evaluate the expression sin(2sin^-1(1/sqrt 26)) without a calculator
How would I do this?
Trigonometry homework help?
2013-03-23 5:48 pm
回答 (7)
2013-03-23 5:56 pm
✔ 最佳答案
Let theta = arcsin ( 1 / sqrt 26 ) sin theta = 1 / sqrt ( 26 )
cos theta = 5 / sqrt ( 26 )
sin ( 2 theta ) = 2 sin theta cos theta
sin ( 2 theta = 2 ( 1 / sqrt 26 ) ( 5 / sqrt 26 )
sin ( 2 theta ) = 10 / 26
sin ( 2 theta ) = 5 / 13
參考: my brain
2013-03-23 6:04 pm
sin(2θ) = 2sinθcosθ
So we have 2sin(sin^-1(1/√26))cos(sin^-1(1/√26)) = 2(1/√26)(5/√26) = 10/26 = 5/13
So we have 2sin(sin^-1(1/√26))cos(sin^-1(1/√26)) = 2(1/√26)(5/√26) = 10/26 = 5/13
2013-03-23 6:01 pm
Let x = sin^-1(1/sqrt(26))
We want to find sin(2x) = 2sin(x)cos(x)
sin(x)=1/sqrt(26)
cos(x) = sqrt(1 -sin^2(x)) = sqrt(1 - 1/26) = sqrt(25/26) = 5/sqrt(26)
2sin(x)cos(x) = 2 x 1/sqrt(26) x 5/sqrt(26) = 10/26 = 5/13
We want to find sin(2x) = 2sin(x)cos(x)
sin(x)=1/sqrt(26)
cos(x) = sqrt(1 -sin^2(x)) = sqrt(1 - 1/26) = sqrt(25/26) = 5/sqrt(26)
2sin(x)cos(x) = 2 x 1/sqrt(26) x 5/sqrt(26) = 10/26 = 5/13
2013-03-23 6:01 pm
sin2θ
= 2sinθcosθ
sin(2sin^-1(1/√26))
= 2sin(sin^-1(1/√26))cos(sin^-1(1/√26))
= 2 * 1/√26 * √((√26)^2 - 1^2)/√26
= 10/26
= 5/13
= 2sinθcosθ
sin(2sin^-1(1/√26))
= 2sin(sin^-1(1/√26))cos(sin^-1(1/√26))
= 2 * 1/√26 * √((√26)^2 - 1^2)/√26
= 10/26
= 5/13
2013-03-23 5:58 pm
Let sin^-1 (1/sqrt 26) =x.
So, sin(x)= 1/ sqrt(26) & cos(x)= 5/ sqrt(26).
So the main expression to be evaluated is sin(2x) which is equal to 2.sin(x).cos(x).
Answer comes out to be 10/26 or 5/13.
Hope this helps, all the best :)
So, sin(x)= 1/ sqrt(26) & cos(x)= 5/ sqrt(26).
So the main expression to be evaluated is sin(2x) which is equal to 2.sin(x).cos(x).
Answer comes out to be 10/26 or 5/13.
Hope this helps, all the best :)
2013-03-23 5:58 pm
put sin^-1(1/sqrt26) =x
Therefore sin(x) 1/sqrt(26)..............(i)
and cos (x) = sqrt[1-sin^2(x)]
= sqrt[1-1/26]
= sqrt[25/26) =5/sqrt(26).....(ii)
Therefore sin(2sin^-1(1/sqrt 26)) = sin 2x = 2 sin(x) cos(x) =2{1/sqrt(26)}{5/sqrt(26)}
= 10/26 = 5/13 ...........Ans
Therefore sin(x) 1/sqrt(26)..............(i)
and cos (x) = sqrt[1-sin^2(x)]
= sqrt[1-1/26]
= sqrt[25/26) =5/sqrt(26).....(ii)
Therefore sin(2sin^-1(1/sqrt 26)) = sin 2x = 2 sin(x) cos(x) =2{1/sqrt(26)}{5/sqrt(26)}
= 10/26 = 5/13 ...........Ans
2013-03-23 5:56 pm
sin^-1(1/√26) is the same thing as saying sin(θ) = 1/√26. sin^-1 determines an angle measure. Draw a right triangle such that sin(θ) = 1/√26, this means that the adjacent side is 5. You have a right triangle whose sides are 1, 5, and √26 (hypotenuse). You're being asked to find sin(2θ).
sin(2θ) = 2sin(θ)cos(θ)
sin(θ) = 1/√26
cos(θ) = 5/√26
sin(2θ) = 2*1/√26*5/√26
sin(2θ) = 10/26
sin(2θ) = 5/13
This means that sin(2sin^-1(1/sqrt 26)) = 5/13.
sin(2θ) = 2sin(θ)cos(θ)
sin(θ) = 1/√26
cos(θ) = 5/√26
sin(2θ) = 2*1/√26*5/√26
sin(2θ) = 10/26
sin(2θ) = 5/13
This means that sin(2sin^-1(1/sqrt 26)) = 5/13.
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