CameraIcon
CameraIcon
SearchIcon
MyQuestionIcon
MyQuestionIcon
1
You visited us 1 times! Enjoying our articles? Unlock Full Access!
Question

Find the values of the following trigonometric ratios:
(i) sin5π3

(ii) sin 17π

(iii) tan11π6

(iv) cos-25π4

(v) tan 7π4

(vi) sin17π6

(vii) cos19π6

(viii) sin-11π6

(ix) cosec-20π3

(x) tan-13π4

(xi) cos19π4

(xii) sin41π4

(xiii) cos39π4

(xiv) sin151π6

Open in App
Solution

i We have:5π3 = 53×180° = 300° = 90°×3 + 30°300° lies in the fourth quadrant in which the sine function is negative.Also, 3 is an odd integer. sin5π3 = sin 300° =sin 90°×3 + 30° = -cos30° = -32

ii We have:sin 17π=sin 3060°3060° = 90°×34 + 0°Clearly 3060° is in the negative direction of the x-axis, i.e. on the boundary line of the II and III quadrants.Also, 34 is an even integer. sin3060° =sin 90°×34 + 0° =-sin 0° =0

iii We have:11π6 = 116×180° = 330° = 90°×3 + 60°330° lies in the fourth quadrant in which the tangent function is negative.Also, 3 is an odd integer. tan11π6 = tan 330° =tan90°×3 + 60° = -cot 60° = -13

iv We have:cos-25π4=cos1125°cos -1125° =cos 1125°= cos 90°×12 + 45°1125° lies in the first quadrant in which the cosine function is positive.Also, 12 is an even integer. cos-1125° = cos1125° =cos90°×12 + 45° = cos 45° = 12

v We have:tan 7π4=tan 315°315° = 90°×3 + 45°315° lies in the fourth quadrant in which the tangent function is negative.Also, 3 is an odd integer. tan 315° = tan 90°×3 + 45° = -cot 45° = -1


vi We have:sin17π6=sin 510°510° = 90°×5 + 60°510° lies in the second quadrant in which the sine function is positive.Also, 5 is an odd integer. sin510° = sin 90°×5 + 60° =cos 60° = 12

vii We have:cos19π6=cos 570°570° = 90°×6 + 30° 570° lies in the third quadrant in which the cosine function is negative.Also, 6 is an odd integer. cos570° = cos 90°×6 + 30° =-cos 30° = -32

viii We have:sin-11π6=sin -330°sin -330° =-sin 330°= -sin 90°×3 + 60°330° lies in the fourth quadrant in which the sine function is negative.Also, 3 is an odd integer.sin-330° =-sin330° =-sin90°×3 + 60° = --cos60° =-- 12 =12

ix We have:cosec-20π3=cosec -1200°cosec -1200° =-cosec 1200°= -cosec 90°×13 + 30°1200° lies in the second quadrant in which the cosec function is positive.Also, 13 is an odd integer. cosec-1200° = -cosec1200° =-cosec90°×13 + 30° = -sec30° = -23

x We have:tan -13π4=tan -585°tan -585° =-tan 585°= -tan 90°×6 + 45°585° lies in the third quadrant in which tangent function is positive.Also, 6 is an even integer. tan -585° = -tan 585° =-tan 90°×6 + 45° = -tan 45° = -1

xi We have:cos19π4=cos 855°855° = 90°×9 + 45°855° lies in the second quadrant in which the cosine function is negative.Also, 9 is an odd integer. cos855° = cos90°×9 + 45° =-sin45° = -12

xii We have:sin41π4=sin 1845°1845° = 90°×20 + 45°1845° lies in the first quadrant in which the sine function is positive.Also, 20 is an even integer. sin 1845° = sin 90°×20 + 45° =sin 45° = 12

xiii We have:cos39π4=cos 1755°1755° = 90°×19 + 45°1755° lies in the fourth quadrant in which the cosine function is positive.Also, 19 is an odd integer. cos1755° = cos90°×19 + 45° =sin 45° = 12

xiv We have:sin151π6=sin 4530°4530° = 90°×50 + 30°4530° lies in the third quadrant in which the sine function is negative.Also, 50 is an even integer. sin 4530° = sin 90°×50 + 30° =-sin 30° = -12

flag
Suggest Corrections
thumbs-up
4
Join BYJU'S Learning Program
similar_icon
Related Videos
thumbnail
lock
Trigonometric Functions in a Unit Circle
MATHEMATICS
Watch in App
Join BYJU'S Learning Program
CrossIcon