Trigonometric Value Calculation
We are given the equation: $ \sin \theta + \text{cosec } \theta = \sqrt{5} $ We need to find the value of $\sin^3 \theta + \text{cosec}^3 \theta$.
Applying Algebraic Identity
Let $a = \sin \theta$. Then $\text{cosec } \theta = \frac{1}{\sin \theta} = \frac{1}{a}$. The given equation becomes: $ a + \frac{1}{a} = \sqrt{5} $ We need to find the value of $a^3 + \frac{1}{a^3}$. We can use the algebraic identity: $ x^3 + y^3 = (x+y)^3 - 3xy(x+y) $ Substitute $x = a$ and $y = \frac{1}{a}$: $ a^3 + \frac{1}{a^3} = \left(a + \frac{1}{a}\right)^3 - 3 \left(a \cdot \frac{1}{a}\right) \left(a + \frac{1}{a}\right) $
Step-by-Step Calculation
- Substitute the known values into the identity. We know $a + \frac{1}{a} = \sqrt{5}$ and $a \cdot \frac{1}{a} = 1$. $ a^3 + \frac{1}{a^3} = (\sqrt{5})^3 - 3(1)(\sqrt{5}) $
- Calculate $(\sqrt{5})^3$: $ (\sqrt{5})^3 = \sqrt{5} \times \sqrt{5} \times \sqrt{5} = 5\sqrt{5} $
- Substitute this back into the equation: $ a^3 + \frac{1}{a^3} = 5\sqrt{5} - 3\sqrt{5} $
- Simplify the expression: $ a^3 + \frac{1}{a^3} = 2\sqrt{5} $
Therefore, the value of $\sin^3 \theta + \text{cosec}^3 \theta$ is $2\sqrt{5}$.
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