Trigonometric Functions of Sum and Difference of Two Angles: Trigonometry has been a very useful branch of Mathematics, helping us in many practical life situations, from finding the height of a tree without actually measuring it to finding the distance between two celestial objects! However, these calculations can be done only with the help of various identities and formulae derived using the properties and facts of the six basic trigonometric functions. One of the most important types of formulae is that for the trigonometric functions of the sum and difference of two angles. 

These formulae further help us to form many other standard formulae that are useful in the field. Let us now learn the trigonometric functions of sum and difference of two angles, their derivations, and applications using some examples.

Basic Trigonometric Functions

Consider an acute angle in a right triangle \(ABC.\) There are six basic trigonometric ratios.

They are:

1. \( \sin \theta = \frac{{{\rm{ opposite}}\,{\rm{side }}}}{{{\rm{ hypotenuse }}}} = \frac{{AB}}{{BC}}\)
2. \( \cos \theta = \frac{{{\rm{ adjacent}}\,{\rm{side }}}}{{{\rm{ hypotenuse }}}} = \frac{{AC}}{{BC}}\)
3. \( \tan \theta = \frac{{{\rm{ opposite}}\,{\rm{side }}}}{{{\rm{ adjacent}}\,{\rm{side }}}} = \frac{{AB}}{{AC}}\)
4. \({\rm{cosec}}\,\theta = \frac{1}{{\sin \,\theta }} = \frac{{{\rm{hypotenuse}}}}{{{\rm{opposite}}\,{\rm{side}}}} = \frac{{BC}}{{AB}}\)
5. \({\rm{sec}}\,\theta = \frac{1}{{\cos \,\theta }} = \frac{{{\rm{hypotenuse}}}}{{{\rm{adjacent}}\,{\rm{side}}}} = \frac{{BC}}{{AC}}\)
6. \(\cot \,\theta = \frac{1}{{\tan \,\theta }} = \frac{{{\rm{adjacent}}\,{\rm{side}}}}{{{\rm{opposite}}\,{\rm{side}}}} = \frac{{AC}}{{AB}}\)

Trigonometric Functions on a Unit Circle

Consider a point \(P\) in a unit circle with the coordinates \((a, b)\) that makes an angle \(x\) with the positive \(x-\)axis in the anti-clockwise direction, as shown in the diagram.

It is possible to define all the six trigonometric ratios for the angle \(x\) in the right triangle \(OPM.\)

Here, for \(∠x,\)

Adjacent side\(=a\)
Opposite side\(=b\)
Hypotenuse\(=\)radius\(=1\)

Therefore, the six trigonometric ratios for \(x\) are defined as follows.

1. \( \sin x = a\)
2. \(\cos x = b\)
3. \( \tan x = \frac{a}{b}\)
4. \({\rm{cosec}}\,x = \frac{1}{a}\)
5. \( \sec x = \frac{1}{b}\)
6. \(\cot x = \frac{b}{a}\)

If a point \(Q\) falls in the fourth quadrant with the same angle but in the clockwise direction, the coordinates would be \((a, -b)\)

Just as in the case of \(P,\) the trigonometric ratios for the angle \(–x\) can also be defined.

Here, for \(∠-x,\)

Adjacent side\(=a\)
Opposite side\(=-b\)
Hypotenuse\(=\)radius\(=1\)

1. \( \sin ( – x) = \frac{{ – b}}{1} =\, – b = \,- \sin x\)
2. \(\cos \,\left( { – x} \right) = \frac{a}{1} = a = \cos \,x\)
3. \( \tan ( – x) = \frac{{ – b}}{a} =\, – \frac{b}{a} = \, – \tan x\)

Need for Sum and Difference Formulae

If the trigonometric ratio of any angle can be calculated using its definition, why is the sum and difference formulae required? Or what is its significance? The answer is, it can make the calculations for multiple angles a lot simpler.The trigonometric ratios of standard angles like \({30^{\rm{o}}},{45^{\rm{o}}},{60^{\rm{o}}},{90^{\rm{o}}}\) etc. can easily be derived and memorised.

