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Trigonometric Identities
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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
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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 =
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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 ) =
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We have sin (180 ) = sin cos (180 ) = cos
tan = tan (180 ) = sin (180 ) = sin cos (180 ) = cos tan (180 ) = tan
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In fact, the relationships between the trigonometric ratios of and (180 ) are true for any acute angle .
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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
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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
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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:
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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.
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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
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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:
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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.
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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)
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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:
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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.
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What about the trigonometric ratios of ?
What do you observe about the terminal sides of and (360 ) ?
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The terminal sides of - and (360 - ) are coincident.
y 360 - x O - The terminal sides of - and (360 - ) are coincident.
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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.
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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
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Let’s summarize using the ‘CAST’ diagram.
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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.
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(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
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(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
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(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 =
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(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
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Follow-up question Simplify
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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 - =
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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 - =
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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.
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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
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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 =
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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
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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 – =
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Hence, for any acute angle , we have
sin (270 - ) = -cos sin (270 + ) = -cos cos (270 - ) = -sin cos (270 + ) = sin tan (270 - ) = tan (270 + ) =
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Let’s summarize using the ‘CAST’ diagram.
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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.
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Follow-up question Simplify cos q sin q tan =
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