Question

If sinØ=4/5 in quadrant ll, what is cosØ?

Answer 1

Answer: cosφ=-3/5

Explanation:

`\displaystyle\\sin\phi=(4)/(5) \ \ \ \ \ 90^0 < \phi < 180^0\\\\sin^2\phi+cos^2\phi=1\\\\\cos^2\phi=1-sin^2\phi\\\\cos^2\phi=1-((4)/(5))^2\\\\ cos^2\phi=1-(16)/(25) \\\\cos^2\phi=(1(25)-16)/(25)\\\\ cos^2\phi=(25-16)/(25) \\\\cos^2\phi=(9)/(25) \\\\cos\phi=б\sqrt{(9)/(25) } \\\\cos\phi=б\sqrt{(3^2)/(5^2) }\\\\cos\phi=б\sqrt{((3)/(5))^2} \\\\cos\phi=б(3)/(5) \\\\90^0 < \phi < 180^0\\\\Hence,\ \\\\cos\phi=-(3)/(5)`