Trigonometric Identities - ppt download

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Trigonometric Ratios of (180   ) y The circle is centred at the origin O. It cuts the positive x-axis at A. A x O
y. The circle is centred at the origin O. It cuts the positive x-axis at A. A. x. O.
y. P(3, 4) is a point on the circle.  P(3, 4) AOP =  A. x. O. By the definitions of trigonometric ratios, sin  = cos  = tan  =
y. P is reflected about the y-axis to Q. Q(3, 4) Q.  P(3, 4) Coordinates of Q = (3, 4) 180º - q.  A. x. O. AOQ = 180   By the definitions of trigonometric ratios, sin (180   ) = cos (180   ) = tan (180   ) =
tan  = tan (180   ) = sin (180   ) = sin  cos (180   ) = cos  tan (180   ) = tan 
In fact, the relationships between the trigonometric ratios of  and (180   ) are true for any acute angle .
Suppose P(a, b) is a point on a circle of radius r centred at the origin O. y.  P(a, b) By the definitions of trigonometric ratios, r. B. A. x. O. sin  = r. b. cos  = r. a. tan  = a. b.
Q(a, b) Q.  P(a, b) r. r. Coordinates of Q = (a, b) 180º – q. B.  A. x. O. OQ = r. AOQ = 180  
Q(a, b) P(a, b) sin (180   ) = r. b. = sin  r. 180º – q. B.  A. x. O. r. a. cos (180   ) = - = -cos  a. b. tan (180   ) = - = -tan  cos  = r. a. tan  = b. sin  = Note:
sin (180   ) = sin  cos (180   ) = cos  tan (180   ) = tan  If  is an acute angle, then (180   ) lies in quadrant II, and only sin (180   ) is positive.
y. R is obtained by rotating P(a, b) through 180 about O. P(a, b) r. 180º + q. 180 Coordinates of R. = (a, b)  A. x. O. OR = r. r. Reflex AOR. = 180 +  R(a, b) R.
y. By the definitions of trigonometric ratios, P(a, b) sin (180 +  ) = r. b.  = -sin  180º + q.  A. x. O. r. a. cos (180 +  ) =  = -cos  r. R(a, b) a. b. tan (180 +  ) = = tan  cos  = r. a. tan  = b. sin  = Note:
sin (180 +  ) = –sin  cos (180 +  ) = –cos  tan (180 +  ) = tan  If  is an acute angle, then (180 +  ) lies in quadrant III, and only tan (180  +  ) is positive.
y. S is obtained by reflecting P(a, b) about the x-axis. P(a, b) Coordinates of S = (a, b) r. 360º - q.  A. x. OS = r. O. q. r. Reflex AOS. = 360   S. (a, b)
y. By the definitions of trigonometric ratios, P(a, b) sin (360 -  ) = r. b. - = -sin  360º - q.  A. x. O. r. a. cos (360 -  ) = = cos  r. S. (a, b) a. b. tan (360 -  ) = = -tan  - cos  = r. a. tan  = b. sin  = Note:
sin (360 –  ) = –sin  cos (360 –  ) = cos  tan (360 –  ) = –tan  If  is an acute angle, then (360 –  ) lies in quadrant IV, and only cos (360 –  ) is positive.
What do you observe about the terminal sides of  and (360   )
y. 360 -  x. O. - The terminal sides of - and (360 -  ) are coincident.
sin (360 –  ) = –sin  cos (– ) = cos (360 –  ) = cos  tan (– ) = tan (360 –  ) = –tan  If  is an acute angle, then – lies in quadrant IV, and only cos (– ) is positive.
x. y.  360 +  The terminal sides of  and (360 +  ) are coincident. We have. sin (360 +  ) = sin  cos (360 +  ) = cos  tan (360 +  ) = tan 
Let’s summarize using the ‘CAST’ diagram.
