Ratios of the Difference of Two Angles
The difference of two angles has its own formulae, very similar to those for the sum but with the signs swapped. With them we can calculate exact values such as $\sin(15º)$, prove identities that have been in textbooks for decades (such as $\cos(180º-\alpha)=-\cos\alpha$) and solve problems involving two signals, two rays of light or two angles of observation.
The difference formulae: sine, cosine and tangent
They are obtained from the sum formulae by replacing $\beta$ with $-\beta$ and using the facts that $\sin(-\beta)=-\sin\beta$ and $\cos(-\beta)=\cos\beta$ (sine is an odd function, cosine is even).
Simplify the following trigonometric expression as much as possible: $\cos(x) \cos(y) + \sin(x) \sin(y)$. Then, use the simplified result to evaluate the expression…
Simplify the following trigonometric expression as much as possible: $\cos(x) \cos(y) + \sin(x) \sin(y)$. Then, use the simplified result to evaluate the expression exactly when $x = 75^\circ$ and $y = 15^\circ$.
Special cases: complementary and supplementary angles
When one of the two angles is $90º$ or $180º$, the difference formulae simplify a great deal and give very widely used identities.
Simplify the expression $\sin(90^\circ - x)$ by applying the formula for the sine of the difference of two angles and verify that it is equivalent to one of the basic…
Simplify the expression $\sin(90^\circ - x)$ by applying the formula for the sine of the difference of two angles and verify that it is equivalent to one of the basic trigonometric ratios of the angle $x$.
Prove step by step that $\cos(180^\circ - \alpha) = -\cos \alpha$ using the formula for the cosine of the difference of two angles.
Prove step by step that $\cos(180^\circ - \alpha) = -\cos \alpha$ using the formula for the cosine of the difference of two angles.
Calculating exact values by combining special angles
Just as with the sum, an angle such as $15º$ is written as the difference of two special angles ($45º-30º$, or alternatively $60º-45º$) so that its ratios can be calculated without a calculator.
Calculate the exact value of $\cos(15^\circ)$ without using a calculator, applying the formula for the cosine of the difference of two angles and knowing that…
Calculate the exact value of $\cos(15^\circ)$ without using a calculator, applying the formula for the cosine of the difference of two angles and knowing that $15^\circ = 45^\circ - 30^\circ$.
Find the exact value of $\sin(15^\circ)$ by expressing the angle as the difference $60^\circ - 45^\circ$ and applying the corresponding identity.
Find the exact value of $\sin(15^\circ)$ by expressing the angle as the difference $60^\circ - 45^\circ$ and applying the corresponding identity.
Determine the exact value of $\tan(15^\circ)$ from the tangents of the special angles $45^\circ$ and $30^\circ$. Rationalise the result if necessary.
Determine the exact value of $\tan(15^\circ)$ from the tangents of the special angles $45^\circ$ and $30^\circ$. Rationalise the result if necessary.
Solving equations and problems with known ratios
When the question gives the ratios of $\alpha$ and $\beta$ separately (rather than the angles), you first calculate the missing ratios and then substitute into the difference formula; if the unknown is an angle, you solve a trigonometric equation.
Given that $\sin \alpha = \frac{4}{5}$ and $\cos \beta = \frac{12}{13}$, and that both $\alpha$ and $\beta$ are angles in the first quadrant, calculate the exact value…
Given that $\sin \alpha = \frac{4}{5}$ and $\cos \beta = \frac{12}{13}$, and that both $\alpha$ and $\beta$ are angles in the first quadrant, calculate the exact value of $\sin(\alpha - \beta)$.
In a maritime navigation simulation, a ship must adjust its current heading. The necessary correction angle is given by the expression $\alpha - \beta$. Given that…
In a maritime navigation simulation, a ship must adjust its current heading. The necessary correction angle is given by the expression $\alpha - \beta$. Given that $\alpha$ and $\beta$ are acute angles with $\tan \alpha = 3$ and $\tan \beta = \frac{1}{2}$, calculate the exact value of $\tan(\alpha - \beta)$ and deduce the size of the correction angle.
Solve the following trigonometric equation, finding all possible solutions in the interval $0^\circ \leq x < 360^\circ$: $\sin(x - 30^\circ) = \frac{\sqrt{3}}{2}$
Solve the following trigonometric equation, finding all possible solutions in the interval $0^\circ \leq x < 360^\circ$:
$\sin(x - 30^\circ) = \frac{\sqrt{3}}{2}$
Applications: optics, surveying and signals
Two rays of light, two lines of sight from an observation point or two out-of-phase signals can all be described by a single difference angle, even though the context (physics, surveying, electricity) changes completely.
A light ray passes through two different media with angles of refraction $\alpha$ and $\beta$. If, from the material properties, we know that $\cos \alpha = \frac{3}{5}$…
A light ray passes through two different media with angles of refraction $\alpha$ and $\beta$. If, from the material properties, we know that $\cos \alpha = \frac{3}{5}$ (where $\alpha$ is an angle in the fourth quadrant) and $\sin \beta = \frac{5}{13}$ (where $\beta$ is an angle in the second quadrant), determine the exact value of $\cos(\alpha - \beta)$.
In a market study, the seasonal sales cycles of two products are modelled using trigonometric functions. The phase difference that determines direct competition is given…
In a market study, the seasonal sales cycles of two products are modelled using trigonometric functions. The phase difference that determines direct competition is given by the value of $\cos(x - y)$. If $\sin x = \frac{1}{3}$ and $\cos y = \frac{2}{5}$, with $x$ located in the first quadrant and $y$ in the fourth quadrant, find the exact value of $\cos(x - y)$.
In an alternating current circuit, the potential difference between two nodes is given by the superposition of two out-of-phase signals:…
In an alternating current circuit, the potential difference between two nodes is given by the superposition of two out-of-phase signals: $V(t) = 10 \sin(\omega t) - 10 \sin(\omega t - \frac{\pi}{3})$. Use the trigonometric difference angle formulas to express the potential difference $V(t)$ in the simplified form $A \sin(\omega t + \phi)$, determining the values of $A$ and $\phi$.
Back to the beginning: To calculate the exact value of $\sin(15º)$ mentioned at the start, we apply the difference formula: $\sin(45º-30º) = \sin(45º)\cos(30º) - \cos(45º)\sin(30º) = \frac{\sqrt{2}}{2}\cdot\frac{\sqrt{3}}{2} - \frac{\sqrt{2}}{2}\cdot\frac{1}{2} = \frac{\sqrt{6}-\sqrt{2}}{4}$.