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:


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


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

  • $\begingroup$ Please consider using \sin and \cos in "Math mode" to get $\sin$ and $\cos$ instead of simply sin and cos which give $sin$ and $cos$. $\endgroup$ – Fly by Night Sep 4 '15 at 22:43

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

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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


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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