Prove (3 A − 3 A)/( A − A) = 1 + 1/2 2A and solve 42 theta = 3 for 0^

Question

Prove (3 A − 3 A)/( A − A) = 1 + 1/2 2A and solve 42 theta = 3 for 0^ <= theta <= (pi^c)/(2).

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Here are the solutions to the problems.

1. Prove that: $\frac{\sin^3 A - \cos^3 A}{\sin A - \cos A} = 1 + \frac{1}{2} \sin 2A$

Start with the Left Hand Side (LHS).
$LHS = \frac{\sin^3 A - \cos^3 A}{\sin A - \cos A}$
Apply the algebraic identity $a^3 - b^3 = (a - b)(a^2 + ab + b^2)$ .
Here, $a = \sin A$ and $b = \cos A$ .
$LHS = \frac{(\sin A - \cos A)(\sin^2 A + \sin A \cos A + \cos^2 A)}{\sin A - \cos A}$
Cancel out the common term $(\sin A - \cos A)$ from the numerator and denominator.
$LHS = \sin^2 A + \sin A \cos A + \cos^2 A$
Use the trigonometric identity $\sin^2 A + \cos^2 A = 1$ .
$LHS = 1 + \sin A \cos A$
Use the double angle identity $\sin 2A = 2 \sin A \cos A$ , which implies $\sin A \cos A = \frac{1}{2} \sin 2A$ .
$LHS = 1 + \frac{1}{2} \sin 2A$
Compare with the Right Hand Side (RHS).
$RHS = 1 + \frac{1}{2} \sin 2A$
Since $LHS = RHS$ , the identity is proven.
$\boxed{\frac{\sin^3 A - \cos^3 A}{\sin A - \cos A} = 1 + \frac{1}{2} \sin 2A}$
2. Solve: $4\cos^2 \theta = 3 \quad \left(0^\circ \leq \theta \leq \frac{\pi^c}{2}\right)$
Isolate $\cos^2 \theta$ .
$4\cos^2 \theta = 3$ $\cos^2 \theta = \frac{3}{4}$
Take the square root of both sides.
$\cos \theta = \pm \sqrt{\frac{3}{4}}$ $\cos \theta = \pm \frac{\sqrt{3}}{2}$
Consider the given range for $\theta$ .
The range is $0^\circ \leq \theta \leq \frac{\pi^c}{2}$ , which means $0^\circ \leq \theta \leq 90^\circ$ . In this first quadrant, the cosine function is positive.
Therefore, we only consider the positive value:
$\cos \theta = \frac{\sqrt{3}}{2}$
Find the value of $\theta$ .
We know that $\cos 30^\circ = \frac{\sqrt{3}}{2}$ .
In radians, $30^\circ = \frac{\pi}{6}$ radians.
So, $\theta = 30^\circ$ or $\theta = \frac{\pi}{6}$ .

The value of $\theta$ is $\boxed{30^\circ \text{ or } \frac{\pi}{6}}$ .

3. The position vectors of the points A and B are $2\vec{i} + 3\vec{j}$ and $-3\vec{i} + 8\vec{j}$ respectively. If M divides the line segment AB externally in the ratio 5:2, then find the position vector of the point M.
Write down the given position vectors and the ratio.
Position vector of A: $\vec{a} = 2\vec{i} + 3\vec{j}$
Position vector of B: $\vec{b} = -3\vec{i} + 8\vec{j}$
Ratio of external division $m:n = 5:2$ . So, $m=5$ and $n=2$ .
Use the formula for the position vector of a point M that divides a line segment AB externally in the ratio $m:n$ .
$\vec{m} = \frac{m\vec{b} - n\vec{a}}{m - n}$
Substitute the given values into the formula.
$\vec{m} = \frac{5(-3\vec{i} + 8\vec{j}) - 2(2\vec{i} + 3\vec{j})}{5 - 2}$
Simplify the expression.
$\vec{m} = \frac{(-15\vec{i} + 40\vec{j}) - (4\vec{i} + 6\vec{j})}{3}$ $\vec{m} = \frac{-15\vec{i} + 40\vec{j} - 4\vec{i} - 6\vec{j}}{3}$ $\vec{m} = \frac{(-15 - 4)\vec{i} + (40 - 6)\vec{j}}{3}$ $\vec{m} = \frac{-19\vec{i} + 34\vec{j}}{3}$ $\vec{m} = -\frac{19}{3}\vec{i} + \frac{34}{3}\vec{j}$
The position vector of point M is $\boxed{-\frac{19}{3}\vec{i} + \frac{34}{3}\vec{j}}$ .

