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forsale [732]
3 years ago
8

Write the product as a sum: 14cos(39x)sin(19x)

Mathematics
1 answer:
charle [14.2K]3 years ago
4 0

Recall that

sin(<em>a</em> + <em>b</em>) = sin(<em>a</em>) cos(<em>b</em>) + cos(<em>a</em>) sin(<em>b</em>)

sin(<em>a</em> - <em>b</em>) = sin(<em>a</em>) cos(<em>b</em>) - cos(<em>a</em>) sin(<em>b</em>)

Adding these together gives

sin(<em>a</em> + <em>b</em>) + sin(<em>a</em> - <em>b</em>) = 2 sin(<em>a</em>) cos(<em>b</em>)

To get 14 cos(39<em>x</em>) sin(19<em>x</em>) on the right side, multiply both sides by 7 and replace <em>a</em> = 19<em>x</em> and <em>b</em> = 39<em>x</em> :

7 (sin(19<em>x</em> + 39<em>x</em>) + sin(19<em>x</em> - 39<em>x</em>)) = 14 cos(39<em>x</em>) sin(19<em>x</em>)

7 (sin(58<em>x</em>) + sin(-20<em>x</em>)) = 14 cos(39<em>x</em>) sin(19<em>x</em>)

7 (sin(58<em>x</em>) - sin(20<em>x</em>)) = 14 cos(39<em>x</em>) sin(19<em>x</em>)

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Since we know that all angles in a triangle add up to 180^{\circ}, we can solve for x and substitute it back into (x-5)^{\circ} to find m\angle O.

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\begin{aligned}x=16\end{aligned}

Now that we have the value of x, we can substitute it back into (x-5)^{\circ} to find m\angle O.

\begin{aligned}m\angle O&=(x-5)\\&=16-5\\&=\boxed{11}\end{aligned}

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