Finding sin(theta) + cos(theta) Value
We are given the equation:
$\sin\theta - \cos\theta = \frac{\sqrt{3}}{2}$
We need to find the positive value of $\sin\theta + \cos\theta$. Let's denote this value as $x$.
$x = \sin\theta + \cos\theta$
Algebraic Manipulation
Square both the given equation and the equation for $x$:
- Squaring the given equation: $(\sin\theta - \cos\theta)^2 = \left(\frac{\sqrt{3}}{2}\right)^2$ $\sin^2\theta - 2\sin\theta\cos\theta + \cos^2\theta = \frac{3}{4}$ Using the identity $\sin^2\theta + \cos^2\theta = 1$: $1 - 2\sin\theta\cos\theta = \frac{3}{4}$ $2\sin\theta\cos\theta = 1 - \frac{3}{4} = \frac{1}{4}$
- Squaring the equation for $x$: $x^2 = (\sin\theta + \cos\theta)^2$ $x^2 = \sin^2\theta + 2\sin\theta\cos\theta + \cos^2\theta$ Using the identity $\sin^2\theta + \cos^2\theta = 1$: $x^2 = 1 + 2\sin\theta\cos\theta$
Calculating the Final Value
Substitute the value of $2\sin\theta\cos\theta$ found earlier into the equation for $x^2$:
$x^2 = 1 + \frac{1}{4}$
$x^2 = \frac{5}{4}$
Now, solve for $x$:
$x = \pm\sqrt{\frac{5}{4}}$
$x = \pm\frac{\sqrt{5}}{2}$
The question asks for the positive value.
Therefore, the positive value of $\sin\theta + \cos\theta$ is $\frac{\sqrt{5}}{2}$.
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Important Questions from Trigonometry
The given equation can be reduced to
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