Now, what if an angle can be written as a sum or difference of these angles? There comes the application of the sum and the difference between two angles formulae.

For example, to calculate the cosine of the angle of measure \({75^{\rm{o}}},{75^{\rm{o}}}\)can be written as the sum of \({45^{\rm{o}}}\) and \({30^{\rm{o}}}\)

So the cosine of the angle of measure of \({75^{\rm{o}}}\) is equal to the cosine of the sum of \({45^{\rm{o}}}\) and \({30^{\rm{o}}}\)

\( \cos {75^{\rm{o}}} = \cos \left( {{{45}^{\rm{o}}} + {{30}^{\rm{o}}}} \right)\)

Trigonometric Functions of Sum and Difference of Two Angles Formulae Derivation

Consider the points \({P_1},{P_2},{P_3},\) and \({P_4}\) on a unit circle centred at origin such that \(\angle {P_1}O{P_4} = x,\angle {P_1}O{P_2} = y,\angle {P_4}O{P_3} = – y\) and \({P_4}\) at \((1, 0).\)

Then, the coordinates of \({P_1}\) are \(( \cos x, \sin x),{P_2}\) are

\(( \cos (x + y), \sin (x + y)\) and \({P_3}\) are \(( \cos ( – y), \sin ( – y))\)

Consider the triangles \({P_1}O{P_3}\) and \({P_2}O{P_4}.\)

Now,  \({P_1}O \cong {P_2}O\) and \({P_3}O \cong {P_4}O,\) since they are radii of the same unit circle. Also, each of the included angles of the sides considered measures \(y + \angle {P_2}O{P_3}.\) So, by the Side-Angle-Side Congruence, the triangles \({P_1}O{P_3}\) and \({P_2}O{P_4}\) are congruent.

Corresponding parts of congruent triangles are congruent, hence \({P_1}{P_3} \cong {P_2}{P_4}.\)

So, \({P_1}{P_3} = {P_2}{P_4}\)

Therefore, \({\left( {{P_1}{P_3}} \right)^2} = {\left( {{P_2}{P_4}} \right)^2}\)

Using the distance formula,

\({[ \cos x – \cos ( – y)]^2} + {[\sin x – \sin ( – y)]^2} = {[1 – \cos (x + y)]^2} + {[ 0- \,\sin \,\left( {x + y} \right)]^2}\)

Since \( \sin ( – x) = \, – \sin x,\) and \(\cos ( – x) = \cos x\) the above equation can be written as

\({[\cos x – \cos y]^2} + {[\sin x + \sin y]^2} = {[1 – \cos (x + y)]^2} + {[ 0- \sin (x + y)]^2}\)

Expanding the equation using the identity \({(a + b)^2} = {a^2} + 2ab + {b^2}\) and

\({(a – b)^2} = {a^2} – 2ab + {b^2}:\)

\({\cos ^2}x + {\cos ^2}y – 2 \cos x \cos y + {\sin ^2}x + {\sin ^2}y + 2 \sin x \sin y = \)
\(1 + {\cos ^2}(x + y) – 2 \cos (x + y) + {\sin ^2}(x + y)\)

Since \({\sin ^2}x + {\cos ^2}x = 1\) for any \(x,\) in the above equation:

1. \({\sin ^2}x + {\cos ^2}x = 1\)
2. \({\sin ^2}y + {\cos ^2}y = 1\) 
3. \({\sin ^2}\left( {x + y} \right) + {\cos ^2}\left( {x + y} \right) = 1\)

Then, the equation becomes:

\(2 + 2 \sin x\sin y – 2 \cos x \cos y = 2 – 2 \cos (x + y)\)

Formula for cos(x+y):

Subtracting \(2\) from either side:

\(2 \sin x \sin y – 2 \cos x \cos y = \,- 2 \cos (x + y)\)
\( – 2( \cos x \cos y – \sin x\sin y) = \, – 2 \cos (x + y)\)

