Title: Default for 4.3 Trigonometric Identities
Course: HELM Workbooks 2023-2024

Trigonometric identities

An identity is a relation which is always true. To emphasise this the symbol ‘\(\equiv\)’ is often used rather than ‘\(=\)’. For example, \((x+1)^2\equiv x^2+2x+1\) (always true) but \((x+1)^2=0\) (only true for \(x=-1\)).

Key Point

For any value of \(\theta\) \[\sin^2\theta+\cos^2\theta\equiv1\]

One way of proving the result in the Key Point is to use the definitions of \(\sin\theta\) and \(\cos\theta\) obtained from the circle of unit radius.

Recall that \(\cos\theta=OQ\), \(\ \sin\theta=OR=QP\). By Pythagoras’ theorem:\((OQ)^2+(QP)^2=(OP)^2=1\)

hence \(\cos^2\theta+\sin^2\theta=1.\)

We have demonstrated the Key Point using an angle \(\theta\) in the first quadrant but the result is true for any \(\theta\) i.e. it is indeed an identity.

Sum of angles identity

This Key Point introduces two further important identities.

Key Point

\[\begin{aligned} \sin (A+B) &\equiv \sin A\cos B+\cos A\sin B \qquad \qquad (1) \\ \cos(A+B) &\equiv \cos A\cos B-\sin A\sin B \qquad \qquad (2) \end{aligned}\]

Note carefully the addition sign in \((1)\) but the subtraction sign in \((2)\). Further identities can readily be obtained from \((1)\) and \((2)\).

Dividing \((1)\) by \((2)\) we obtain

\[\tan(A+B)\equiv \dfrac{\sin(A+B)}{\cos(A+B)}\equiv \dfrac{\sin A\cos B+\cos A\sin B}{\cos A\cos B-\sin A\sin B}\]

Dividing every term by \(\cos A\ \cos B\) we obtain:

\[\tan (A+B)\equiv \dfrac{\tan A+\tan B}{1-\tan A\tan B}\]

Replacing \(B\) by \(-B\) in \((1)\) and \((2)\) and remembering that \(\cos (-B)\equiv \cos B\) and \(\sin(-B)\equiv -\sin B\) we find

\[\begin{aligned} \sin(A-B) &\equiv \sin A\cos B-\cos A\sin B \\ \cos (A-B) &\equiv \cos A\cos B+\sin A\sin B. \end{aligned}\]

Key Point

\[\begin{aligned} \tan (A+B) &\equiv \dfrac{\tan A+\tan B}{1-\tan A\tan B} \\ \sin(A-B) &\equiv \sin A\cos B-\cos A\sin B \\ \cos (A-B) &\equiv \cos A\cos B+\sin A\sin B \\ \tan(A-B) &\equiv \dfrac{\tan A-\tan B}{1+\tan A\tan B} \end{aligned}\]

Example

Amplitude modulation

Introduction

Amplitude Modulation (the AM in AM radio) is a method of sending electromagnetic signals of a certain frequency (signal frequency) at another frequency (carrier frequency) which may be better for transmission. Modulation can be represented by the multiplication together of the carrier and modulating signals. To demodulate the signal the carrier frequency must be removed from the modulated signal.

Problems in words

1. A single frequency of \(200\ \mathrm{Hz}\) (message signal) is amplitude modulated with a carrier frequency of \(2\ \mathrm{MHz}\). Show that the modulated signal can be represented by the sum of two frequencies at \(2 \times 10^6 \pm 200\ \mathrm{Hz}\)
2. Show that the modulated signal can be demodulated by using a locally generated carrier and applying a low-pass filter.

Mathematical statement of problem

1. Express the message signal as \(m=a\cos(\omega_mt)\) and the carrier as \(c=b\cos(\omega_c t)\).
Assume that the modulation gives the product \(mc=ab\cos(\omega_ct)\cos(\omega_mt)\).
Use trigonometric identities to show that
\[mc=ab\cos(\omega_ct)\cos( \omega_mt)=k_1\cos((\omega_c- \omega_m)t)+k_2\cos((\omega_c+ \omega_m)t),\]
where \(k_1\) and \(k_2\) are constants.
Then substitute \( \omega_c= 2 \times 10^6 \times 2\pi \) and \( \omega_m= 200 \times 2\pi \) to calculate the two resulting frequencies.
2. Use trigonometric identities to show that multiplying the modulated signal by \(b\cos ( \omega_ct)\) results in the lowest frequency component of the output having a frequency equal to the original message signal.

