, when π is 405 degrees, sin of π and cos of π are the same when π is equal to 45 degrees and so on and this leads us to a definition.
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We say that sin of π and cos of π are periodic functions; that is, they repeat.
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They do so every 360 degrees.
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So, we say their period is 360 degrees.
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Letβs sketch the graphs.
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Weβve seen that both graphs oscillate; that is, they move between the values of one and negative one.
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Those are their maximums and minimums.
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So, weβll start with the graph of sin of π or, in this case, π¦ is equal to sin of π₯.
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It passes through the origin; that is, when π is zero, sin of π is zero.
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So, when π₯ is zero, sin of π₯ is zero.
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Then, when π₯ is 45, we get sin of π₯ to be root two over two.
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Thatβs roughly here.
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When π₯ is 90, sin of π₯ is one.
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Then, when π₯ is a 135, sin of π₯ is root two over two.
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And when π₯ is 180, sin of π₯ is zero.
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We could continue in this manner.
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Letβs join these up with a smooth curve.
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And when we do, we find the graph of π¦ equals sin of π₯ between the values of zero and 360 looks a little bit like this.
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Of course, we said that these graphs are periodic.
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They repeat, so we could continue in this manner, repeating this exact graph every 360 degrees.
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Similarly, the graph of π¦ equals cos π₯ in the interval from zero to 360 looks like this.
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Since it is also periodic, we could continue either side in the same manner.
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And so, to recap a few features of our graphs, theyβre periodic; they have a period of 360 degrees.
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They both have maxima at one and minima at negative one.
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Both of them, in fact, have a little bit of symmetry.
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The graph of π¦ equals cos of π₯ has reflectional symmetry about the line π₯ equals 180 or the line π₯ equals zero.
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The graph of π¦ equals sin of π₯ has rotational symmetry about the origin.
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But if we narrow it down and just look at, say, a part of the graph, that is, the interval from zero to 180, we see that π¦ equals sin of π₯ does have a little bit of reflectional symmetry about the line π₯ equals 90 degrees.
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These features can help us to solve problems involving cos and sin of π₯.
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The graph of π¦ equals tan of π₯ is a little bit stranger than this.
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And rather than using the unit circle method, weβre going to plot a table of values using our calculator.
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For values of π of zero, 45, 90, 135, and so on, we get the following table of values.
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Notice we have an error at π equals 90 degrees and π equals 270 degrees.
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This will continue every 180 degrees.
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But whatβs actually happening here?
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Well, we know tan π is opposite over adjacent, but we canβt actually draw a right-angled triangle at π equals 90 degrees.
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And so, we say that as π approaches 90 degrees, tan of π approaches β.
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We canβt evaluate it, and so we represent this fact using vertical asymptotes on our graph.
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The graph of π¦ equals tan π gets closer and closer to these asymptotes but will never quite touch them.
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And so, the graph of π¦ equals tan of π₯ looks a little something like this.
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Once again, we see that the function tan of π is periodic.
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But this time, its period is 180 degrees.
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It has rotational symmetry about the origin.
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And we canβt define its maximums or minimums because we saw that as π approaches 90 and then multiples of 180 degrees, tan of π approaches β.
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Itβs really important that we can identify and sketch the graphs of π¦ equals sin of π₯, π¦ equals cos of π₯, and π¦ equals tan of π₯.
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And we also need to be able to transform them such that for the graph of π¦ equals π of π₯, π¦ equals π of π₯ plus π is a translation by the vector negative π, zero, whereas π¦ equals π of π₯ plus π is a translation by the vector zero, π.
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π¦ equals π of π times π₯ is a horizontal stretch with a scale factor one over π.
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And π¦ equals π times π of π₯ is a vertical stretch by a scale factor of π.
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Then, π¦ equals negative π of π₯ is a reflection in the π₯-axis, and π¦ equals π of negative π₯ is a reflection in the π¦-axis.
