trigo qns help urgent

[helpmaths]

tan @ = x
@ is acute angle
sin^2(2@)=?


please help solve asap tyvm


the answer is (4x^2) / (x^2+1)^2
but
I need to know the working.

 

[Deleted member 4993]

tan @ = x
@ is acute angle
sin^2(2@)=?


please help solve asap tyvm


the answer is (4x^2) / (x^2+1)^2
but
I need to know the working.

Hint:

tan(Θ) = x

1 + tan²(Θ) = 1 + x² → sec²(Θ) = ? → cos(Θ) = ? → sin(Θ) = ? → sin²(2Θ) = ?

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[helpmaths]

Hint:

tan(Θ) = x

1 + tan²(Θ) = 1 + x² → sec²(Θ) = ? → cos(Θ) = ? → sin(Θ) = ? → sin²(2Θ) = ?

i'm stuck as i didnt know what formula i can use for sin²(2Θ).

i followed your hint and
i'm still stuck at the part where i have to change from cos(Θ) → sin(Θ)
i got 1 / 1+x^2 when i reach cos(Θ).
how do i change from cos to sin??


my original method was to get rid of the square in the sin.
by using 2sin^2@ = 1-cos2@
im stuck at the cos 2@ part how to find cos 2@

 

[helpmaths]

i tried getting rid of the squre in the sin^2@

using the formula cos2x = 1 - 2sin^2
and im stuck here as i dont know how to find cos2@(for cos2x)

 

[Deleted member 4993]

Hint:

tan(Θ) = x

1 + tan²(Θ) = 1 + x² → sec²(Θ) = ? → cos(Θ) = ? → sin(Θ) = ? → sin²(2Θ) = ?

i'm stuck as i didnt know what formula i can use for sin²(2Θ). ← sin(2Θ) = 2 * sin (Θ) * cos(Θ)

i followed your hint and
i'm still stuck at the part where i have to change from cos(Θ) → sin(Θ)

cos²(Θ) + sin²(Θ) = 1 → sin(Θ) = ± √[1 - cos²(Θ)]

i got 1 / 1+x^2 when i reach cos(Θ).
how do i change from cos to sin??


my original method was to get rid of the square in the sin.
by using 2sin^2@ = 1-cos2@
im stuck at the cos 2@ part how to find cos 2@

These are preliminary trigonometric identities - you must be able to recall these!!