1. ## Trig

The displacement of a certain object is given as a function of the time $t$ (in seconds) by $s = sin \;t + cos\; 2t$. Find the smallest positive value of $t$ for which the displacement is zero.

Is the answer $t = 90$

2. Originally Posted by TI-84
The displacement of a certain object is given as a function of the time $t$ (in seconds) by $s = sin \;t + cos\; 2t$. Find the smallest positive value of $t$ for which the displacement is zero.

Is the answer $t = 90$
Use the equation:

$\cos(2t) = 1-2\sin^2(t)$

With s = 0 your equation becomes:

$0 = \sin(t) + 1 - 2\sin^2(t)$ ..... Use the substitution:

$y = \sin(t)$ ..... Then your equation becomes:

$0= y + 1 - 2y^2$..... Solve for y. Afterwards re-substitute and solve for t. Your result is OK if 1 s correspond to 1°.

3. I'm a little confused after working your method out. Here's the way I did it.

$S = Sin\;t + cos\;2t$

$0 = Sin\;t + cos\;2t$

$sin\;t + (1 - 2\;sin^2\;t) = 0$

$-2\;sin^2\;t + sin\;t + 1 = 0$

$(-sin\;t + 1) (2\;sin\;t + 1) = 0$

$-sin\;t = -1\Longrightarrow sin\;t = 1\Longrightarrow t = sin^{-1}(1)$

$2\;sin\;t = -1\Longrightarrow sin\;t = \frac{-1}{2}\Longrightarrow t = sin^{-1}\left(\frac{-1}{2}\right)$

$t = 90\;or\; t = -30$

The $t = -30$ gets crossed out because its negative which leaves $t = 90$ as the final answer. Or did I mess up somewhere?

4. Originally Posted by TI-84
I'm a little confused after working your method out. Here's the way I did it.

$S = Sin\;t + cos\;2t$

$0 = Sin\;t + cos\;2t$

$sin\;t + (1 - 2\;sin^2\;t) = 0$

$-2\;sin^2\;t + sin\;t + 1 = 0$

$(-sin\;t + 1) (2\;sin\;t + 1) = 0$

$-sin\;t = -1\Longrightarrow sin\;t = 1\Longrightarrow t = sin^{-1}(1)$

$2\;sin\;t = -1\Longrightarrow sin\;t = \frac{-1}{2}\Longrightarrow t = sin^{-1}\left(\frac{-1}{2}\right)$

$t = 90\;or\; t = -30$

The $t = -30$ gets crossed out because its negative which leaves $t = 90$ as the final answer. Or did I mess up somewhere?
Believe it or not, your method (which by the way is correct) is more or less the same as what earboth proposed ....!

The correct answer is $t = {\color{red}\frac{\pi}{2}}$ seconds ...... Capisce?

Radians, NOT degrees is the implied unit .....

5. Originally Posted by TI-84
I'm a little confused after working your method out. Here's the way I did it.

$S = Sin\;t + cos\;2t$

$0 = Sin\;t + cos\;2t$

$sin\;t + (1 - 2\;sin^2\;t) = 0$

$-2\;sin^2\;t + sin\;t + 1 = 0$

$(-sin\;t + 1) (2\;sin\;t + 1) = 0$

$-sin\;t = -1\Longrightarrow sin\;t = 1\Longrightarrow t = sin^{-1}(1)$

$2\;sin\;t = -1\Longrightarrow sin\;t = \frac{-1}{2}\Longrightarrow t = sin^{-1}\left(\frac{-1}{2}\right)$

$t = 90\;or\; t = -30$

The $t = -30$ gets crossed out because its negative which leaves $t = 90$ as the final answer. Or did I mess up somewhere?
By the way .....

Even though they are not used here, you do understand how to get positive solutions for t from $\sin t = -\frac{1}{2}$, right?

6. Does it have something to do with reference angles. One more thing. How did you know it was radians as the units I should use? Unless its just common sense. Then my bad.

7. Originally Posted by TI-84
Does it have something to do with reference angles. One more thing. [snip]
Yes. The third and fourth quadrant angles are $t = \frac{7\pi}{6}$ and $t = \frac{11 \pi}{6} ~ ( = -\frac{\pi}{6} \,$ , the 'calculator answer') respectively.

Then $t = \frac{7\pi}{6} + 2n\pi$ and $t = \frac{11 \pi}{6} + 2n\pi$ where n is an integer.

Originally Posted by TI-84
[snip]How did you know it was radians as the units I should use? [snip]

When workng with models involving trigonometric functions, the implied unit for the angle will always be radians. Note that the formulae for the derivative of sin x, cos x, tan x etc are for x measured in radians.