Cancelling the \(-2\) on either side, the formula for the cosine of sum of two angles can be written as:

\(\cos \left( {x + y} \right) = \cos x\cos y – \sin x\sin y\)

Formula for cos(x-y):

Replacing \(y\) with \(–y\) in equation \((1),\) the formula for the cosine of difference of two angles can be written as:

\(\cos \left( {x + \left( { – y} \right)} \right) = \cos x\cos \left( { – y} \right) – \sin x\sin \left( { – y} \right)\)

\( \cos (x – y) = \cos x\cos y + \sin x \sin y……..\left( 2 \right)\)

Formula for sin(x+y):

Replacing \(x\) with \(\frac{\pi }{2}\) and \(y\) with \(x\) in \((2):\)

\( \cos \left( {\frac{\pi }{2} – x} \right) = \sin \frac{\pi }{2}\sin x + \cos \frac{\pi }{2} \cos x = \sin x…….\left( 3 \right)\)

Similarly, replacing \(x\) with \(\frac{\pi }{2} – x\) in \( \sin x = \cos \left( {\frac{\pi }{2} – x} \right):\)

\( \sin \left( {\frac{\pi }{2} – x} \right) = \cos \left( {\frac{\pi }{2} – \left[ {\frac{\pi }{2} – x} \right]} \right) = \cos x…..\left( 4 \right)\)

Now, consider the angle \((x+y)\) instead of \(x\), in \((3):\)

\( \sin (x + y) = \cos \left[ {\frac{\pi }{2} – (x + y)} \right] = \cos \left[ {\left( {\frac{\pi }{2} – x} \right) + y} \right]\)

Using \((1):\)

\( = \cos \left( {\frac{\pi }{2} – x} \right)\cos \,y – \sin \left( {\frac{\pi }{2} – x} \right)\sin \,y\)

Using \((3)\) and \((4):\)

\( = \sin x\cos y + \cos x\sin y\)

Thus, the formula for the sine of sum of two angles can be obtained as:

\(\sin \left( {x + y} \right) = \sin \,x\cos \,y + \cos \,x\sin \,y…..(5)\)

Formula for sin(x-y):

Replacing \(y\) with \(-y\) in \((5),\) the formula for the sine of difference of two angles can be written as:

\(\sin \left( {x + \left( { – y} \right)} \right) = \sin \,x\cos \,\left( { – y} \right) + \cos \,x\sin \,\left( { – y} \right)\)

\(\sin \left( {x – y} \right) = \sin x\cos y – \cos x\sin y…(6)\)

Sum and Difference Formula for Tangent Ratio

Similar sum and difference formulae for the cotangent ratio can be derived like the tangent ratio if none of the angles \(x, y,\) or \((x+y)\) is a multiple of \(\frac{\pi }{2}\) so that the sines of these angles are non-zero. \(\tan \,\left( {x + y} \right) = \frac{{\sin \,\left( {x + y} \right)}}{{\cos \left( {x + y} \right)}} = \frac{{\sin x\cos y + \cos x\sin y}}{{\cos x\cos y – \sin x\sin y}}\)

Dividing the numerator and the denominator by \({\cos x\cos y}:\)

\(\frac{{\sin x\cos y + \cos x\sin y}}{{\cos x\cos y – \sin x\sin y}} = \frac{{\left( {\frac{{\sin x\cos y}}{{\cos x\cos y}}} \right) + \left( {\frac{{\cos x\sin y}}{{\cos x\cos y}}} \right)}}{{\left( {\frac{{\cos x\cos y}}{{\cos x\cos y}}} \right) – \left( {\frac{{\sin x\sin y}}{{\cos x\cos y}}} \right)}}\)
\( = \frac{{\tan x + \tan y}}{{1 – \tan x\tan y}}\)
\(\tan \left( {x + y} \right) = \frac{{\tan x + \tan y}}{{1 – \tan x\tan y}}\)
Replacing \(y\) with \(-y\) where none of the angles \(x,\,y\) or \(\left( {x – y} \right)\) is a multiple of \(\frac{\pi }{2},\) the formula for the tangent of difference of two angles can be written as:
\(\tan \left( {x + \left( { – y} \right)} \right) = \frac{{\tan x + \tan \left( { – y} \right)}}{{1 – \tan x\tan \left( { – y} \right)}} = \frac{{\tan x – \tan y}}{{1 + \tan x\tan y}}\) as \(\tan \left( { – y} \right) = \, – \tan y.\)

