06Dsymmetry

= Symmetry =


 * We have previously established that the trig functions are positive in different quadrants depending on this table:




 * Recall the unit circle:


 * 2nd Quadrant **
 * Because a circle is symmetrical, we can draw a __ **congruent** __ triangle in the second quadrant


 * Remember that in trigonometry, we measure angles anticlockwise from the positive x-axis
 * So the point B can be described as at an angle of 180 – q from the positive x-axis


 * From this diagram, we can see that
 * sin(180º – q ) = sin( q )
 * sine is the y coordinate of the point on the unit circle
 * recall that in quadrant 2, sine is positive
 * cos(180º – q ) = –cos( q )
 * cosine is the x coordinate of the point on the unit circle
 * recall that in quadrant 2, cosine is negative


 * And because tan = sin / cos
 * tan(180º – q ) = –tan( q )
 * in quadrant 2, tangent is negative


 * for example (correct to 3 decimal places)
 * sin(130º) = sin(180º – 50º) = +sin(50º) = +0.766
 * cos(165º) = cos(180º – 15º) = –cos(15º) = –0.966


 * 3rd Quadrant **
 * In the same way, we can draw a __**congruent**__ triangle in the third quadrant




 * From this diagram, we can see that
 * sin(180º + q ) = –sin( q )
 * recall that in quadrant 3, sine is negative
 * cos(180º + q ) = –cos( q )
 * recall that in quadrant 3, cosine is negative


 * And because tan = sin / cos
 * tan(180º + q ) = tan( q )
 * in quadrant 3, tangent is positive


 * 4th Quadrant **
 * We could do the same in the fourth quadrant, to get
 * sin(360º – q ) = –sin( q )
 * recall that in quadrant 4, sine is negative
 * cos(360º – q ) = cos( <span style="font-family: Symbol,sans-serif; font-size: 17px;">q )
 * recall that in quadrant 4, cosine is positive


 * And because tan = sin / cos
 * tan(360º – <span style="font-family: Symbol,sans-serif; font-size: 17px;">q ) = tan( <span style="font-family: Symbol,sans-serif; font-size: 17px;">q )
 * in quadrant 4, tangent is negative


 * Summary **


 * You could use the list of rules from the above sections
 * But it is worth noting they are essentially all the same
 * As long as the angle is written in terms of 180 ± q, or 360 ± q
 * this works for any multiple of 180, (ie 180n where n is an integer)
 * ** Notice that this is the angle from the nearest part of the x-axis **
 * Then
 * sin( 180n ± <span style="font-family: Symbol,sans-serif; font-size: 120%; line-height: 1.5;">q ) = | sin( <span style="font-family: Symbol,sans-serif; font-size: 120%; line-height: 1.5;">q ) |
 * cos(180n ± <span style="font-family: Symbol,sans-serif; font-size: 16px;">q ) = | cos( <span style="font-family: Symbol,sans-serif; font-size: 16px;">q ) |
 * tan(180n ± <span style="font-family: Symbol,sans-serif; font-size: 16px;">q ) = | tan( <span style="font-family: Symbol,sans-serif; font-size: 16px;">q ) |
 * all you have to consider is the sign, which depends on the quadrant


 * For example (correct to 3 decimal places)
 * sin(260º)=sin(180º + 80º) = –sin(80º) = –0.985
 * cos(300º) = cos(360º – 60º) = +cos(60º) = +0.5
 * tan(–160º) = tan(–180º + 20º) = +tan(20º) = +0.364


 * Radians **


 * Use the same rules when operating in radians
 * Make sure you express the angle as the difference from a multiple of <span style="font-family: Symbol,sans-serif; font-size: 120%;">p
 * ** Remember this is the angle from the nearest part of the x-axis **
 * sin( <span style="font-family: Symbol,sans-serif; font-size: 120%; line-height: 1.5;">p n ± <span style="font-family: Symbol,sans-serif; font-size: 16px;">q ) = | sin( <span style="font-family: Symbol,sans-serif; font-size: 16px;">q ) |
 * cos( <span style="font-family: Symbol,sans-serif; font-size: 16px;">p n ± <span style="font-family: Symbol,sans-serif; font-size: 16px;">q ) = | cos( <span style="font-family: Symbol,sans-serif; font-size: 16px;">q ) |
 * tan( <span style="font-family: Symbol,sans-serif; font-size: 16px;">p n ± <span style="font-family: Symbol,sans-serif; font-size: 16px;">q ) = | tan( <span style="font-family: Symbol,sans-serif; font-size: 16px;">q ) |

math . \qquad \qquad \sin \Big( \dfrac{2\pi}{3} \Big) =\sin \Big( \pi - \dfrac{\pi}{3} \Big) = +\sin\Big(\dfrac{\pi}{3} \Big) = \dfrac{\sqrt{3}}{2} \\.\\ math math . \qquad \qquad \cos \Big( \dfrac{7\pi}{4} \Big) = \cos \Big( 2\pi - \dfrac{\pi}{4} \Big) = +\cos\Big(\dfrac{\pi}{4} \Big) = \dfrac{1}{\sqrt{2}} \\.\\ math math . \qquad \qquad \tan \Big( -\dfrac{\pi}{3} \Big) = \tan \Big( 0 - \dfrac{\pi}{3} \Big) = -\tan \Big( \dfrac{\pi}{3} \Big) = -\sqrt{3} math

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