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Question

Find the values of the following trigonometric ratios:

(i)sin5π3

(ii)sin3060

(iii)tan11π6

(iv)cos(1125)

(V)tan315

(vi)sin510

(vii)cos570

(viii)sin(330)

(ix)cosec(1200)

(X)tan(585)

(xi)cos855

(xii)sin1845

(xiii)cos1755

(xiv)sin4530


Solution

(i)sin5π3=sin(2ππ3)

=sinπ3[sin(2πθ)=sinθ]

=32

(ii)3060=17π[π=180]

sin3060=sin17π

=0

[sinnπ=0forallnz]

(iii)tan11π6=tan(2ππ6)

=tanπ6

[tan(2πθ)=tanθ]

=13

(iv)1125=6π+π4(π=180)

cos(1125)=cos((6π+π4))

=cos(6π+π4)

[cos(θ)=cosθ]

=cos(2×3π+π4)

=cosπ4

[cos(2kπ+θ)=cosθ,kn]

=12

(v)tan315=tan(2ππ4)

=tanπ4

[tan(2πθ)=tanθ]

=-1

(vi)sin510=sin(3ππ6)

=sin(π)6

[3ππ6 lies in second quadrant]

=12

Alternative solution 

sin510=sin(3ππ6)

=sin(2π+(ππ6))

=sin(ππ6)

[sin(2π+θ)=sinθ, as sine is periodic with period 2π]

=sinπ6[sin(πθ)=sinθ]

=12

(vii)cos570=cos(3π+π6)

=cos(2π+(π+π6))

=cos(π+π6)

[cos(2π+θ)=cosθ, as cosine is periodoc with period 2π]

=cosπ6

[cos(π+θ)=cosθ]

=32

(viii)sin(330)=sin((2ππ6))

=sin(2ππ6)[sin(θ)=sinθ]

=(sinπ6)

[sin(2πθ)=sinθ]

=sinπ6

=12

(ix)cosec(1200)=cosec((7ππ3))

=cosec(7ππ3)

[cosec(θ)=cosecθ]

=cosec(2×3π+(ππ3))

=cosec(ππ3)

[ cosec is periodic of period 2π, cosec (2π+θ)=cosec(2nπ+θ)

=cosecθforallnN]

=cosecπ3

[cosec(πθ)cosecθ]

=23

(x)tan(585)=tan(585)

[tan(θ)=tanθ]

=tan(3π+π4)

=tan(2π+(π+π4))

[tan(2π+θ)=tanθ]

=tanπ4

[tan(π+θ)=tanθ]

=-1

(xi)cos(855)=cos(5ππ4)

=cos(2×2π+(ππ4))

=cos(ππ4)

[cos(2kπ+θ)=cosθ for all kN]

=cosπ4

[cos(πθ)=cosθ]

=12

(xii)sin1845=sin(10π+π4)

=(2π5π+π4)

=sinπ

[sin(2kπ+θ)=sinθ, for all kN]

=12

(xiii)cos1755=cos(10ππ4)

=cos(2π5ππ4)

=cosπ4

[cos(2kπθ)=cosθ,kN]

=12

(xiv)4530=(25π+π6)

sin4530=sin(25π+π6)

=sin2×12π+(π+π6)

=sin(ππ6)

[sin(2kπ+θ)=sinθ,kN]

=sinπ6[sin(π+θ)=sinθ]

=12

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