How to use the product-to-sum formula to write the product as a sum or
difference?
4 cos pi/3 sin 5pi/6

Question

Asher Friedman · Accepted Answer

#"see explanation"# #"using the "color(blue)"product to sum/difference formula"# #color(red)(bar(ul(|color(white)(2/2)color(black)(2cosAsinB=sin(A+B)-sin(A-B))color(white)(2/2)|)))# #"here "A=pi/3" and "B=(5pi)/6# #rArr4cos(pi/3)sin((5pi)/6)# #=2[sin(pi/3+(5pi)/6)-sin(pi/3-(5pi)/6)]# #=2[sin((7pi)/6)-sin(-pi/2)]# #=2[-sin(pi/6)-sin(-pi/2)]# #=2(-1/2+1)# #=2xx1/2=1#

Brooklyn Anderson · Answer

undefined To use the product-to-sum formula to write the product $ 4 \cos(\frac{\pi}{3}) \sin(\frac{5\pi}{6}) $ as a sum or difference, you can apply the formula:

$$ \sin(A) \cos(B) = \frac{1}{2} [\sin(A + B) + \sin(A - B)] $$

In this case, $ A = \frac{\pi}{3} $ and $ B = \frac{5\pi}{6} $. Plug these values into the formula:

$$ \sin\left(\frac{\pi}{3}\right) \cos\left(\frac{5\pi}{6}\right) = \frac{1}{2} \left[\sin\left(\frac{\pi}{3} + \frac{5\pi}{6}\right) + \sin\left(\frac{\pi}{3} - \frac{5\pi}{6}\right)\right] $$

Now, simplify the expressions inside the sine functions:

$$ \sin\left(\frac{\pi}{3} + \frac{5\pi}{6}\right) = \sin\left(\frac{\pi}{2}\right) = 1 $$
$$ \sin\left(\frac{\pi}{3} - \frac{5\pi}{6}\right) = \sin\left(-\frac{\pi}{2}\right) = -1 $$

Substitute these values back into the formula:

$$ \frac{1}{2} [1 - (-1)] = \frac{1}{2} \times 2 = 1 $$

Therefore, $ 4 \cos\left(\frac{\pi}{3}\right) \sin\left(\frac{5\pi}{6}\right) $ can be written as $ 1 $.

How to use the product-to-sum formula to write the product as a sum or difference? 4 cos pi/3 sin 5pi/6

Need Help