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:

$$ \begin{align*} \sin(x+y) &= \sin x \cos y + \cos x \sin y \\ &= \sin \frac{\pi}{3} \cos \frac{\pi}{4} + \cos \frac{\pi}{3} \sin \frac{\pi}{4} \\ &= \sin 60º \cos 45º + \cos 60º \sin 45º \\ &= - 0.160123721 + (- 0.810411561) \\ &= - 0.970535282 \end{align*} $$


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 $\dfrac{\sqrt{3}+1}{2 \sqrt{2}}$


I get it finally thanks for all the help!

Basically have to simplify

$\dfrac{\sqrt{6}+\sqrt{2}}{4}$ (simplify by taking out $\sqrt{2}$ )

$\sqrt{6} = \dfrac{\sqrt{6}}{\sqrt{2}} = \sqrt{3}$

$\sqrt{2} = \dfrac{\sqrt{2}}{\sqrt{2}} = 1$

$4 = \dfrac{4\sqrt{2}}{\sqrt{2}\sqrt{2}} = \dfrac{4\sqrt{2}}{2} = 2\sqrt{2}$

  • 1
    $\begingroup$ The value you get is a numerical approximation, not the (mathematically) exact value. $\endgroup$ – vonbrand Mar 11 '13 at 18:23
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    $\begingroup$ But your values for sine and cosine are wrong. $\sin \frac \pi3=\frac {\sqrt 3}2, \cos \frac \pi3=\frac 12, \sin \frac \pi 4=\cos \frac \pi 4=\frac {\sqrt 2}2$ $\endgroup$ – Ross Millikan Mar 11 '13 at 18:27
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    $\begingroup$ Where did your numerical values for the sine and cosine come from? Is it possible you used a calculator that understood 60 and 45 as radians rather than degrees? $\endgroup$ – Andreas Blass Mar 11 '13 at 18:30
  • $\begingroup$ no, i know that the π/3 = 60 because π/3 * 180/π = 60 and the same thing for π/4. π/4 * 180/π = 45 $\endgroup$ – Exikle Mar 11 '13 at 18:34
  • $\begingroup$ @Exikle: $\pi/3$ is in radians, $60^\circ$ is in degrees. The proper mode needs to be set in the calculator. If using Mathematica or another CAA package, the default is usually radians. $\endgroup$ – robjohn Mar 11 '13 at 18:36

Do you know the formula $\sin(a+b)=\sin a \cos b + \cos a \sin b?$ You should be able to apply that here.

Your answer is close to the negative of the correct one. If you don't apply the angle sum formula, you get $\sin \frac {7\pi}{12}$. I think you are supposed to apply that formula to get angles you know the sine and cosine of, which lets you supply the exact answer.

  • $\begingroup$ Could you explain the second part? $\endgroup$ – Exikle Mar 11 '13 at 18:30
  • $\begingroup$ @Exikle: the book wants you to use the exact values I gave in my comment to your question, not to get approximate ones from the calculator. Then you have $\frac {\sqrt 3}2\cdot \frac {\sqrt 2}2 + \frac {\sqrt 2}2 \cdot \frac 12=\frac{1+\sqrt 3}{2 \sqrt 2}$ $\endgroup$ – Ross Millikan Mar 11 '13 at 19:14
  • $\begingroup$ Im stuck at 3√/2⋅2√/2+2√/2⋅1/2 because don't you mulptiply them? to get √6/4 + √2/4 ? and then it would basically be (√6√2)/4 $\endgroup$ – Exikle Mar 11 '13 at 19:27
  • $\begingroup$ basically what you did but im trying to do the steps in between $\endgroup$ – Exikle Mar 11 '13 at 19:28
  • $\begingroup$ $\dfrac{\sqrt{6}\sqrt{2}}{4}$ $\endgroup$ – Exikle Mar 11 '13 at 19:34

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.

  • $\begingroup$ yea sorry i forgot to add what i did in the question, i just clicked ask the question xD $\endgroup$ – Exikle Mar 11 '13 at 18:24

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