Mathematical analysis

1. The message signal has a frequency of \(f_m=200\ \mathrm{Hz}\) so \( \omega_m = 2\pi f_c= 2\pi \times 200 = 400\pi \) radians per second.
The carrier signal has a frequency of \(f_c=2 \times 10^6\ \mathrm{Hz}\).
Hence \(\omega_c = 2\pi f_c=2\pi\times 2 \times10^6 = 4\times 10^6\pi\) radians per second.
So \(mc=ab\cos(4\times 10^6\pi t)\cos(400\pi t)\).
We know the identity
\[\cos(A+B)+\cos(A-B)\equiv 2\cos(A)\cos(B)\]
Rearranging gives the identity:
\[\cos(A)\cos(B)\equiv \frac{1}{2}(\cos(A+B)+\cos(A-B)) \tag{1}\]
Using \((1)\) with \(A=4\times 10^6\pi t\) and \(B= 400\pi t\) gives
\[\begin{aligned} mc =& ab(\cos(4\times10^6\pi t)\cos (400\pi t))\\ =&\frac{1}{2}ab(\cos(4\times 10^6\pi t+400\pi t)+ \cos(4\times 10^6\pi t-400\pi t))\\ =&\frac{1}{2}ab(\cos(4000400\pi t)+\cos(3999600\pi t)). \end{aligned}\]
So the modulated signal is the sum of two waves with angular frequency of \(4000400\)\(\pi\) and \(3999600\)\(\pi\) radians per second corresponding to frequencies of \(4000400\pi/(2\pi)\) and \(39996000\pi /(2\pi)\), that is \(2000200\ \mathrm{Hz}\) and \(1999800\ \mathrm{Hz}\), i.e. \(2 \times 10^6 \pm 200\ \mathrm{Hz}\).
2. Taking identity \((1)\) and multiplying through by \(\cos(A)\) gives
\[\cos(A)\cos(A)\cos(B)\equiv \frac{1}{2}\cos (A)(\cos(A+B)+\cos(A-B))\]
so
\[\cos(A)\cos(A)\cos(B)\equiv \frac{1}{2}(\cos (A)\cos(A+B)+\cos (A)\cos(A-B)) \tag{2}\]
Identity \((1)\) can be applied to both expressions in the right-hand side of \((2)\). In the first expression, using ‘\(A+B\)’ instead of ‘\(B\)’, gives
\[\cos(A)\cos(A+B)\equiv \frac{1}{2}(\cos(A+A+B)+\cos(A-A-B))\equiv \frac{1}{2}(\cos (2A+B)+\cos (B)),\]
where we have used \(\cos(-B)\equiv \cos(B).\).
Similarly, in the second expression, using ‘\(A-B\)’ instead of ‘\(B\)’, gives
\[\cos(A)\cos(A-B)\equiv \frac{1}{2}(\cos(2A-B)+\cos(B))\]
Together these give:
\[\begin{aligned} \cos(A)\cos(A)\cos(B) \equiv & \textstyle\frac{1}{4}(\cos(2A+B)+\cos(B)+\cos(2A-B)+\cos(B))\\ \equiv&\textstyle\frac{1}{2}\cos(B)+\textstyle\frac{1}{4}(\cos(2A+B)+\cos(2A-B))\qquad\qquad\qquad\qquad \end{aligned}\]
With \(A=4\times 10^6\pi t\) (carrier) and \(B=400\pi t\) (message) and substituting for the given frequencies, the modulated signal multiplied by the original carrier signal gives
\[ab^2\cos(4\times 10^6\pi t)\cos(4\times 10^6\pi t)\cos(400\pi t)=\] \[\frac{1}{2}ab^2\cos(2\pi\times 200t)+\frac{1}{4}ab^2(\cos(2\times 4\times10^6\pi t+400\pi t) + \cos(2\times 4\times 10^6\pi t-400\pi t))\]
The last two terms have frequencies of \(4\times 10^6 \pm 200\ \mathrm{Hz}\) which are sufficiently high that a low-pass filter would remove them and leave only the term
\[ab^2\cos(2\pi\times 200 t)\]
which is the original message signal multiplied by a constant term.