Thus:

\(\tan \left( {x – y} \right) = \frac{{\tan x – \tan y}}{{1 + \tan x\tan y}}\)

Sum and Difference Formulae for Cotangent Ratio:

Similar sum and difference formulae for the cotangent ratio can be derived like the tangent ratio if none of the angles \(x,\,y\) or \(\left( {x + y} \right)\) is a multiple of \(\pi \) so that the sines of these angles are non-zero.
\(\cot \left( {x + y} \right) = \frac{{\cos \left( {x + y} \right)}}{{\sin \left( {x + y} \right)}} = \frac{{\cos x\cos y – \sin x\sin y}}{{\sin x\cos y + \cos x\sin y}}\)
Dividing the numerator and the denominator by \({\sin x\sin y}:\)

\(\frac{{\cos x\cos y – \sin x\sin y}}{{\sin x\cos y + \cos x\sin y}} = \frac{{\left( {\frac{{\cos x\cos y}}{{\sin x\sin y}}} \right) – \left( {\frac{{\sin x\sin y}}{{\sin x\sin y}}} \right)}}{{\left( {\frac{{\sin x\cos y}}{{\sin x\sin y}}} \right) + \left( {\frac{{\cos x\sin y}}{{\sin x\sin y}}} \right)}}\)

\( = \frac{{\cot x\cot y – 1}}{{\cot x + \cot y}}\)

\(\cot(x + y) = \frac{{ \cot x\cot y – 1}}{{\cot x + \cot y}}\)

Replacing \(y\) with \(-y\) where none of the angles \(x, y,\) or \((x-y)\) is a multiple of \(\pi \) the formula for the cotangent of difference of two angles can be written as:

\(\cot \left( {x + \left( { – y} \right)} \right) = \frac{{\cot x\cot \left( { – y} \right) – 1}}{{\cot x + \cot \left( { – y} \right)}} = \frac{{\cot x\cot y + 1}}{{\cot y – \cot x}}\) as \(\cot \left( { – y} \right) = \, – \cot y.\)

Thus:

\(\cot (x – y) = \frac{{{\mathop{\rm cot}\nolimits} x\cot y + 1}}{{\cot y – \cot x}}\)

Application of Sum and Difference Formulae

The basic sum and difference formulae can be tabulated as:

Sum Formulae	Difference Formulae
\(\sin \left( {x + y} \right) = \cos x\sin y + \sin x\cos y\)	\(\sin \left( {x – y} \right) = \sin x\cos y – \cos x\sin y\)
\(\cos \left( {x + y} \right) = \cos x\cos y – \sin x\sin y\)	\(\cos \left( {x – y} \right) = \cos x\cos y + \sin x\sin y\)
\(\tan \left( {x + y} \right) = \frac{{\tan x + \tan y}}{{1 – \tan x\tan y}},\) if angles \(x,\,y\) and \(\left( {x + y} \right)\) are not multiples of \(\frac{\pi }{2}\)	\(\tan \left( {x – y} \right) = \frac{{\tan x – \tan y}}{{1 + \tan x\tan y}},\) if angles \(x,\,y\) and \(\left( {x – y} \right)\) are not multiples of \(\frac{\pi }{2}\)

Now, when these angles are substituted with desired angles, other useful formulae can be derived:

Basic Formula	Value of \(x\)	Value of \(y\)	Derived Formula
\(\cos \left( {x + y} \right) =\) \( \cos x\cos y – \sin x\sin y\)	\(x\)	\(x\)	\(\cos \,2x = {\cos ^2}x – {\sin ^2}x\) or \(\cos \,2x = 2{\cos ^2}x – 1\) or \(\cos \,2x = 1 – 2{\sin ^2}x\) or \(\cos \,2x = \frac{{1 – {{\tan }^2}x}}{{1 + {{\tan }^2}x}}\) \(\forall x \ne n\pi  + \frac{\pi }{2}\)
\(\sin \left( {x + y} \right) =\) \( \cos x\sin y + \sin x\cos y\)	\(x\)	\(x\)	\(\sin \,2x = 2\sin x\cos x\) or \(\sin \,2x = \frac{{2\tan x}}{{1 + {{\tan }^2}x}}\) \(\forall x \ne n\pi  + \frac{\pi }{2}\)
\(\tan \left( {x + y} \right) = \frac{{\tan x + \tan y}}{{1 – \tan x\tan y}}\)	\(x\)	\(x\)	\(\tan \,2x = \frac{{2\tan x}}{{1 – {{\tan }^2}x}}\) \(\forall 2x \ne n\pi  + \frac{\pi }{2}\)
\(\cos \left( {x + y} \right) = \) \(\cos x\cos y – \sin x\sin y\)	\(2x\)	\(2x\)	\(\cos 3x = 4{\cos ^3}x – 2\cos x\)
\(\sin \left( {x + y} \right) =\) \( \cos x\sin y + \sin x\cos y\)	\(2x\)	\(x\)	\(\sin 3x = 3\sin x – 4{\sin ^3}x\)
\(\tan \left( {x + y} \right) = \frac{{\tan x + \tan y}}{{1 – \tan x\tan y}}\)	\(2x\)	\(x\)	\(\tan \,3x = \frac{{3\tan x – {{\tan }^3}x}}{{1 – 3\,{{\tan }^2}x}}\) \(\forall 3x \ne n\pi  + \frac{\pi }{2}\)

Solved Examples – Trigonometric Functions of Sum and Difference of Two Angles

Q.1. Find the exact value of \( \sin \left( {\frac{\pi }{3} – \frac{\pi }{4}} \right)\)
Ans: \( \sin (x – y) = \sin x \cos y – \cos x \sin y\)
When \(x = \frac{\pi }{3}\) and \(y = \frac{\pi }{4}:\)
\( \sin \left( {\frac{\pi }{3} – \frac{\pi }{4}} \right) = \sin \left( {\frac{\pi }{3}} \right) \cos \left( {\frac{\pi }{4}} \right) – \cos \left( {\frac{\pi }{3}} \right) \sin \left( {\frac{\pi }{4}} \right)\)
\( = \frac{{\sqrt 3 }}{2} \times \frac{1}{{\sqrt 2 }} – \frac{1}{2} \times \frac{1}{{\sqrt 2 }}\)
\( = \frac{{\sqrt 3 – 1}}{{2\sqrt 2 }}\)

Q.2. Evaluate \(\cos {75^{\rm{o}}}.\)
Ans: The measure \({75^{\rm{o}}}\) can be written as \({45^{\rm{o}}} + {30^{\rm{o}}}.\)
\( \cos {75^{\rm{o}}} = \cos \left( {{{45}^{\rm{o}}} + {{30}^{\rm{o}}}} \right)\)
\(\cos \left( {x + y} \right) = \cos x\cos y – \sin x\sin y\)
\(\cos {75^{\rm{o}}} = \cos \left( {{{45}^{\rm{o}}} + {{30}^{\rm{o}}}} \right)\)
\( = \cos {45^{\rm{o}}} \cos {30^{\rm{o}}} – \sin {45^{\rm{o}}}\sin \sin {30^{\rm{o}}}\)
\( = \frac{1}{{\sqrt 2 }} \times \frac{{\sqrt 3 }}{2} – \frac{1}{{\sqrt 2 }} \times \frac{1}{2}\)
\( = \frac{{\sqrt 3 – 1}}{{2\sqrt 2 }}\)

