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Find the exact value of each of the remaining trigonometric functions of θ.

role="math" localid="1646521470980" ²õ¾±²Ôθ=23,³Ù²¹²Ôθ<0

Short Answer

Expert verified

The exact values are,

³¦´Ç²õθ=-53³Ù²¹²Ôθ=-255³¦²õ³¦Î¸=32²õ±ð³¦Î¸=-355³¦´Ç³Ùθ=-52

Step by step solution

01

Step 1. Given Information

Given that the function is ²õ¾±²Ôθ=23,³Ù²¹²Ôθ<0.To find the exact values of θ.

02

Step 2. Using theorem and signs

There is a difference in sign due to the quadrant of II or IV. When y=2>0,the quadrant is in II. The angle lies to the point P=(x,y)in the standard position. The radius of the circle has r2=x2+y2.The values are ²õ¾±²Ôθ=yr,y=2,r=3.On solving and substituting,

x2=r2-y2x2=(3)2-(2)2 x2=9-4x2=5x=-5

03

Substitution to find exact values

On substituting the values for exact values are,

³¦´Ç²õθ=xr³¦´Ç²õθ=-53³Ù²¹²Ôθ=yx³Ù²¹²Ôθ=-25.55³Ù²¹²Ôθ=-255

04

Substitution to find exact values

On finding the exact values are,

³¦²õ³¦Î¸=ry³¦²õ³¦Î¸=32²õ±ð³¦Î¸=rx²õ±ð³¦Î¸=-35.55²õ±ð³¦Î¸=-355³¦´Ç³Ùθ=xy³¦´Ç³Ùθ=-52

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