Question

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)

sin(V-U)

cos(U-V)

Answer 1

cosU:-

`\\ \rm\longmapsto √(1-sin^2U)=√(1-49/625)=24/25`

As U lies in Q3

• cosU=-24/25

sinV

`\\ \rm\longmapsto √(1-cos^2V)=√(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=44/125
• cos(U-V)=cosUcosV+sinUsinV=(-24/25)(-4/5)+(-7/25)(-3/5)=96/125+21/125=117/125