Trigonometry & Inverse Trigonometry Question - None

Question

Ex. 60: If $x\cos\alpha + y\sin\alpha = x\cos\beta + y\sin\beta = 2a$ where $0 < \alpha, \beta < \frac{\pi}{2}$, then

(a) $\cos\alpha + \cos\beta = \frac{4ax}{x^2 + y^2}$

(b) $\cos\alpha\cos\beta = \frac{4a^2 - y^2}{x^2 + y^2}$

(c) $\sin\alpha + \sin\beta = \frac{4ay}{x^2 + y^2}$

(d) $\sin\alpha\sin\beta = \frac{4a^2 - x^2}{x^2 + y^2}$

Accepted Answer

Step 1: From the given condition $x\cos heta + y\sin heta = 2a$, square both sides:

$$x^2\cos^2 heta + y^2\sin^2 heta + 2xy\cos heta\sin heta = 4a^2$$

Step 2: Rewrite: $$(x^2 + y^2)\cos^2 heta - 4ax\cos heta + 4a^2 - y^2 = 0$$

Step 3: This is a quadratic in $\cos heta$ with roots $\cos\alpha$ and $\cos\beta$.

Step 4: By Vieta's formulas:

$$\cos\alpha + \cos\beta = \frac{4ax}{x^2 + y^2}$$

$$\cos\alpha\cos\beta = \frac{4a^2 - y^2}{x^2 + y^2}$$

Step 5: Similarly, derive the quadratic in $\sin heta$:

$$(x^2 + y^2)\sin^2 heta - 4ay\sin heta + 4a^2 - x^2 = 0$$

Step 6: By Vieta's formulas:

$$\sin\alpha + \sin\beta = \frac{4ay}{x^2 + y^2}$$

$$\sin\alpha\sin\beta = \frac{4a^2 - x^2}{x^2 + y^2}$$

∴ Answer is (a), (b), (c), and (d).

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