Q.3. Find the value of \(\tan {225^{\rm{o}}}\)
Ans: The measure \({135^{\rm{o}}}\) can be written as \({180^{\rm{o}}} + {45^{\rm{o}}}.\)
\(\tan {225^{\rm{o}}} = \tan \left( {{{180}^{\rm{o}}} + {{45}^{\rm{o}}}} \right)\)
\(\tan \left( {x + y} \right) = \frac{{\tan x + \tan y}}{{1 – \tan x\tan y}}\)
\( \tan \left( {{{180}^{\rm{o}}} + {{45}^{\rm{o}}}} \right) = \frac{{{\mathop{\rm tan}\nolimits} {{180}^{\rm{o}}} + \tan {{45}^{\rm{o}}}}}{{1 – \tan {{180}^{\rm{o}}} \tan {{45}^{\rm{o}}}}}\)
\( = \frac{{0 + 1}}{{1 – (0 \times 1)}} = 1\)
\( \tan {225^{\rm{o}}} = 1\)

Q.4. Prove that \( \cos \left( {\frac{\pi }{4} – x} \right) \cos \left( {\frac{\pi }{4} – y} \right) + \sin \left( {\frac{\pi }{4} – x} \right) \sin \left( {\frac{\pi }{4} – y} \right) = \sin (x + y)\)
Ans: L.H.S.\( = \cos \left( {\frac{\pi }{4} – x} \right) \cos \left( {\frac{\pi }{4} – y} \right) + \sin \left( {\frac{\pi }{4} – x} \right) \sin \left( {\frac{\pi }{4} – y} \right) \)
\(= \cos \left( {\left( {\frac{\pi }{4} – x} \right) + \left( {\frac{\pi }{4} – y} \right)} \right)\)
\( = \cos \left( {2 \times \frac{\pi }{4} – x – y} \right)\)
\( = \cos \left( {\frac{\pi }{2} – (x + y)} \right)\)
\( = \sin (x + y)\)
\(=\)R. H. S
Hence proved.

Q.5. Prove that \(\frac{{\cos \left( {\pi + x} \right)\cos \left( { – x} \right)}}{{\sin \left( {\pi – x} \right)\cos \left( {\frac{\pi }{2} + x} \right)}} = {\cot ^2}x\)
Ans: Using the sum and difference formulae:
\(\cos \left( {\pi + x} \right) = \cos \pi \cos x – \sin \pi \sin x = \, – \cos x\)
\(\sin \left( {\pi – x} \right) = \sin \pi \cos x – \cos \pi \sin x = \,\sin x\)
\(\cos \left( { – x} \right) = \cos \left( {0 – x} \right) = \cos 0\cos x + \sin 0\sin x = \cos x\)
\(\cos \left( {\frac{\pi }{2} + x} \right) = \cos \frac{\pi }{2}\cos x – \sin \frac{\pi }{2}\sin x = \, – \sin x\)
L.H.S.\( = \frac{{\cos \left( {\pi + x} \right)\cos \left( { – x} \right)}}{{\sin \left( {\pi – x} \right)\cos \left( {\frac{\pi }{2} + x} \right)}} = \frac{{ – \cos x\cos x}}{{\sin x\left( { – \sin x} \right)}}\)
\( = \frac{{ – {{\cos }^2}x}}{{ – {{\sin }^2}x}}\)
\( = {\cot ^2}x\)
\(=\)R.H.S.
Hence proved.

Summary

This article would help you understand the basic concept of the trigonometric functions of the sum and difference of two angles. The basic results can be tabulated as:

Sum Formulae	Difference Formulae
\(\sin \left( {x + y} \right) = \cos x\sin y + \sin x\cos y\)	\(\sin \left( {x – y} \right) = \sin x\cos y – \cos x\cos y\)
\(\cos \left( {x + y} \right) = \cos x\cos y – \sin x\sin y\)	\(\cos \left( {x – y} \right) = \cos x\cos y + \sin x\sin y\)
\(\tan \left( {x + y} \right) = \frac{{\tan x + \tan y}}{{1 – \tan x\tan y}},\) if angles \(x,\,y\) and \(\left( {x + y} \right)\) are not multiples of \(\frac{\pi }{2}.\)	\(\tan \left( {x – y} \right) = \frac{{\tan x – \tan y}}{{1 + \tan x\tan y}},\) if angles \(a,\,y\) and \(\left( {x – y} \right)\) are not multiples of \(\frac{\pi }{2}.\)