Interpretation

Amplitude modulation of a single frequency message signal \((f_m)\) with a single frequency carrier signal \((f_c)\) can be shown to be equal to the sum of two cosines with frequencies \(f_c\pm f_m\). Multiplying the modulated signal by a locally generated carrier signal and applying a low-pass filter can reproduce the frequency, \(f_m\), of the message signal.

This is known as double side band amplitude modulation.

Example

Obtain expressions for \(\cos\theta\) in terms of the sine function and for \(\sin\theta\) in terms of the cosine function.

Solution

Using

\[\cos (A-B)\equiv \cos A\cos B+\sin A\sin B\]

with \(A=\theta,\ B=\dfrac{\pi}{2}\) we obtain

\[ \cos\left(\theta-\dfrac{\pi}{2}\right)\equiv\cos\theta\ \cos\left(\dfrac{\pi}{2}\right)+\sin\theta\ \sin \left (\dfrac{\pi}{2}\right)\equiv \cos\theta\ (0)+\sin \theta\ \]

i.e.

\[\sin\theta\equiv\cos\left(\theta-\dfrac{\pi}{2}\right)\equiv\cos\left(\dfrac{\pi}{2}-\theta\right)\]

This result explains why the graph of \(\sin\theta\) has exactly the same shape as the graph of \(\cos\theta\) but it is shifted to the right by \(\dfrac{\pi}{2}\). A similar calculation yields

\[\cos\theta\equiv\sin\left(\theta+\dfrac{\pi}{2}\right).\]

Double angle formulae

If we put \(B=A\) in the identity \(\sin (A+B)\equiv\sin A\cos B+\cos A\sin B\) we obtain

Key Point

\[\sin 2A\equiv\sin A\cos A+\cos A\sin A\]

so

\[\sin 2A\equiv 2\sin A\cos A\]

Half-angle formulae

If we replace \(A\) by \(\dfrac{A}{2}\) and, consequently \(2A\) by \(A\), in the double angle formula, we obtain

\(\sin A\equiv 2\sin\left(\dfrac{A}{2}\right)\cos\left(\dfrac{A}{2}\right) \)

Similarly for the cosine:

\(\cos A\equiv 2\cos^2\left(\dfrac{A}{2}\right)-1\)

These are examples of half-angle formulae. We can obtain a half-angle formula for \(\tan A\) as well. Replacing \(A\) by \(\dfrac{A}{2}\) and \(2A\) by \(A\) in the double angle formula for \(\tan\) we obtain:

\(\tan A\equiv \dfrac{2\tan\left(\dfrac{A}{2}\right)}{1-\tan^2\left(\dfrac{A}{2}\right)} \)

Other formulae, which can be obtained using the half-angle formulae and are useful for integration when trigonometric functions are present:

Key Point

If \(t= \tan\left(\dfrac{A}{2}\right)\) then
\(\sin A= \dfrac{2t}{1+t^2}\)
\(\cos A= \dfrac{1-t^2}{1+t^2}\)
\(\tan A= \dfrac{2t}{1-t^2}\)

Sum of two sines and sum of two cosines

Here we obtain results that are widely used in areas of science and engineering such as vibration theory, wave theory and electric circuit theory.

We return to the identities angle addition formulae:

\[\begin{aligned} \sin(A+B ) &\equiv \sin A\cos B+\cos A\sin B\\ \sin (A-B) &\equiv \sin A\cos B-\cos A\sin B \end{aligned}\]

Adding these identities gives

\[\sin(A+B)+\sin(A-B)\equiv 2\sin A\cos B.\]

Subtracting the identities produces

\[\sin (A+B)-\sin(A-B)\equiv 2\cos A\sin B\]

It is now convenient to let \(C=A+B\) and \(D=A-B\) so that

\[A=\dfrac{C+D}{2} \qquad \text{and} \qquad B=\dfrac{C-D}{2}\]

The above then becomes:

\[\sin C+\sin D\equiv 2\sin\left(\dfrac{C+D}{2}\right)\cos \left(\dfrac{C-D}{2}\right) \]

Similarly:

\[\sin C-\sin D\equiv 2\cos\left(\dfrac{C+D}{2}\right)\sin \left(\dfrac{C-D}{2}\right) \]

Summary

We have covered a large number of trigonometric identities. The most important of them and probably the ones most worth memorising are given in the following Key Point.

