Half Angle Identities, I think

[calc1]

Using half angle identities

Solve the equation sin²(2x) = 1 for 0° < x < 180°

Attempt:

sin²(2x)-1, then

((1/2)(√1-cos4x))-1 . . ?

Or simply, the only position where sin is 1 is at 90°, but that is not the correct answer.

Where do I get an actual angle, the cos, to put under the half angle formula radical? And what to do with that 1? I don't see how to manipulate this.

 

[tkhunny]

That's an awful lot of pain.

\(\displaystyle \sin^{2}(2x) = 1\) leads immediately to \(\displaystyle \sin(2x) = 1\;or\;\sin(2x) = -1\).

What after that?

 

[calc1]

That's an awful lot of pain.

Welcome to my world. Obviously I am inexperienced with the tools.

\sin^{2}(2x) = 1[/tex] leads immediately to \(\displaystyle \sin(2x) = 1\;or\;\sin(2x) = -1\).

What after that?

sin(x)=1/2 is not right.

I tried sin(2x) = 2sin(x)cos(x), but then dividing either of the 1s by any those got me nowhere.

 

[Deleted member 4993]

sin(π/2 + 2*k * π) = 1 .............................k = 0, 1, 2, 3............

sin(3π/2 + 2*k*π) = -1.............................k = 0, 1, 2, 3............

 

[calc1]

sin(π/2 + 2*k * π) = 1 .............................k = 0, 1, 2, 3............

sin(3π/2 + 2*k*π) = -1.............................k = 0, 1, 2, 3............

I don't know how you got there.

Is this translated correctly? My answer is π/2 and 3π/2 and then each successive angle where 2π*k is added?

 

[soroban]

Hello, calc1!

You are not using the correct identity.

Using half angle identities, solve the equation:
. . \(\displaystyle \sin^2\!2x = 1\,\text{ for }0^o < x < 180^o\)

The identity is: .\(\displaystyle \sin^2\!\theta \:=\:\dfrac{1-\cos2\theta}{2}\)

So \(\displaystyle \sin^2\!2x\) becomes \(\displaystyle \dfrac{1-\cos4x}{2}\)

The equation is: .\(\displaystyle \dfrac{1-\cos4x}{2} \:=\:1\)

n . . . . . . . . . . . .\(\displaystyle 1-\cos4x \:=\:2\)

. . . . . . . . . . . . . . . .\(\displaystyle \cos4x \:=\:-1\)

. . . . . . . . . . . . . . . . . . \(\displaystyle 4x \:=\:180^o,\;540^o,\;900^o\;\cdots\)

. . . . . . . . . . . . . . . . . . . \(\displaystyle x \:=\:45^o,\;135^o,\;225^o\;\cdots\)

Therefore: .\(\displaystyle x \;=\;45^o,\;135^o\)

 

[Deleted member 4993]

I don't know how you got there.

Is this translated correctly? My answer is π/2 and 3π/2 and then each successive angle where 2π*k is added?

sin(2x) = sin(π/2) , sin(5π/2)..... → 2x = π/2 , 5π/4...... → x = π/4, 5π/4......

sin(2x) = sin(3π/2) , sin(7π/2)... → 2x = 3π/2 , 7π/2.... → x = 3π/4, 7π/4....

Now choose the correct answers within the given domain.

 

[Deleted member 4993]

I don't know how you got there.

Is this translated correctly? My answer is π/2 and 3π/2 and then each successive angle where 2π*k is added?

Are you saying you don't know that:

sin(90°) = sin(π/2) = 1

 

[calc1]

Thanks, soroban. At least I was on the right track. I should have left the 1 alone at the beginning. I see that I am not pivoting well between using basic algebra and using the identities.