1. Find the value of (i) sin 210 (iii) tan 480° (iv) sec 510 (ii) cos 315 Find the value of​

Question

1. Find the value of
(i) sin 210
(iii) tan 480°
(iv) sec 510
(ii) cos 315
Find the value of​

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Melody 1 month 2021-08-14T14:02:30+00:00 2 Answers 0 views 0

Answers ( )

    0
    2021-08-14T14:03:57+00:00

    Answer:

    Sin 210 = sin (270 – 60) = sin (3 x 90 – 60) = – cos 60 = – 1/2 <=ANS

    Another Method:

    sin 210 = sin (360 – 150)= – sin 150 = – sin (180 – 30) = – sin (2 x 90 – 30)

    = – sin 30 [Since, 150 lies in 2nd quadrant & sin is +ve]

    = – 1/2

    Step-by-step explanation:

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    0
    2021-08-14T14:04:21+00:00

    Answer:

    sin (315)0

    sin (315)0 = sin (90 x 3 + 45)

    Since 315 lies in the 4th quadrant and in this quadrant sine function is negative , also 3 is an odd integer.

    ∴ sin (315)0 = sin (90 x 3 + 45) = – cos 45 = −12√

    cos (210)0

    cos (210)0 = cos(90 x 2 + 30)

    Since 210 lies in the 3rd quadrant and in this quadrant cosine function is negative.Also the multiple of 90 is even.

    ∴ cos (210)0 = cos (90 x 2 + 30) = – cos 30 = −3√2

    cos (480)

    Solution : cos (480)0

    cos (480)0 = cos(90 x 5 + 30)

    Since 480 lies in the 2nd quadrant and in this quadrant cosine function is negative.Also the multiple of 90 is odd.

    ∴ cos (480)0 = cos (90 x 5 + 30) = -sin 30 = −12

    We have,

    cos (510)0 cos (330)0 + sin (390)0 cos (120)0

    = cos(90 x 5 + 60) cos(90 x 3 + 60) + sin(90 x 3 + 30) cos(90 x 1 + 30)

    = (- sin 60) (sin 60) + (sin 30)(- sin 30)

    = −3√2×3√2+12×−12

    = −34−14

    = -1

    ∴ cos (510)0 cos (330)0 + sin (390)0 cos (120)0 = -1

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