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If sin t= 7/9, and t is in quadrant I, find the exact value of sin(2t) , cos(2t) , and tan(2t) algebraically without solving for t .

User Pat Lee
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1 Answer

7 votes

Explanation:

we know :

sin(2t) = 2×sin(t)×cos(t)

cos(2t) = cos²(t) – sin²(t) = 2×cos²(t) – 1 = 1 – 2×sin²(t)

tan(t) = sin(t)/cos(t)

tan(2t) = sin(2t)/cos(2t) = (2×tan(t))/(1 – tan²(t))

sin²(t) + cos²(t) = 1

as t is in quadrant I, 2t must be in either quadrant I or II.

sin(t) = 7/9

sin²(t) = (7/9)² = 49/81

cos²(t) = 1 - sin²(t) = 1 - 49/81 = 81/81 - 49/81 = 32/81

cos(t) = sqrt(32)/9

sin(2t) = 2×sin(t)×cos(t) = 2×7/9 × sqrt(32)/9 =

= 14×sqrt(32)/81 ≈ 0.98

cos(2t) = cos²(t) - sin²(t) = 32/81 - 49/81 = -17/81 ≈

≈ -0.21

tan(2t) = sin(2t)/cos(2t) =

= 14×sqrt(32)/81 / -17/81 =

= 14×sqrt(32)/81 × -81/17 =

= -14×sqrt(32)/17 ≈ -4.66

User Boardtc
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