solving for the left side for a double angle formula

Question

$\cos t-\sin 2t=0$

Solve the left hand side so that it equals zero.

Do I use $(2\sin t \cos {t})$ for $\sin 2t$?

Answer 1

Yes, you can use the double angle formula.

$$\cos{t} - \sin{2t} = 0\\ \cos{t} - 2\sin{t}\cos{t} = 0\\ \cos{t}(1 - 2\sin{t}) = 0\\$$

From this, $\cos{t} = 0$ or $\sin{t} = \frac{1}{2}$. Now, just solve for $t$ for both cases.

Answer 2

Sure, then $$\cos t - \sin(2t) = \cos t - 2\sin t \cos t = \cos t(1 - 2\sin t) = 0$$

Now, $\cos t(1-2\sin t) = 0$ if and only if either

• $\cos t = 0 \implies t = ?$, or else

• $1 - 2\sin t = 0 \iff 2\sin t = 1 \iff \sin t = 1/2$.

For what values of $t$ does $\cos t = 0\;\;?$ For example, we know that $\cos (\pi/2) = 0$, and $\cos (3\pi/2) = 0$.

For what values of $t$ does $\sin t = 1/2\;\;?$ Recall that $\sin(\pi/6) = 1/2$. Consider what other angles, $t$ make $\sin t = 1/2$.

Answer 3

Not sure why you need "Double Angle" Formula

As $\displaystyle\sin2t=\cos t=\sin\left(\frac\pi2-t\right)$

$\displaystyle\implies2t=n\pi+(-1)^n t$ where $n$ is any integer

Now check for the odd & the even cases

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Alternatively, $\displaystyle\cos t=\sin2t=\cos\left(\frac\pi2-2t\right)$

$\displaystyle\implies t=2m\pi\pm\left(\frac\pi2-2t\right) $ where $m$ is any integer

Now check for either signs separately.