Question

What is the smallest possible value of theta for which sin(2(theta)) = cos(4(theta))

Answer 1

@IrishBoy123 @nincompoop @pooja195

Answer 2

@aaronq

Answer 3

why not sub \(\phi = 2 \theta\) first, so you are into simple double angle formulae.....see if it makes sense and if there is an answer....

Answer 4

What? I don't understand... @IrishBoy123

Answer 5

Bloom:
"What is the smallest possible value of theta for which sin(2(theta)) = cos(4(theta))"\[\sin 2 \theta=\cos 4\theta\]

Answer 6

can be solved in several ways. One way would be to graph both functions on the same set of axes and determine the first angle theta at which the two graphs intersect.

There is a "double angle formula" for the cosine function.
supposing we wanted cos 2x, we could write that as cos (x+x).

cos (x+x) = cos x cos x - sin x sin x, or \[\cos^2x-\sin^2x.\]

Answer 7

This, in turn, can be re-written as \[(1-\sin^2 x)-\sin^2x, .or.1-2\sin^2x. \]

Answer 8

This is a variation of what IrishBoy was asking you to do.

Answer 9

See whether you can prove to yourself that \[\cos4 \theta=\cos(2\theta+2\theta)=1-\sin^2 2\theta\]

Answer 10

Now let's go back to the original problem:

\[\sin 2\theta=\cos 4 \theta\]

Answer 11

and replace the right side with its equivalent\[1-\sin^22\theta\]

Answer 12

Then you have sin2theta on the left equal to 1-(sin 2theta)^2.\[\sin2\theta=1-\sin^22\theta\]

Answer 13

Write this in the form of a quadratic equation. (Move the right 2 terms to the left side and replace them with zero on the right side.)

Can you now solve this quadratic equation?

Answer 14

Hi Bloom!

i mean, if you re-write it as:

\( \sin \phi = cos2 \phi\), where \(2 \theta = \phi\), then you can then say stuff like:

\( \sin \phi = 1 - 2 \sin^2 \phi\)

so you can explore it as a quadratic, as @mathmale suggests.......

Answer 15

Well, the issue is, we can't use a calculator .-. it's for a math competion

Answer 16

I'm looking at old problems so yea

Answer 17

now we dig some more

from
\(\sin \phi = 1 - 2 \sin^2 \phi\)

we say that:

\(a = 1 - 2 a^2\)

can you solve that for a??

Answer 18

one moment.

Answer 19

all we've done is say that \(a = \sin \phi = \sin 2 \theta\)

Answer 20

-1, and 1/2

Answer 21

think that's good

so that's it solved for "a"

but we've said \(a = \sin \phi = \sin 2 \theta\)

so \(a = \sin \phi = \sin 2 \theta = -1\) and \(a = \sin \phi = \sin 2 \theta = 1/2\)

Answer 22

you can solve that for \(\phi\), then \(\theta\)....

.... or for \(2 \theta\),...


and you're done.

Answer 23

There's certainly more than one way in which to skin a cat. However, I'd still advocate using my model, which has resulted in a quadratic equation. \[\sin2\theta=1-\sin^22\theta\]

Answer 24

Moving both terms from the right side to the left side,\[\sin^22 \theta+\sin2\theta-1=0\]

Answer 25

Letting\[\sin 2\theta = x\]

Answer 26

this boils down to x^2+x-1=0. Determine whether or not this has any real roots.

If c is a real root, then solve the equation c=sin 2theta. Remember, we want the smallest theta will satisfy the original equation.