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if $\sin(x)=\frac{1}{3}$ and $\sec(y)=\frac{5}{4} $, what is $\sin(x+y)?$

Here is my thought process:

the identity says:

$\sin(x+y)=\sin(x)\cos(y)+\cos(x)\sin(y)$

We know $\sin(x)=1/3$ and $\sec(y)=\frac{1}{\cos(y)} = \frac{4}{5}$, so

$\sin(x+y)=\frac{1}{3}\times\frac{4}{5}+\cos(x)\sin(y)$

But I'm stuck here because I don't know how to find $\cos(x)$ or $\sin(y)$

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  • $\begingroup$ Please consider using \sin and \cos in "Math mode" to get $\sin$ and $\cos$ instead of simply sin and cos which give $sin$ and $cos$. $\endgroup$ Commented Sep 4, 2015 at 22:43
  • $\begingroup$ This problem is badly posed. There is no solution as it is written. This problem comes from James Stewart's verification tests --- I believe it appears at least in the 6th edition. The problem restricts the angle to be between $0$ and $\pi/2$. $\endgroup$ Commented Feb 28, 2023 at 0:21

2 Answers 2

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Draw a right triangle for each angle, label the known sides, then find the missing side. The unknown ratio should be obvious.

enter image description here

Note that the Pythagorean theorem and the Pythagorean identity are equivalent to one another. $$\text{opposite}^2+\text{adjacent}^2 = \text{hypotenuse}^2$$ $$\frac{\text{opposite}^2+\text{adjacent}^2}{\text{hypotenuse}^2} = \frac{\text{hypotenuse}^2}{\text{hypotenuse}^2}$$ $$\bigg(\frac{\text{opposite}}{\text{hypotenuse}}\bigg)^2 + \bigg(\frac{\text{adjacent}}{\text{hypotenuse}}\bigg)^2=\bigg( \frac{\text{hypotenuse}}{\text{hypotenuse}}\bigg)^2$$ $$\sin^2\theta+\cos^2\theta=1^2$$ enter image description here

Lastly, try to derive the sum of angles formula by referring to the last diagram. Hint: use the law of sines.

enter image description here

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You have started off well. To get the values of $\cos x$ and $\sin y$, you should use the identity

$$\sin^2\theta + \cos^2\theta \equiv 1$$

I will show you how to find $\cos x$.

You can use the same method to find $\sin y$ yourself.

For example, if $\sin x = \frac{1}{3}$ then we have

\begin{eqnarray*} \sin^2 x + \cos^2x &\equiv& 1 \\ \\ \left(\frac{1}{3}\right)^{\! 2} + \cos^2x &=& 1 \\ \\ \frac{1}{9} + \cos^2x &=& 1 \\ \\ \cos^2x &=& \frac{8}{9} \\ \\ \cos x&=& \pm\frac{2}{3}\sqrt{2} \end{eqnarray*}

Next you need to decide on the $\pm$. Looking at the graph of $y=\cos x$ we see that $\cos x > 0$ if $0^{\circ} \le x < 90^{\circ}$ or $270^{\circ} < x \le 360^{\circ}$. So, for example: if we know that $x$ is acute then $$\cos x = +\frac{2}{3}\sqrt{2}$$

If we know that $x$ is obtuse then $$\cos x = -\frac{2}{3}\sqrt{2}$$

If $x$ is reflex and less that $270^{\circ}$ then $$\cos x = +\frac{2}{3}\sqrt{2}$$

If $x$ is reflex and more than $270^{\circ}$ then $$\cos x = -\frac{2}{3}\sqrt{2}$$

The same is true modulo $360^{\circ}$.

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  • $\begingroup$ The sine of $x$ is positive, which only ocurrs in the first and second quadrant, but that's where cossine is positive, so $\cos x$ is positive. However, for the sine of $y$, I seem to be in trouble. The cossine of $y$ is positive, which occurs in the first and fourth quadrant. The sine of $y$ is positive in the first, but negative in the fourth. So I cannot decide yet. Any advice? $\endgroup$ Commented Feb 27, 2023 at 2:11

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