I need help understanding why how the textbook got a certain answer but I got a completly different answer.
The question was to apply a compound angle formula, and then determine and exact value for each. It's two questions (a&b) but I think that if someone can help me understand one I'll get the other.
a) $\sin(\pi/3 + \pi/4)$
The back of the text book says $\dfrac{\sqrt{3}+1}{2 \sqrt{2}}$
I got -0.970535282
My method was:
$$
\begin{align*}
\sin(x+y) &= \sin x \cos y + \cos x \sin y \\
&= \sin \frac{\pi}{3} \cos \frac{\pi}{4}
+ \cos \frac{\pi}{3} \sin \frac{\pi}{4} \\
&= \sin 60º \cos 45º + \cos 60º \sin 45º \\
&= – 0.160123721 + (- 0.810411561) \\
&= – 0.970535282
\end{align*}
$$
EDIT:
So so far after all the help I have recieved i have done
$$
\begin{align*}
\sin(x+y) &= \sin x \cos y + \cos x \sin y \\
&= \sin \frac{\pi}{3} \cos \frac{\pi}{4}
+ \cos \frac{\pi}{3} \sin \frac{\pi}{4} \\
&=(\dfrac{\sqrt{3}}{4} * \dfrac{\sqrt{2}}{4}) + (\dfrac1{2} * \dfrac{\sqrt{2}} {2})\\
&=\dfrac{\sqrt{6}+\sqrt{2}}{4}\\
&=\dfrac{\sqrt{3}+1}{2 \sqrt{2}}\\
\end{align*}
$$
But I'm a bit confused as to what i have to do next to get $\dfrac{\sqrt{3}+1}{2 \sqrt{2}}$
EDIT 2:
I get it finally thanks for all the help!
Basically have to simplify
$\dfrac{\sqrt{6}+\sqrt{2}}{4}$
(simplify by taking out $\sqrt{2}$ )
$\sqrt{6} = \dfrac{\sqrt{6}}{\sqrt{2}} = \sqrt{3}$
$\sqrt{2} = \dfrac{\sqrt{2}}{\sqrt{2}} = 1$
$4 = \dfrac{4\sqrt{2}}{\sqrt{2}\sqrt{2}} = \dfrac{4\sqrt{2}}{2} = 2\sqrt{2}$
Best Answer
Hint: use the formula $\sin(a+b)=\sin(a)\cos(b)+\cos(a)\sin(b)$ and the values given in the book for the trigonometric functions of $\pi/3$ and $\pi/4$.
Suggestion: The answer sought in most books is exact rather than plugging into a calculator. Usually books have a table of trigonometric functions for some special values, such as π/3 and π/4. Find those and use them.