The formula for the cosine function of the sum of two angles is derived using various points on a unit circle. The formulae for the trigonometric functions of sine, tangent and cotangent are deduced by applying the basic cosine formula derived. Further, the article procures some important results by substituting special values in the formulae deduced. The solved examples reinforce the concepts as well as explains the major applications of these formulae.

Frequently Asked Questions (FAQs)

Q.1. What are the functions of the sum and difference of two angles?
Ans: The sum and difference trigonometric identities give the formulae to find the trigonometric functions of sum and difference of two angles. This is mainly used to find the exact values of trigonometric functions of certain angles. Moreover, many related formulae can be derived using sum and difference identities.

Q.2. What is the sin of sum of two angles?
Ans: The sine of the sum of two angles is the sum of the cosine of the first angle times the sine of the second angle, and the sine of the first angle times the cosine of the second angle.
That is given as, \(\sin \left( {x + y} \right) = \cos x\sin y + \sin x\cos y\)

Q.3. What are the sum and difference identities?
Ans: The three basic sum and difference formulae can be tabulated as:

Sum Formulae	Difference Formulae
\(\sin \left( {x + y} \right) = \cos x\sin y + \sin x\cos y\)	\(\sin \left( {x – y} \right) = \sin x\cos y – \cos x\cos y\)
\(\cos \left( {x + y} \right) = \cos x\cos y – \sin x\sin y\)	\(\cos \left( {x – y} \right) = \cos x\cos y + \sin x\sin y\)
\(\tan \left( {x + y} \right) = \frac{{\tan x + \tan y}}{{1 – \tan x\tan y}},\) if angles \(x,\,y\) and \(\left( {x + y} \right)\) are not multiples of \(\frac{\pi }{2}.\)	\(\tan \left( {x – y} \right) = \frac{{\tan x – \tan y}}{{1 + \tan x\tan y}},\) if angles \(a,\,y\) and \(\left( {x – y} \right)\) are not multiples of \(\frac{\pi }{2}.\)

Q.4. How do you use angle sum identity?
Ans: The angle sum identity can be used to find the exact value of trigonometric ratios of non-standard angles.
For example:
\( \sin {15^{\rm{o}}} = \sin \left( {{{45}^{\rm{o}}} – {{30}^{\rm{o}}}} \right)\)
\( = \sin {45^{\rm{o}}} \cos {30^{\rm{o}}} – \cos {45^{\rm{o}}} \sin {30^{\rm{o}}}\)
\( = \frac{1}{{\sqrt 2 }} \times \frac{{\sqrt 3 }}{2} – \frac{1}{{\sqrt 2 }} \times \frac{1}{2}\)
\( = \frac{{\sqrt 3 – 1}}{{2\sqrt 2 }}\)

Q.5. How do you remember trigonometric functions of sum and difference of two angles?
Ans: \(\sin \left( {x \pm y} \right) = \sin x\cos y \pm \cos x\sin y\)
\(\cos \left( {x \pm y} \right) = \cos x\cos y \mp \sin x\sin y\)
Points to help remember the sum and difference formulae:
1. The sum and the difference expansion always starts with the same trigonometric ratios as in the LHS. The sum and difference of cosine of two angles start with cosine, and sine starts with sine.
2. The sine expansion has sine and cosine ratios in both terms with a sign similar to the LHS in between.
3. In the cosine expansion, the first term has only cosine ratios, and the second term has only sine ratios with a sign opposite to that of the LHS in between them.

We hope this detailed article on the trigonometric functions of sum and difference of two angles helped you in your studies. If you have any doubts, queries or suggestions regarding this article, feel to ask us in the comment section and we will be more than happy to assist you. Happy learning!