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

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Answer 1

Asher Friedman

#"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#

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Answer 2

Brooklyn Anderson

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 $1$ .

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.