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Question
secondary 3 | A Maths
4 Answers Below
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Help please!!
Date Posted: 4 years ago
Views: 262
(a)
sin 300 deg
= sin (360-60) deg
= -sin 60 deg
= -√3/2
(b)
tan (-210 deg)
= tan (360 -210) deg
= tan 150 deg
= tan (180-30) deg
= -tan 30 deg
= -1/√3
(c)
cos(-π/3)
= cos π/3
= 0.5
(d)
tan 2π/3
= tan (π - 2π/3)
= -tan π/3
= -√3
...
sin 300 deg
= sin (360-60) deg
= -sin 60 deg
= -√3/2
(b)
tan (-210 deg)
= tan (360 -210) deg
= tan 150 deg
= tan (180-30) deg
= -tan 30 deg
= -1/√3
(c)
cos(-π/3)
= cos π/3
= 0.5
(d)
tan 2π/3
= tan (π - 2π/3)
= -tan π/3
= -√3
...
See 4 Answers
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The first one. This gets very difficult to explain on paper. You will need to know how the Quadrants and their angles work and you also need to know the values of some trigonometric ratios (sin, cos, tan) of special angles such as 0, 30, 45, 60 and 90 (and their equivalent in radians).
Date Posted:
4 years ago
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For the second case
Date Posted:
4 years ago
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Done
Date Posted:
4 years ago
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Some useful numbers to know for special angles
Date Posted:
4 years ago
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