Problem Analysis
The question asks for the value of the expression $\sin \theta - \cos \theta$, given that $\sin \theta + \cos \theta = \frac{\sqrt{7}}{2}$.
Solving Trigonometric Expressions
We can solve this problem using algebraic manipulation combined with trigonometric identities.
Let $A = \sin \theta + \cos \theta$ and $B = \sin \theta - \cos \theta$. We are given $A = \frac{\sqrt{7}}{2}$ and need to find $B$.
Step-by-Step Calculation
- Consider the square of the given expression: $A^2 = (\sin \theta + \cos \theta)^2$ $A^2 = \sin^2 \theta + \cos^2 \theta + 2 \sin \theta \cos \theta$
- Using the identity $\sin^2 \theta + \cos^2 \theta = 1$: $A^2 = 1 + 2 \sin \theta \cos \theta$
- Substitute the given value $A = \frac{\sqrt{7}}{2}$: $(\frac{\sqrt{7}}{2})^2 = 1 + 2 \sin \theta \cos \theta$ $\frac{7}{4} = 1 + 2 \sin \theta \cos \theta$
- Calculate the value of $2 \sin \theta \cos \theta$: $2 \sin \theta \cos \theta = \frac{7}{4} - 1$ $2 \sin \theta \cos \theta = \frac{3}{4}$
- Now consider the square of the expression we need to find: $B^2 = (\sin \theta - \cos \theta)^2$ $B^2 = \sin^2 \theta + \cos^2 \theta - 2 \sin \theta \cos \theta$
- Substitute $\sin^2 \theta + \cos^2 \theta = 1$ and the value of $2 \sin \theta \cos \theta$ from step 4: $B^2 = 1 - (\frac{3}{4})$ $B^2 = \frac{1}{4}$
- Solve for B: $B = \pm \sqrt{\frac{1}{4}}$ $B = \pm \frac{1}{2}$
The possible values for $\sin \theta - \cos \theta$ are $\frac{1}{2}$ and $-\frac{1}{2}$. Since $\frac{1}{2}$ is an option, it is the required value.
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Important Questions from Trigonometry
The given equation can be reduced to
If sin2x = a - b√c, where a and b are natural numbers and c is prime number, then what is the value of a - b + 2c ?
If \(\sin θ = \frac{8}{{17}}\) , then find the value of tan θ.
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If 3 tanθ = \(2\sqrt 3 \) sinθ, 0° < θ < 90°, then the value of \(\rm \frac{{\cos e{c^2}2\,\theta + {{\cot }^2}2\,\theta }}{{{{\sin }^2}\,\theta + {{\tan }^2}2\,\theta }}\) is:
