Question -
Answer -
(i) sin 5π/3
5π/3 = (5π/3 × 180)o
= 300o
= (90×3 + 30)o
Since, 300o lies in IV quadrant inwhich sine function is negative.
sin 5π/3 = sin (300)o
= sin (90×3 + 30)o
= – cos 30o
= – √3/2
(ii) sin 17π
Sin 17π = sin 3060o
= sin (90×34 + 0)o
Since, 3060o lies in the negativedirection of x-axis i.e., on boundary line of II and III quadrants.
Sin 17π = sin (90×34 + 0)o
= – sin 0o
= 0
(iii) tan 11π/6
tan 11π/6 = (11/6 × 180)o
= 330o
Since, 330o lies in the IV quadrant inwhich tangent function is negative.
tan 11π/6 = tan (300)o
= tan (90×3 + 60)o
= – cot 60o
= – 1/√3
(iv) cos (-25π/4)
cos (-25π/4) = cos (-1125)o
= cos (1125)o
Since, 1125o lies in the I quadrant inwhich cosine function is positive.
cos (1125)o = cos (90×12 + 45)o
= cos 45o
= 1/√2
(v) tan 7π/4
tan 7π/4 = tan 315o
= tan (90×3 + 45)o
Since, 315o lies in the IV quadrant inwhich tangent function is negative.
tan 315o = tan (90×3 + 45)o
= – cot 45o
= -1
(vi) sin 17π/6
sin 17π/6 = sin 510o
= sin (90×5 + 60)o
Since, 510o lies in the II quadrant inwhich sine function is positive.
sin 510o = sin (90×5 + 60)o
= cos 60o
= 1/2
(vii) cos 19π/6
cos 19π/6 = cos 570o
= cos (90×6 + 30)o
Since, 570o lies in III quadrant inwhich cosine function is negative.
cos 570o = cos (90×6 + 30)o
= – cos 30o
= – √3/2
(viii) sin (-11π/6)
sin (-11π/6) = sin (-330o)
= – sin (90×3 + 60)o
Since, 330o lies in the IV quadrant inwhich the sine function is negative.
sin (-330o) = – sin (90×3 + 60)o
= – (-cos 60o)
= – (-1/2)
= 1/2
(ix) cosec (-20π/3)
cosec (-20π/3) = cosec (-1200)o
= – cosec (1200)o
= – cosec (90×13 + 30)o
Since, 1200o lies in the II quadrantin which cosec function is positive.
cosec (-1200)o = – cosec (90×13 + 30)o
= – sec 30o
= -2/√3
(x) tan (-13π/4)
tan (-13π/4) = tan (-585)o
= – tan (90×6 + 45)o
Since, 585o lies in the III quadrantin which the tangent function is positive.
tan (-585)o = – tan (90×6 + 45)o
= – tan 45o
= -1
(xi) cos 19π/4
cos 19π/4 = cos 855o
= cos (90×9 + 45)o
Since, 855o lies in the II quadrant inwhich the cosine function is negative.
cos 855o = cos (90×9 + 45)o
= – sin 45o
= – 1/√2
(xii) sin 41π/4
sin 41π/4 = sin 1845o
= sin (90×20 + 45)o
Since, 1845o lies in the I quadrant inwhich the sine function is positive.
sin 1845o = sin (90×20 + 45)o
= sin 45o
= 1/√2
(xiii) cos 39π/4
cos 39π/4 = cos 1755o
= cos (90×19 + 45)o
Since, 1755o lies in the IV quadrantin which the cosine function is positive.
cos 1755o = cos (90×19 + 45)o
= sin 45o
= 1/√2
(xiv) sin 151π/6
sin 151π/6 = sin 4530o
= sin (90×50 + 30)o
Since, 4530o lies in the III quadrantin which the sine function is negative.
sin 4530o = sin (90×50 + 30)o
= – sin 30o
= -1/2