For angles \(A\) and \(B\) it can be proven geometrically that

\(\sin ( A + B ) = \sin A \cos B + \cos A \sin B\) and \(\sin ( A - B ) = \sin A \cos B - \cos A \sin B\)

It can also be geometrically verified that

\(\cos ( A + B ) = \cos A \cos B - \sin A \sin B\) and \(\cos ( A - B ) = \cos A \cos B + \sin A \sin B\).

Using these results it can be shown that

\(\tan ( A + B ) = \dfrac{ \tan A + \tan B }{ 1 - \tan A \tan B }\) and \(\tan ( A - B ) = \dfrac{ \tan A - \tan B}{1 + \tan A \tan B} \).

These are referred to as the **addition** and **subtraction** identities.

Use the exact values of the trigonometric functions of \(30^\circ\) and \(45^\circ\) and the addition and subtraction identities to find exact values of the trigonometric functions of \(75^\circ\) and \(15^\circ\).

See SolutionProve the cofunction identity \(\cos ( 90^\circ - A ) = \sin A\) using the difference identity for cosine.

See SolutionFind the exact value of \(\sin 17^\circ \cos 13^\circ + \cos 17^\circ \sin 13^\circ\) without using a calculator.

See SolutionFind an identity for \(\tan \left( x + \frac{\pi}{4} \right)\) in terms of \(\tan x\).

See Solution
A **sum of multiples** of two functions is called a **linear combination** of the two functions.

Thus, if \(f ( x ) = a \sin x + b \cos x\), then \(f ( x )\) is a linear combination of \(\sin x\) and \(\cos x\).

We are going to see an example of the general principle that every linear combination of \(\sin x\) and \(\cos x\) is a sinusoidal function.

Let \(f ( x ) = 3 \sin x + 4 \cos x\). Show that \(f\) is a sinusoidal function. Find the amplitude, period and phase shift.

.The first step is to find the square root of the sum of the squares of the coefficients. In this case, that is\(\sqrt{3^2+4^2}=\sqrt{9+16}=\sqrt{25}= 5\). Factor \(5\) out of the expression to get \(f ( x ) = 5 \left( \frac{3}{5} \sin x + \frac{4}{5}\cos x\right) \)

Next, regard the coefficient of \(\sin x\) as the cosine of some angle \(\phi\) and the coefficient of \(\cos x\) as the sine of the same angle \(\phi\). We can make this assumption since the sum of the squares of the coefficients equals \(1\) as does the sum of the squares of \(\cos\phi\) and \(\sin\phi\) for any angle \(\phi\)..

This gives \(f ( x ) = 5 ( \cos\phi \sin x + \sin\phi \cos x) \)

Using the sum formula for sine, this can be re-written \(f ( x ) = 5 \sin ( x + \phi )\).

The amplitude is \(5\), the period is \(2 \pi\) and the phase shift is \(-\phi\).

What is the value of \(\phi\) ?

Since both the sine and the cosine of \(\phi\) are positive, \(\phi\) is an angle in quadrant I. So we can calculate \(\phi=\arccos \left( \frac{3}{5} \right)\) or alternately, \(\phi=\arcsin \left( \frac{4}{5} \right)\). So \(\phi=0.9273\) radians or \(53.1^\circ\).

Find the amplitude, period and phase shift of \(f ( x ) = - 2 \sin ( x ) + 3 \cos ( x )\)

Sketch the graph.

See Solution