Key Point

\[\begin{aligned} \cos^2\theta+\sin^2\theta&\equiv 1\\ \sin 2\theta&\equiv 2\sin\theta\cos\theta\\ \cos 2\theta&\equiv \cos^2\theta-\sin^2\theta\\ &\equiv 2\cos^2\theta -1\\ &\equiv 1-2\sin^2\theta\\ \sin (A\pm B)&\equiv \sin A\cos B\pm\cos A\sin B\\ \cos (A\pm B)&\equiv \cos A\cos B\mp\sin A\sin B \end{aligned}\]

Exercise

Using the exact values, evaluate \(\sin^2 \theta + \cos^2 \theta\) for the following angles:

1. For \(\theta = \displaystyle{[value]}\),   \(\sin^2 \theta + \cos^2 \theta =\)
2. For \(\theta = \displaystyle{[value]}\),   \(\sin^2 \theta + \cos^2 \theta=\)
3. For \(\theta = \displaystyle{[value]}\),   \(\sin^2 \theta + \cos^2 \theta =\)
Feedback: Using a table or otherwise, we can find the exact values of \(\sin \theta\) and \(\cos \theta\) for each of these angles. 1. For \(\theta = [value]\), we have \(\sin\theta = [value]\) and \(\cos\theta = [value]\). Then \[\sin^2 \theta + \cos^2 \theta = \left([value]\right)^2 + \left([value]\right)^2 = 1.\] 2. For \(\theta = [value]\), we have \(\sin\theta = [value]\) and \(\cos\theta = [value]\). Then \[\sin^2 \theta + \cos^2 \theta = \left([value]\right)^2 + \left([value]\right)^2 = 1.\] 3. For \(\theta = [value]\), we have \(\sin\theta = [value]\) and \(\cos\theta = [value]\). Then \[\sin^2 \theta + \cos^2 \theta = \left([value]\right)^2 + \left([value]\right)^2 = 1.\]

Exercise

1. Dividing the identity \(\sin^2\theta + \cos^2\theta\) by \(\sin^2\theta\), obtain an expression for \(\csc^2 \theta\) in terms of \(\cot \theta\).
\(\csc^2 \theta=\)
2. Dividing the identity \(\sin^2\theta + \cos^2\theta\) by \(\cos^2\theta\), obtain an expression for \(\sec^2\theta\) in terms of \(\tan\theta\).
\(\sec^2 \theta=\)
Feedback: 1. \( \dfrac{\sin^2\theta}{\sin^2\theta}+\dfrac{\cos^2\theta}{\sin^2\theta}=\dfrac{1}{\sin^2\theta}\) and thus \(1+\ {\rm cot}^2\theta\equiv{\rm cosec}^2\theta\) 2. \(\dfrac{\sin^2\theta}{\cos^2\theta}+\dfrac{\cos^2\theta}{\cos^2\theta}=\dfrac{1}{\cos^2\theta}\) and thus \(\tan^2\theta+1\equiv{\rm sec}^2\theta\)

Exercise

Using the identities

\[\begin{aligned} \sin (A-B) &\equiv \sin A\cos B-\cos A\sin B \\ \cos (A -B) &\equiv \cos A\cos B +\sin A\sin B, \end{aligned}\]

obtain an expansion for \(\tan (A-B)\) in terms of \(\tan A\) and \(\tan B\):

\(\tan (A-B) =\)
Feedback: We have \[\tan(A-B) \equiv \frac{\sin(A - B)}{\cos(A - B)} \equiv \dfrac{\sin A\cos B-\cos A\sin B}{\cos A\cos B+\sin A\sin B}.\] Dividing every term by \(\cos A\ \cos B\) gives \[\tan(A-B)\equiv \dfrac{\tan A-\tan B}{1+\tan A\tan B}.\]