Sine is positive. All are positive. O. y. x. S. A. sin (180 –  ) = + sin q. sin (360 +  ) = + sin q. cos (180 –  ) = – cos q. 180 -  cos (360 +  ) = + cos q. 180 +  360 -  tan (180 –  ) = – tan q.  tan (360 +  ) = + tan q. T. C. sin (180 +  ) = – sin q. sin (360 –  ) = – sin q. cos (180 +  ) = – cos q. cos (360 –  ) = + cos q. tan (180 +  ) = + tan q. tan (360 –  ) = – tan q. Tangent is positive. Cosine is positive. These identities are also true even if  is not an acute angle.
Express each of the following trigonometric ratios in terms of the same trigonometric ratio of an acute angle. (a) cos 125 = cos (180  55) = cos 55  cos (180   ) = cos 
Express each of the following trigonometric ratios in terms of the same trigonometric ratio of an acute angle. (b) sin (250) = sin [360 + (250)]  sin (360 +  ) = sin  = sin 110 = sin (180  70) = sin 70  sin (180   ) = sin 
Find the values of the following trigonometric ratios. (a) sin 210 = sin (180 + 30) = –sin 30  sin (180 +  ) = sin  =
Find the values of the following trigonometric ratios. (b) tan 315 = tan (360 – 45) = –tan 45  tan (360   ) = tan  = –1.
Follow-up question Simplify .
By using the trigonometric ratios of (180 - q) and (90 - q), we have. ) 90 sin( q. + ) 90 180 sin( q. + - = )] 90 ( 180 sin[ q. - = ) 90 sin( q. - = sin (180 - f) = sin f. q. cos. = ) 90 cos( q. + ) 90 180 cos( q. + - = )] 90 ( 180 cos[ q. - = ) 90 cos( q. - = cos (180 - f) = -cos f. q. sin. - =
By using the trigonometric ratios of (180 - q) and (90 - q), we have. ) 90 tan( q. + ) 90 180 tan( q. + - = )] 90 ( 180 tan[ q. - = ) 90 tan( q. - = tan (180 - f) = -tan f. q. tan =
sin (90 +  ) = cos  cos (90 +  ) = -sin  tan (90 +  ) = These identities are also true even if  is not an acute angle.
By using the trigonometric ratios of (180 + ) and (90 - q), we have. sin (270 – q) = sin [180 + (90 – )] = –sin(90 – ) = –cos  sin (180 + f) = –sin f. cos (270 – q) = cos [180 + (90 – )] = –cos (90 – ) = –sin  cos (180 + f) = –cos f.
By using the trigonometric ratios of (180 + ) and (90 - q), we have. tan (270 – q) = tan [180 + (90 – )] = tan (90 – ) q. tan. 1. tan (180 + f) = tanf. =
By using the trigonometric ratios of (360 – ) and (90 – q), we have. sin (270 + q) = sin [360 – (90 – )] = –sin (90 – ) = –cos  sin (360 – f) = –sin f. cos (270 + q) = cos [360 – (90 – )] = cos (90 – ) = sin  cos (360 – f) = cos f.
By using the trigonometric ratios of (360 – ) and (90 – q), we have. tan (270 + q) = tan [360 – (90 – )] = –tan (90 – ) tan (360 – f) = –tanf. q. tan. 1. – =
sin (270 -  ) = -cos  sin (270 +  ) = -cos  cos (270 -  ) = -sin  cos (270 +  ) = sin  tan (270 -  ) = tan (270 +  ) =
Sine is positive. All are positive. O. y. x. S. A. sin (90 +  ) = + cos q. sin (90 –  ) = + cos q. cos (90 +  ) = – sin q. cos (90 –  ) = + sin q. 90 +  270 –  270 +  tan (90 +  ) = – 90 –  tan (90 –  ) = + T. C. sin (270 –  ) = – cos q. sin (270 +  ) = – cos q. cos (270 –  ) = – sin q. cos (270 +  ) = + sin q. tan (270 –  ) = + tan (270 +  ) = – Tangent is positive. Cosine is positive. These identities are also true even if  is not an acute angle.
Follow-up question Simplify . cos q sin q  tan  =