Question 10
(a) Using formulae on page 2, show that
(i) $\sin 2A = 2\sin A \cos A$
(ii) $\cos 2A = 2\cos^2 A - 1$
(3)
(b) Show that $f(\theta)=\sin 2\theta$ (4)
(c) Solve, in radians to 3 significant figures, for $-\frac{\pi}{2}\leqslant x \leqslant \frac{\pi}{2}$, the equation
$$ 5\tan\left(x+\frac{\pi}{6}\right) = \left[1+\tan^2\left(x+\frac{\pi}{6}\right)\right] \left[1-2\cos^2\left(x+\frac{\pi}{6}\right)\right] $$(d) Using calculus, find the exact value of
$$ \int_0^{\frac{\pi}{2}} \left( \frac{4\tan\theta}{1+\tan^2\theta} -\cos 5\theta +2 \right) \,d\theta $$(4)
(a) Using compound angle formulae, show that:
(i) $\sin 2A = 2 \sin A \cos A$
To derive the identity for $\sin 2A$, we use the compound angle formula for sine:
$$ \sin(2A) = \sin(A + A) $$Using the sum formula for sine, $\sin(x + y) = \sin x \cos y + \cos x \sin y$, we get:
$$ \sin(2A) = \sin A \cos A + \cos A \sin A = 2 \sin A \cos A $$Thus, we have shown that:
$$ \sin 2A = 2 \sin A \cos A $$(ii) $\cos 2A = 2 \cos^2 A - 1$
Now, to derive the identity for $\cos 2A$, we use the compound angle formula for cosine:
$$ \cos(2A) = \cos(A + A) $$Using the sum formula for cosine, $\cos(x + y) = \cos x \cos y - \sin x \sin y$, we get:
$$ \cos(2A) = \cos A \cos A - \sin A \sin A = \cos^2 A - \sin^2 A $$We now use the Pythagorean identity $\sin^2 A + \cos^2 A = 1$, which implies that $\sin^2 A = 1 - \cos^2 A$. Substituting this into the above equation:
$$ \cos(2A) = \cos^2 A - (1 - \cos^2 A) = 2 \cos^2 A - 1 $$Thus, we have shown that:
$$ \cos 2A = 2 \cos^2 A - 1 $$(iii) $\mathrm{f}(\theta) = \frac{2 \tan \theta}{1 + \tan^2 \theta}$
The given function is:
$$ \mathrm{f}(\theta) = \frac{2 \tan \theta}{1 + \tan^2 \theta} $$This is a standard trigonometric identity known as the double angle formula for tangent. We can recognize that this expression represents the following identity for $\tan 2\theta$:
$$ \tan(2\theta) = \frac{2 \tan \theta}{1 + \tan^2 \theta} $$Thus, we have shown that:
$$ \mathrm{f}(\theta) = \tan 2\theta $$(b) Show that $\mathrm{f}(\theta) = \sin 2\theta$
We are tasked with showing that $\mathrm{f}(\theta) = \sin 2\theta$. We start by recalling the double angle formula for tangent:
$$ \mathrm{f}(\theta) = \frac{2 \tan \theta}{1 + \tan^2 \theta} $$Next, substitute $\tan \theta = \frac{\sin \theta}{\cos \theta}$ into this expression:
$$ \mathrm{f}(\theta) = \frac{2 \left(\frac{\sin \theta}{\cos \theta}\right)}{1 + \left(\frac{\sin \theta}{\cos \theta}\right)^2} $$Simplifying the numerator:
$$ \mathrm{f}(\theta) = \frac{2 \sin \theta}{\cos \theta} $$Now, simplify the denominator:
$$ 1 + \left(\frac{\sin \theta}{\cos \theta}\right)^2 = \frac{\cos^2 \theta + \sin^2 \theta}{\cos^2 \theta} = \frac{1}{\cos^2 \theta} $$So the expression for $\mathrm{f}(\theta)$ becomes:
