im not sure what is under the square root. can you add parentheses?
I'll interpret it the best i can…
so (√ 5/5, 2√ 5/5) is your point on the terminal side of an angle?
ughh… i hope so.
anyway, i set it up using a reference angle. t being my angle.
im guessing "a real number t is given. find sin t, cos t, tan t," means that t is an angle.
i used Pythagorean theorem to find my third side.
(√ 5/5)^2 + (2√ 5/5)^2 = c^2
1/5 + 4/5 = c^2
1 = c
c = 1 (this point is on the unit circle)
sin t = (2√(5) / 5
cos t = √(5) / 5
tan t = 2
i hope this helps.
November 30th, 2008 at 8:07 am
sin t = 2sqrt 5 /5 (hypotenuse is 1)
cos t =sqrt 5 /5 (hypotenuse is 1)
tan t =2
References :
November 30th, 2008 at 8:28 am
Adjacent=rt(5)/5
Opposite=2rt(5)/5
Hypotenuse^2=[2rt(5)/5]^2 + [rt(5)/5]^2
Hypotenuse^2= 20/25 + 5/25, =25/25, =1
sin t=2rt(5)/5 / 1, =2rt(5)/5
cos t=rt(5)/5 / 1, =rt(5)/5
tan t = 2rt(5)/5 / rt(5)/5, = 2/1, =2
References :
November 30th, 2008 at 8:46 am
im not sure what is under the square root. can you add parentheses?
I'll interpret it the best i can…
so (√ 5/5, 2√ 5/5) is your point on the terminal side of an angle?
ughh… i hope so.
anyway, i set it up using a reference angle. t being my angle.
im guessing "a real number t is given. find sin t, cos t, tan t," means that t is an angle.
i used Pythagorean theorem to find my third side.
(√ 5/5)^2 + (2√ 5/5)^2 = c^2
1/5 + 4/5 = c^2
1 = c
c = 1 (this point is on the unit circle)
sin t = (2√(5) / 5
cos t = √(5) / 5
tan t = 2
i hope this helps.
References :