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Finally, sometimes we measure angles in radians such that two π radians is equal to 360 degrees and π radians is 180 and so on.
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Now, donβt worry if youβve not encountered these yet, theyβre just another way of representing an angle.
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So, letβs have a look at some questions on trigonometric graphs.
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Assign each plot shown in the graph below to the function it represents.
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Here, we see we have two very similar-looking graphs.
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These graphs are clearly periodic.
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They appear to repeat every two π radians.
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Remember, thatβs repeating every 360 degrees.
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We see they have maximums at one and minimums at negative one, respectively.
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In fact, we know that we can describe the graphs of the sine and cosine functions in the same way.
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The main difference is where these graphs intersect the π¦-axis.
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π¦ equals sin of π₯ passes through the π¦-axis at zero, whereas π¦ equals cos of π₯ passes through at one.
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And we said, of course, that both have a period of 360 degrees or two π radians, maxima at one, and minima at negative one.
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This means that the red plot which intersects the π¦-axis at one must be the cosine curve, whereas the blue plot must be the sine curve.
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Now, this actually shows a really interesting feature of these functions.
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Letβs define π of π₯ to be equal to sin of π₯.
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Then, we see that the function π of π₯ plus π over two or π of π₯ plus 90, which represents a translation by the vector negative 90, zero, maps the sine curve onto the cosine curve.
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And of course, the converse is also true.
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So, weβve seen that thereβs a relationship between the sine and cosine graphs by a horizontal translation.
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Now, letβs look at a stretch.
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Find the maximum value of the function π of π is equal to 11 sin π.
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Firstly, weβre going to recall what the graph of π of π equals sin of π looks like.
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It has maxima and minima at one and negative one, respectively.
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We know that it passes through the origin and that itβs periodic and it has a period that repeats every 360 degrees.
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So, π of π equals sin of π has a graph that looks a little something like this.
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But of course, we were actually interested in the graph of the function π of π is 11 sin π.
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And so, we recall that for a function π¦ equals π of π₯, π¦ equals π times π of π₯ represents a vertical stretch by a scale factor of π.
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In this case, we can see our scale factor is 11.
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And so, π of π equals 11 sin π looks something like this.
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It still intersects the π₯- and π¦-axes at the same places, but now it travels as high as 11 and as low as negative 11.
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And so, the maximum value of the function π of π equals 11 sin π is 11.
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Weβll now have a look at a reflection.
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Which of the following is the graph of π¦ equals negative tan of π₯?
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Letβs just begin by recalling what the graph of π¦ equals tan of π₯ looks like.
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Itβs periodic, and it repeats every 180 degrees.
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It passes through the origin, the point zero, zero.
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It has vertical asymptote at π₯ equals 90 degrees but also every 180 degrees either side of this, in other words, π₯ equals negative 90, π₯ equals 270, and so on.
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In fact, the graph of π¦ equals tan of π₯ is this one.
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Itβs (A).
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Notice that the graph approaches the asymptotes but never actually quite touches them.
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But of course, we were interested in the graph of π¦ equals negative tan of π₯.
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So, we recall that for a function π¦ equals π of π₯, π¦ equals negative π of π₯ is a reflection in the π₯-axis.
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And we see that the only graph that matches this is (D).
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(D) is the graph of π¦ equals negative tan of π₯.
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In this video, weβve learned what the graphs of π¦ equals sin of π₯, π¦ equals cos of π₯, and π¦ equals tan of π₯ looks like.
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We saw that the graphs of π¦ equals sin of π₯ and π¦ equals cos of π₯ are periodic; they have a period of 360 degrees, whereas the graph of π¦ equals tan of π₯ repeats every 180 degrees.
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Finally, we saw that π¦ equals sin of π₯ and cos of π₯ have maxima and minima at one and negative one, respectively, whereas the graph of π¦ equals tan of π₯ has asymptotes at π₯ equals 90 and then multiples of 180 degrees.