Exercise

1. Substitute \(B = A\) in
\[\cos(A + B) = \cos A \cos B - \sin A \sin B\]
to obtain an identity for \(\cos 2A\) in terms of \(\cos A\) and \(\sin A\).
\(\cos(2A)=\)
2. Using \(\sin^2 A + \cos^2 A \equiv 1\), find an expression involving only \(\cos A\).
\(\cos(2A)=\)
3. Now use \(\sin^2 A + \cos^2 A \equiv 1\), to find an expression involving only \(\sin A\).
\(\cos(2A)=\)
Feedback: 1. We are given the identity \(\cos(A + B) = \cos A \cos B - \sin A \sin B\). By substituting \(B=A\) we get: \(\cos(2A)\equiv (\cos A)(\cos A)-(\sin A)(\sin A) = \cos^2 A - \sin^2 A.\) 2. By using \(\sin^2 A + \cos^2 A \equiv 1\) we get: \(\cos2A \equiv \cos^2A-(1-\cos^2A) \equiv 2\cos^2A-1.\) 3. \(\cos2A \equiv (1-\sin^{2}A)-\sin^2A \equiv 1-2\sin^2A.\)

Exercise

Use \(\cos(2A) = 2\cos^2A-1\) and \(\cos(2A) = 1- 2\sin^2 A \) to write \(\cos^2 A\) and \(\sin^2(A)\) in terms of \(\cos(2A)\).

1. \(\cos^2 A =\)
2. \(\sin^2 A =\)
Feedback: 1. \(\cos(2A) = 2 \cos^2 A - 1\) so \(\cos^2 A = \dfrac{1}{2}(1+\cos(2A))\) 2. \(\cos(2A) = 1 - 2\sin^2 A\) so \(\sin^2 A = \dfrac{1}{2}(1-\cos(2A))\)

Exercise

Use the double angle formulae for \(\sin(2A)\) and \(\cos(2A)\) to obtain an expression for \(\tan(2A)\) in terms of \(\tan A\).

\(\tan 2A =\)
Feedback: \[\tan 2A\equiv \dfrac{\sin 2A}{\cos 2A}\equiv \dfrac{2\sin A\cos A}{\cos^2A-\sin^2A}\] Dividing the numerator and denominator by \(\cos^2 A\) we obtain: \[ \tan 2A\equiv \dfrac{2\dfrac{\sin A}{\cos A}}{1-\dfrac{\sin^2A}{\cos^2A}}\equiv \dfrac{2\tan A}{1-\tan^2A} \]

Exercise

Use \(\cos (A\pm B) \equiv \cos A\cos B\mp\sin A\sin B\) to write:

1. \(\cos C + \cos D\) as a product of cosines.
\(\cos C + \cos D=\)
2. \(\cos C - \cos D\) as a product of sines.
\(\cos C - \cos D=\)
Feedback: We are given: \[\begin{aligned} \cos(A+B) &\equiv \cos A\cos B -\sin A\sin B \\ \cos(A-B) &\equiv \cos A\cos B+\sin A\sin B \end{aligned}\] We can add and subtract these two equations to get: \[\begin{aligned} \cos(A+B)+\cos (A-B)&\equiv 2\cos A\cos B\\ \cos (A+B)-\cos (A-B)&\equiv -2\sin A\sin B \end{aligned}\] Hence with \(C=A+B\) and \(D=A-B\), \[\cos C+\cos D\equiv 2\cos\left(\dfrac{C+D}{2}\right)\cos \left(\dfrac{C-D}{2}\right) \] \[\cos C-\cos D\equiv -2\sin\left(\dfrac{C+D}{2}\right)\sin\left(\dfrac{C-D}{2}\right)\]

Exercise

A projectile is fired from the ground with an initial speed \(u\ \mathrm{ms}^{-1}\) at an angle of elevation \(\alpha^\circ\). If air resistance is neglected, the vertical height, \(y\ \mathrm{m}\), is related to the horizontal distance, \(x\ \mathrm{m}\), by the equation

\[y= x\tan\alpha -\dfrac{gx^2{\rm sec}^2\alpha}{2u^2},\]

where \(g\ \mathrm{ms^{-2}}\) is the gravitational constant.

