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Minchanka [31]
2 years ago
15

Find the exact value of the trigonometric function given that sinU=-7/25 and cosV=-4/5 (Both U and V are in Quadrant III)

Mathematics
1 answer:
Contact [7]2 years ago
3 0

cosU:-

\\ \rm\longmapsto \sqrt{1-sin^2U}=\sqrt{1-49/625}=24/25

As U lies in Q3

  • cosU=-24/25

sinV

\\ \rm\longmapsto \sqrt{1-cos^2V}=\sqrt{1-16/25}=3/4

As V lies in Q3

  • sinV=-3/5

So

  • sin(V-U)=sinVcosU-cosVsinU=(-3/5)(-24/25)-(-4/5)(-7/25)=72/125-28/125=72-28/125=<u>4</u><u>4</u><u>/</u><u>1</u><u>2</u><u>5</u>
  • cos(U-V)=cosUcosV+sinUsinV=(-24/25)(-4/5)+(-7/25)(-3/5)=96/125+21/125=117/125
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