If sin(x) + cos(x) = 1/sqrt(2), find the value of sin(2x).
Understand the Problem
The question provides an equation involving sin(x) and cos(x), and asks to find the value of sin(2x). To solve this, we can square the given equation and use trigonometric identities to relate it to sin(2x).
Answer
$-\frac{3}{4}$
Answer for screen readers
$-\frac{3}{4}$
Steps to Solve
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Square both sides of the given equation Given $\sin(x) + \cos(x) = \frac{1}{2}$, square both sides of the equation: $$(\sin(x) + \cos(x))^2 = \left(\frac{1}{2}\right)^2$$
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Expand the left side of the equation Expanding the left side gives: $$\sin^2(x) + 2\sin(x)\cos(x) + \cos^2(x) = \frac{1}{4}$$
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Use the Pythagorean identity Recall the Pythagorean identity: $\sin^2(x) + \cos^2(x) = 1$. Substitute this into the equation: $$1 + 2\sin(x)\cos(x) = \frac{1}{4}$$
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Isolate the term $2\sin(x)\cos(x)$ Subtract 1 from both sides: $$2\sin(x)\cos(x) = \frac{1}{4} - 1 = \frac{1}{4} - \frac{4}{4} = -\frac{3}{4}$$
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Use the double angle identity Recall the double angle identity: $\sin(2x) = 2\sin(x)\cos(x)$. Substitute this into the equation: $$\sin(2x) = -\frac{3}{4}$$
$-\frac{3}{4}$
More Information
The value of $\sin(2x)$ is $-\frac{3}{4}$. This falls within the range of the sine function, which is between -1 and 1.
Tips
A common mistake is not squaring the entire binomial $(\sin(x) + \cos(x))$ correctly. Remember that $(a+b)^2 = a^2 + 2ab + b^2$. Another common mistake is not remembering the trigonometric identities $\sin^2(x) + \cos^2(x) = 1$ and $\sin(2x) = 2\sin(x)\cos(x)$.
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