1. Find the height of the projectile when it has not yet travelled any horizontal distance.
\(y(x=0) = \)
2. For what value of \(x\) other than \(x=0\) do we have \(y=0\)?
Simplify your answer to get a function of \(\sin(2\alpha)\).
\(x=\)
3. Find the value \(x_\text{max}\) for which the value of \(y\) would be a maximum.
Simplify your answer to get a function of \(\sin(2\alpha)\).
\(x_\text{max}=\)
4. Simplify the expression for the maximum height to obtain a function of \(\sin \alpha\).
\(y_\text{max} =\)
5. Assuming \(u = [value]\ \mathrm{ms}^{-1}\), \(\alpha = [value]^{\circ}\) and \(g = 10\ \mathrm{ms}^{-2}\), find the two values \(x_1 \le x_2\) of \(x\) such that \(y = \displaystyle{[value]}\).
Give your answer with three significant figures.
\(x_1=\)   and   \(x_2=\)
Feedback: 1. Since \(x\) appears in both of the terms on the right-hand side, when \(x=0\), the right-hand side is zero. Hence, \(y(x = 0) = 0\). 2. When \(y=0\), the equation can be written as \[\dfrac{gx^2{\sec}^2\alpha}{2u^2}-x\tan\alpha=0.\] If \(x=0\) is excluded from consideration, we can divide through by \(x\) and rearrange to give \[\dfrac{gx{\sec}^2\alpha}{2u^2}=\tan\alpha.\] To solve for \(x\) we need to multiply both sides by \(2u^2/g{\sec}^2\alpha\). Given that \[\begin{aligned} \frac{1}{\sec^2\alpha} &\equiv \cos^2 \alpha, \\ \tan \alpha &\equiv \sin\alpha \cos \alpha, \\ \sin 2 \alpha &\equiv 2\sin\alpha \cos \alpha, \end{aligned}\] this results in \[x=\dfrac{2u^2\sin\alpha\cos\alpha}{g}=\dfrac{u^2\sin 2\alpha}{g}\] This represents the maximum range. 3. As air resistance is neglected, the path of the projectile is a parabola, which is symmetrical about the vertical line through its highest point. Therefore, the maximum height will occur at half the maximum range, i.e. where \[x=\dfrac{u^2\sin 2\alpha}{2g}.\] 4. Substituting this expression for \(x\) in the equation for \(y\) gives \[y=\left(\dfrac{u^2\sin2\alpha}{2g}\right)\tan\alpha-\left(\dfrac{u^2\sin2\alpha}{2g}\right)^2\dfrac{g{\sec}^2\alpha}{2u^2}.\] Using the same trigonometric identities as before, \[\qquad y=\dfrac{u^2\sin^2\alpha}{g}-\dfrac{u^2\sin^2\alpha}{2g}=\dfrac{u^2\sin^2\alpha}{2g}.\] This represents the maximum height. 5. Substitution of \(u=[value]\), \(\alpha=[value]\), \(g=10\) and \(y=[value]\) in the original equation gives a quadratic: \[[value] = 0\] Solving this quadratic yields \(x_1 = [value]\) and \(x_2 = [value]\) as the two horizontal ranges at which \(y = [value]\).

Exercise

Reduce the expression \( \sin t \sec t\) to a formula just involving \(\tan t\).

\( \sin t \sec t =\)
Feedback: Recall that \(\sec t \equiv 1/\cos t\) and \(\tan t \equiv \sin t / \cos t\). Then \[\sin t \sec t \equiv \sin t \dfrac{1}{\cos t} \equiv \dfrac{\sin t}{\cos t}\equiv \tan t.\]

Exercise

Express \((1+\sin t)(1+ \sin(-t))\) as a function of just \(\cos t\):

\((1+\sin t)(1+ \sin(-t))=\)
Feedback: Solution \((1+\sin t)(1+\sin (-t))\equiv (1+\sin t)(1-\sin t)\equiv 1-\sin^2t\equiv \cos^2t\)

Exercise

Express \(\dfrac{1}{\tan \theta + \cot \theta}\) as a function of \(\sin(2 \theta)\):

\(\dfrac{1}{\tan \theta + \cot \theta}=\)
Feedback: Solution \(\dfrac{1}{\tan\theta+\cos\theta}\equiv \dfrac{1}{\dfrac{\sin\theta}{\cos\theta} +\dfrac{\cos\theta}{\sin\theta}}\equiv \dfrac{1}{\dfrac{\sin^2\theta+\cos^2\theta}{\sin\theta\cos\theta}}\equiv \dfrac{\sin\theta\cos\theta}{\sin^2\theta+\cos^2\theta}\equiv \sin\theta\cos\theta\equiv \dfrac{1}{2}\sin 2\theta\)

Exercise

Express \(\sin^2(A+B)-\sin^2(A-B)\) as a function of just \(\sin(2A)\) and \(\sin(2B)\).