4. In a series, the first quartile is 45 and the quartile deviation is 15. Find the coefficient of quartile deviation.
Write down the given values.
First quartile ( $Q_1$ ) = 45
Quartile Deviation (QD) = 15
Use the formula for Quartile Deviation to find the third quartile ( $Q_3$ ).
$QD = \frac{Q_3 - Q_1}{2}$ $15 = \frac{Q_3 - 45}{2}$ $15 \times 2 = Q_3 - 45$ $30 = Q_3 - 45$ $Q_3 = 30 + 45$ $Q_3 = 75$
Use the formula for the coefficient of quartile deviation.
$\text{Coefficient of QD} = \frac{Q_3 - Q_1}{Q_3 + Q_1}$
Substitute the values of $Q_1$ and $Q_3$ into the formula.
$\text{Coefficient of QD} = \frac{75 - 45}{75 + 45}$ $\text{Coefficient of QD} = \frac{30}{120}$ $\text{Coefficient of QD} = \frac{1}{4}$ $\text{Coefficient of QD} = 0.25$
The coefficient of quartile deviation is $\boxed{0.25}$ .

Prove (3 A - 3 A)/( A − A) = 1 + (1)/(2) 2A and solve 42 theta = 3 for 0^ <= theta <= (pi^c)/(2).

Step-by-step solution

Start with the Left Hand Side (LHS).

Apply the algebraic identity $a^3 - b^3 = (a - b)(a^2 + ab + b^2)$ .

Cancel out the common term $(\sin A - \cos A)$ from the numerator and denominator.

Use the trigonometric identity $\sin^2 A + \cos^2 A = 1$ .

Use the double angle identity $\sin 2A = 2 \sin A \cos A$ , which implies $\sin A \cos A = \frac{1}{2} \sin 2A$ .

Compare with the Right Hand Side (RHS).

Isolate $\cos^2 \theta$ .

Take the square root of both sides.

Consider the given range for $\theta$ .

Find the value of $\theta$ .

Write down the given position vectors and the ratio.

Use the formula for the position vector of a point M that divides a line segment AB externally in the ratio $m:n$ .

Substitute the given values into the formula.

Simplify the expression.

Write down the given values.

Use the formula for Quartile Deviation to find the third quartile ( $Q_3$ ).

Use the formula for the coefficient of quartile deviation.

Substitute the values of $Q_1$ and $Q_3$ into the formula.

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More Mathematics Questions

Prove (3 A - 3 A)/( A − A) = 1 + (1)/(2) 2A and solve 42 theta = 3 for 0^ <= theta <= (pi^c)/(2).

Step-by-step solution

Start with the Left Hand Side (LHS).

Apply the algebraic identity a3−b3=(a−b)(a2+ab+b2)a^3 - b^3 = (a - b)(a^2 + ab + b^2)a3−b3=(a−b)(a2+ab+b2).

Cancel out the common term (sin⁡A−cos⁡A)(\sin A - \cos A)(sinA−cosA) from the numerator and denominator.

Use the trigonometric identity sin⁡2A+cos⁡2A=1\sin^2 A + \cos^2 A = 1sin2A+cos2A=1.

Use the double angle identity sin⁡2A=2sin⁡Acos⁡A\sin 2A = 2 \sin A \cos Asin2A=2sinAcosA, which implies sin⁡Acos⁡A=12sin⁡2A\sin A \cos A = \frac{1}{2} \sin 2AsinAcosA=21​sin2A.