Further Mathematics Questions and Answers

\(2\sin^{2}\theta = 1 + \cos \theta \implies 2(1 - \cos^{2}\theta) = 1 + \cos \theta\)

\(2 - 2\cos^{2}\theta = 1 + \cos \theta\)

\(2 - 2\cos^{2}\theta - 1 - \cos \theta = 0\)

\(2\cos^{2}\theta + \cos \theta - 1 = 0\)

\(2\cos^{2}\theta + 2\cos\theta - \cos \theta - 1 = 0 \implies 2\cos \theta(\cos \theta + 1) - 1(\cos \theta + 1) = 0\)

\((2\cos \theta - 1)(\cos \theta + 1) = 0 \implies \cos \theta = \frac{1}{2} \)

\(\theta = \cos^{-1} \frac{1}{2} = 60°\)

\(s = 3i - j; t = 2i + 3j\)

\( t - 3s = (2i + 3j) - 3(3i - j) = 2i + 3j - 9i + 3j = -7i + 6j\)

\(t + 3s = (2i + 3j) + 3(3i - j) = 2i + 3j + 9i - 3j = 11i\)

\((t - 3s).(t + 3s) = (-7i + 6j).(11i) = -77\)

No explanation has been provided for this answer.

The equation of a circle is given as \((x - a)^{2} + (y - b)^{2} = r^{2}\).

Expanding, we have: \(x^{2} - 2ax + a^{2} + y^{2} - 2by + b^{2} = r^{2}\)

\(x^{2} + y^{2} - 2ax - 2by + a^{2} + b^{2} = r^{2}\)

Comparing with the equation, \(x^{2} + y^{2} - 8x + 9y = -15\), we have

\(2a = 8; 2b = -9; r^{2} - a^{2} - b^{2} = -15\)

\(a = 4; b = \frac{-9}{2}\)

\(\therefore  r^{2} = -15 + 4^{2} + (\frac{-9}{2})^{2}\)

= \(-15 + 16 + \frac{81}{4} = \frac{85}{4}\)

\(r = \sqrt{\frac{85}{4} = \frac{1}{2}\sqrt{85}\)

Since the bodies are in an opposite direction, one takes the negative velocity.

\(m_{1}v_{1} + m_{2}v_{2} = (m_{1} + m_{2})v\) (Momentum when the two bodies move in the same direction after collision)

\(3(-2) + 5(V) = (3 + 5)3.5\)

\(-6 + 5V = 28 \implies 5V = 34; V = 6.8 m/s\)