(Hint: the left-hand side is the difference of two squared quantities.)

\(\sin^2(A+B)-\sin^2(A-B)=\)
Feedback: Using the hint and the identity \(x^2-y^2\equiv (x-y)(x+y)\) we have \[\sin^2(A+B)-\sin^2(A-B)\equiv (\sin(A+B)-\sin(A-B))(\sin (A+B)+\sin (A-B)).\] The first bracket gives \[\sin A\cos B+\cos A\sin B-(\sin A\cos B-\cos A\sin B)\equiv 2\cos A\sin B.\] Similarly the second bracket gives \(2\sin A\cos B\). Multiplying we obtain \[(2\cos A\sin A)(2\cos B\sin B)\equiv \sin 2A \sin 2B.\]

Exercise

Express \(\dfrac{\sin 4\theta+\sin 2 \theta}{\cos 4 \theta+\cos 2\theta}\) as a function of \( \tan(3\theta) \)
Feedback: Solution \( \dfrac{\sin 4\theta+\sin 2\theta}{\cos 4\theta+\cos 2 \theta}\equiv \dfrac{2\sin 3\theta\cos\theta}{2\cos3\theta\cos\theta} \equiv \dfrac{\sin 3\theta}{\cos 3\theta}\equiv \tan 3\theta \)

Exercise

Express \( \cos^4A-\sin^4 A\) as a function of \(\cos(2A)\)

\( \cos^4A-\sin^4 A=\)
Feedback: Solution \[\begin{eqnarray*}\cos^4 A-\sin^4 A\equiv (\cos A)^4-(\sin A)^4 &\equiv &(\cos^2A)^2-(\sin^2A)^2\\ &\equiv &(\cos^2A-\sin^2A)(\cos^2A+\sin^2A)\qquad\qquad\qquad\qquad\\ &\equiv &\cos^2A-\sin^2A\equiv \cos 2A\end{eqnarray*}\]

Exercise

Express \(\sin([value])\cos([value])\) as the sum (or difference) of two sines.

\(\sin([value])\cos([value])=\)
Feedback: We use \[\sin A+\sin B\equiv 2\sin \left(\dfrac{A+B}{2}\right)\cos\left(\dfrac{A-B}{2}\right).\] Clearly here \(\dfrac{A+B}{2}=[value]\), \(\dfrac{A-B}{2}=[value]\) giving \(A=[value]\), \(B=[value]\).

Exercise

Express \([value]\) as the sum (or difference) of two cosines.

\([value]=\)
Feedback: Solution Using \(\cos A+\cos B\equiv 2\cos \left(\dfrac{A+B}{2}\right)\cos\left(\dfrac{A-B}{2}\right)\). Clearly here \(\dfrac{A+B}{2}=[value]\), \(\dfrac{A-B}{2}=[value]\) giving \(A=[value]\), \(B=[value]\)

Exercise

1. Express \(\sin(3t)\) in terms of \(\sin t\).
\(\sin(3t) =\)
2. Express \(\cos(3t)\) in terms of \(\cos t\).
\(\cos(3t)=\)
Feedback: 1. We use the sum and double angle formulas. \[\begin{aligned} \sin 3\theta &\equiv \sin (2\theta+\theta)=\sin 2\theta\cos\theta+\cos 2\theta\sin\theta \\ &\equiv 2\sin\theta\cos^2\theta+(\cos^2\theta-\sin^2\theta)\sin\theta\\ &\equiv 3\sin\theta\cos^2\theta-\sin^3\theta\\ &\equiv 3\sin\theta(1-\sin^2\theta)-\sin^3\theta\equiv 3\sin\theta-4\sin^3\theta \end{aligned}\] 2. We proceed simiarly. \[\begin{aligned} \cos 3\theta &\equiv \cos (2\theta+\theta)\equiv \cos 2\theta\cos\theta-\sin 2\theta\sin\theta\\ &\equiv (\cos^2\theta-\sin^2\theta)\cos\theta-2\sin\theta\cos\theta\sin\theta\\ &\equiv \cos^3\theta-3\sin^2\theta\cos\theta\\ &\equiv \cos^3\theta-3(1-\cos^2\theta)\cos\theta\\ &\equiv 4\cos^3\theta-3\cos\theta \end{aligned}\]