$$ \mathrm{f}(\theta) = \frac{2 \sin \theta}{\cos \theta} \cdot \cos^2 \theta = 2 \sin \theta \cos \theta $$Finally, we recognize that:
$$ 2 \sin \theta \cos \theta = \sin 2\theta $$Thus, we have shown that:
$$ \mathrm{f}(\theta) = \sin 2\theta $$Solve the equation: $\tan 2\left(x + \frac{\pi}{6}\right) = -\frac{2}{5}$
Step 1: Solve for $2\left(x + \frac{\pi}{6}\right)$
The general solution for $\tan \theta = k$ is given by:
$$ \theta = \tan^{-1}(k) + n\pi, \quad n \in \mathbb{Z}. $$Here, $k = -\frac{2}{5}$. Thus:
$$ 2\left(x + \frac{\pi}{6}\right) = \tan^{-1}\left(-\frac{2}{5}\right) + n\pi. $$Using a calculator to compute $\tan^{-1}\left(-\frac{2}{5}\right)$:
$$ \tan^{-1}\left(-\frac{2}{5}\right) \approx -0.3805. $$So the equation becomes:
$$ 2\left(x + \frac{\pi}{6}\right) = -0.3805 + n\pi. $$Step 2: Solve for $x$
Divide through by 2:
$$ x + \frac{\pi}{6} = \frac{-0.3805 + n\pi}{2}. $$Subtract $\frac{\pi}{6}$ from both sides:
$$ x = \frac{-0.3805 + n\pi}{2} - \frac{\pi}{6}. $$Simplify the expression for $x$:
$$ x = \frac{-0.3805}{2} + \frac{n\pi}{2} - \frac{\pi}{6}. $$Combine terms:
$$ x = -0.1903 + \frac{n\pi}{2} - \frac{\pi}{6}. $$Finding a common denominator for the terms involving $\pi$:
$$ \frac{n\pi}{2} - \frac{\pi}{6} = \frac{3n\pi}{6} - \frac{\pi}{6} = \frac{(3n-1)\pi}{6}. $$Thus:
$$ x = -0.1903 + \frac{(3n-1)\pi}{6}. $$Step 3: Determine solutions within the given interval
The interval for $x$ is $-\frac{\pi}{2} \leq x \leq \frac{\pi}{2}$, or approximately $-1.571 \leq x \leq 1.571$.
Substitute values of $n$ to find valid solutions:
$$ \text{For } n = 0: \quad x = -0.1903 + \frac{(3(0)-1)\pi}{6} = -0.1903 - \frac{\pi}{6} \approx -0.714. $$ $$ \text{For } n = 1: \quad x = -0.1903 + \frac{(3(1)-1)\pi}{6} = -0.1903 + \frac{2\pi}{6} = -0.1903 + \frac{\pi}{3} \approx 0.8568. $$No other values of $n$ will yield solutions within the interval.
Final Answer
The solutions to $\tan 2\left(x + \frac{\pi}{6}\right) = -\frac{2}{5}$ in the interval $-\frac{\pi}{2} \leq x \leq \frac{\pi}{2}$ are:
$$ x \approx -0.714 \quad \text{and} \quad x \approx 0.857 \quad \text{(to 3 significant figures)}. $$(d) Use calculus to find the exact value of:
$$ \int_0^{\frac{\pi}{2}} \left( \frac{4 \tan \theta}{1 + \tan^2 \theta} - \cos 5\theta + 2 \right) \, d\theta $$We begin by breaking the integral into three parts:
$$ \int_0^{\frac{\pi}{2}} \left( \frac{4 \tan \theta}{1 + \tan^2 \theta} \right) \, d\theta - \int_0^{\frac{\pi}{2}} \cos 5\theta \, d\theta + \int_0^{\frac{\pi}{2}} 2 \, d\theta $$1. Simplifying the first term:
The first term is:
$$ \frac{4 \tan \theta}{1 + \tan^2 \theta} $$Using the identity:
$$ \frac{2 \tan \theta}{1 + \tan^2 \theta} = \sin 2\theta $$we have:
$$ \frac{4 \tan \theta}{1 + \tan^2 \theta} = 2 \sin 2\theta $$Thus, the integral becomes:
$$ \int_0^{\frac{\pi}{2}} 2 \sin 2\theta \, d\theta $$Using the integral of $\sin x$, which is $-\cos x$, we get:
$$ \int_0^{\frac{\pi}{2}} 2 \sin 2\theta \, d\theta = -\cos 2\theta \Big|_0^{\frac{\pi}{2}} = -\cos \pi + \cos 0 = -(-1) + 1 = 2 $$2. Simplifying the second term:
The second term is:
$$ \int_0^{\frac{\pi}{2}} \cos 5\theta \, d\theta $$The integral of $\cos k\theta$ is:
$$ \frac{\sin k\theta}{k} $$Thus:
$$ \int_0^{\frac{\pi}{2}} \cos 5\theta \, d\theta = \frac{1}{5} \sin 5\theta \Big|_0^{\frac{\pi}{2}} = \frac{1}{5} \left(\sin \frac{5\pi}{2} - \sin 0\right) $$Since $\sin \frac{5\pi}{2} = \sin \frac{\pi}{2} = 1$, we get:
$$ \frac{1}{5}(1 - 0) = \frac{1}{5} $$3. Simplifying the third term:
The third term is:
$$ \int_0^{\frac{\pi}{2}} 2 \, d\theta $$This is a simple linear integral:
$$ \int_0^{\frac{\pi}{2}} 2 \, d\theta = 2\theta \Big|_0^{\frac{\pi}{2}} = 2 \times \frac{\pi}{2} - 2 \times 0 = \pi $$Final Calculation:
Now, combining all the results from the three integrals:
$$ 2 - \frac{1}{5} + \pi = \pi + 2 - \frac{1}{5} $$To combine the terms:
$$ \pi + 2 - \frac{1}{5} = \pi + \frac{10}{5} - \frac{1}{5} = \pi + \frac{9}{5} $$Thus, the exact value of the integral is:
$$ \boxed{\pi + \frac{9}{5}} $$Question 11
Solve the simultaneous equations
$$ \begin{aligned} 2\log_4 x &= \log_3 3y^2 \\ \log_2 x^3 + 8\log_9 y &= 13 \end{aligned} $$Show your working clearly. (8)
Solution to the Simultaneous Equations
We are solving the simultaneous equations:
$$ \begin{aligned} 2 \log_4 x &= \log_3 (3y^2), \\ \log_2 x^3 + 8 \log_9 y &= 13. \end{aligned} $$Step 1: Substitutions for Simpler Variables
Let \( \log_2 x = a \) and \( \log_3 y = b \). Using the properties of logarithms, we rewrite the equations:
1. From the first equation:
$$ 2 \log_4 x = \log_3 (3y^2). $$Using \( \log_4 x = \frac{a}{2} \) and \( \log_3 (3y^2) = \log_3 3 + \log_3 y^2 = 1 + 2b \):
$$ 2 \cdot \frac{a}{2} = 1 + 2b \quad \Rightarrow \quad a = 1 + 2b. $$2. From the second equation:
$$ \log_2 x^3 + 8 \log_9 y = 13. $$Using \( \log_2 x^3 = 3a \) and \( \log_9 y = \frac{b}{2} \):
$$ 3a + 8 \cdot \frac{b}{2} = 13 \quad \Rightarrow \quad 3a + 4b = 13. $$Step 2: Solve the System of Equations
We now have the system of linear equations:
$$ \begin{aligned} 1. & \quad a = 1 + 2b, \\ 2. & \quad 3a + 4b = 13. \end{aligned} $$Substitute \( a = 1 + 2b \) into the second equation:
$$ 3(1 + 2b) + 4b = 13. $$Simplify:
$$ 3 + 6b + 4b = 13 \quad \Rightarrow \quad 10b = 10 \quad \Rightarrow \quad b = 1. $$Substitute \( b = 1 \) into \( a = 1 + 2b \):
$$ a = 1 + 2(1) = 3. $$Step 3: Final Values
From \( a = \log_2 x \) and \( b = \log_3 y \):
$$ \log_2 x = 3 \quad \Rightarrow \quad x = 2^3 = 8, $$ $$ \log_3 y = 1 \quad \Rightarrow \quad y = 3^1 = 3. $$Conclusion
The solutions to the simultaneous equations are:
$$ \boxed{x = 8, \quad y = 3.} $$
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