Trigonometric Identities

Trigonometric Ratios of (180   ) y The circle is centred at the origin O. It cuts the positive x-axis at A. A x O

Trigonometric Ratios of (180   ) 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  =

Trigonometric Ratios of (180   ) 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   ) =

We have sin (180   ) = sin  cos (180   ) = cos  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 .

By the definitions of trigonometric ratios, 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 is obtained by reflecting P about the y-axis. Q(a, b) Q  P(a, b) r r Coordinates of Q = (a, b) 180º – q B  A x O OQ = r AOQ = 180  

By the definitions of trigonometric ratios, 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:

Hence, for any acute angle , we have 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.

Trigonometric Ratios of (180 +  ) 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

Trigonometric Ratios of (180 +  ) 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:

Hence, for any acute angle , we have 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.

Trigonometric Ratios of (360   ) &  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)

Trigonometric Ratios of (360   ) &  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:

Hence, for any acute angle , we have 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 about the trigonometric ratios of  ? What do you observe about the terminal sides of  and (360   ) ?

The terminal sides of - and (360 -  ) are coincident. y 360 -  x O - The terminal sides of - and (360 -  ) are coincident.

Since the terminal sides of  and (360   ) are coincident, we have 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.

Trigonometric Ratios of (360 +  ) 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.

For any acute angle , we have 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.

(a) cos 125 = cos (180  55) = cos 55 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 

(b) sin (250) = sin [360 + (250)] = sin 110 = sin (180  70) 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 

(a) sin 210 = sin (180 + 30) = –sin 30 = Find the values of the following trigonometric ratios. (a) sin 210 = sin (180 + 30) = –sin 30  sin (180 +  ) = sin  =

(b) tan 315 = tan (360 – 45) = –tan 45 = –1 Find the values of the following trigonometric ratios. (b) tan 315 = tan (360 – 45) = –tan 45  tan (360   ) = tan  = –1

Follow-up question Simplify .

Trigonometric Ratios of (90 +  ) 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 - =

Trigonometric Ratios of (90 +  ) 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 1 - =

Hence, for any acute angle , we have sin (90 +  ) = cos  cos (90 +  ) = -sin  tan (90 +  ) = These identities are also true even if  is not an acute angle.

Trigonometric Ratios of (270 –  ) & (270 +  ) 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

Trigonometric Ratios of (270 –  ) & (270 +  ) 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 =

Trigonometric Ratios of (270 –  ) & (270 +  ) 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

Trigonometric Ratios of (270 –  ) & (270 +  ) 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 – =

Hence, for any acute angle , we have sin (270 -  ) = -cos  sin (270 +  ) = -cos  cos (270 -  ) = -sin  cos (270 +  ) = sin  tan (270 -  ) = tan (270 +  ) =

Let’s summarize using the ‘CAST’ diagram.

For any acute angle , we have 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  =