Exercise

By writing \(\cos 4x\) as \(\cos 2(2x)\), or otherwise, express \(\cos 4x\) in terms of \(\cos x\).

\(\cos 4x =\)
Feedback: Solution \(\begin{eqnarray*}\cos 4x=\cos 2(2x)&\equiv &2\cos^2(2x)-1\\ &\equiv &2 (\cos 2x)^2-1\\ &\equiv &2(2\cos^2 x-1)^2-1\\ &\equiv &2(4\cos^4 x-4\cos^2 x+1)-1\equiv 8\cos^4 x-8\cos^2 x+1 \end{eqnarray*}\)

Exercise

Express \(\dfrac{2\tan t}{2-\sec^2 t}\) as \(\tan(nt)\) for some integer \(n\).

\(\dfrac{2\tan t}{2-\sec^2 t}=\)
Feedback: Solution \(\dfrac{2\tan t}{2-{\rm sec}^2t} \equiv \dfrac{2\tan t}{1-({\rm sec}^2t-1)}\equiv \dfrac{2\tan t}{1-\tan^2t}\equiv \tan 2t \)

Exercise

Express \(\dfrac{\cos([value]x)-\cos([value]x)}{\sin([value]x)+\sin([value]x)}\) as \(\tan(nx)\) for an integer \(n\).

\( \dfrac{\cos([value]x)-\cos([value]x)}{\sin([value]x)+\sin([value]x)}=\)
Feedback: We first rewrite the numerator and denominator using the sum of sines and cosines formula. \[\begin{aligned} \cos([value]x)-\cos([value]x) &\equiv 2 \sin([value]x)\sin([value]x) \\ \sin([value]x)+\sin([value]x) &\equiv 2\sin([value]x)\cos(-[value]x) \end{aligned}\] Using these equations we can simplify the original expression into the required form. \[\dfrac{\cos([value]x)-\cos([value]x)}{\sin([value]x)+\sin([value]x)}\equiv \dfrac{\sin([value]x)}{\cos(-[value]x)}\equiv \dfrac{\sin([value]x)}{\cos([value]x)}\equiv \tan([value]x)\]

Exercise

What is the area \(A\) of an isosceles triangle with two equal sides of length \(x\) and apex angle \(\theta\)? Express your answer as a function of \(x\) and \(\sin \theta\).

\(A=\)

Hint: Use the following diagram.

[Image: Isosceles triangle with apex angle theta and bisector at apex]
Feedback: We first find the area of the right-angled triangle on the right. The area of the isosceles triangle will then simply be twice this area. The area of the right-angled triangle is given by \[\frac{1}{2}(CD)(AD).\] Using the definitions of \(\sin\) and \(\cos\), we can conclude that: \[\begin{aligned} \sin\left(\frac{\theta}{2}\right) &= \frac{CD}{x} \\ \cos\left(\frac{\theta}{2}\right) &= \frac{AD}{x} \end{aligned}\] Solving for the length of the segment we are interested in in each case gives: \[\begin{aligned} CD &= x \sin\left(\frac{\theta}{2}\right) \\ AD &= x \cos\left(\frac{\theta}{2}\right) \end{aligned}\] Therefore, the area of the right-angled triangle is \[\triangle ACD = \frac{1}{2} x^2 \sin\left(\frac{\theta}{2}\right)\cos\left(\frac{\theta}{2}\right) = \frac{1}{4} x^2 \sin\theta,\] which we rewrote using the half-angle formula. The total area of the isosceles triangle is then \[\triangle ABC = 2 \cdot \triangle ACD = \frac{1}{2} x^2 \sin\theta.\]