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Trigonometric Functions - Trigonometric Functions of Sum and Difference of Two Angles

Grade 11CBSE

Review the key concepts, formulae, and examples before starting your quiz.

🔑Concepts

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Compound angles are defined as the algebraic sum or difference of two or more angles, such as x+yx + y or x−yx - y. Geometrically, this represents the net rotation of a terminal ray in the Cartesian plane after two successive rotations.

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The identity for cos⁡(x−y)\cos(x - y) can be visualized using a unit circle. If we place two points P1P_1 and P2P_2 on the circle at angles xx and yy from the positive x-axis, the coordinates are (cos⁡x,sin⁡x)(\cos x, \sin x) and (cos⁡y,sin⁡y)(\cos y, \sin y). The distance formula between these points, when simplified, yields the identity cos⁡(x−y)=cos⁡xcos⁡y+sin⁡xsin⁡y\cos(x - y) = \cos x \cos y + \sin x \sin y.

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The sum and difference formulas for Sine, sin⁡(x±y)\sin(x \pm y), describe the vertical component of a point on the unit circle. A key visual observation is that the sine function 'mixes' sine and cosine terms (sin⁡xcos⁡y±cos⁡xsin⁡y\sin x \cos y \pm \cos x \sin y), whereas the cosine function 'pairs' like-terms (cos⁡xcos⁡y∓sin⁡xsin⁡y\cos x \cos y \mp \sin x \sin y).

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Tangent identities tan⁡(x±y)\tan(x \pm y) represent the slope of the line forming the compound angle. Since tan⁡θ=sin⁡θcos⁡θ\tan \theta = \frac{\sin \theta}{\cos \theta}, these formulas are derived by dividing the sine sum/difference identities by the cosine sum/difference identities and then dividing both numerator and denominator by cos⁡xcos⁡y\cos x \cos y.

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Sign changes in Cosine: A crucial visual and algebraic rule is that for cos⁡(x+y)\cos(x + y), the operator in the expansion is a minus sign (−-), and for cos⁡(x−y)\cos(x - y), the operator is a plus sign (++). This 'opposite sign' rule is a common area where students make mistakes.

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Application to non-standard angles: These formulas are used to find exact values for angles that are not on the standard unit circle (like 15∘,75∘,105∘15^\circ, 75^\circ, 105^\circ). For example, 75∘75^\circ is visualized as the sum of 45∘45^\circ and 30∘30^\circ, allowing us to use known values to find the result.

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Co-function identities: The sum/difference formulas can be used to prove that sin⁡(π2−x)=cos⁡x\sin(\frac{\pi}{2} - x) = \cos x and cos⁡(π2−x)=sin⁡x\cos(\frac{\pi}{2} - x) = \sin x. On a graph, this shows that the sine and cosine waves are horizontal shifts (phase shifts) of each other by π2\frac{\pi}{2} radians.

📐Formulae

sin⁡(A+B)=sin⁡Acos⁡B+cos⁡Asin⁡B\sin(A + B) = \sin A \cos B + \cos A \sin B

sin⁡(A−B)=sin⁡Acos⁡B−cos⁡Asin⁡B\sin(A - B) = \sin A \cos B - \cos A \sin B

cos⁡(A+B)=cos⁡Acos⁡B−sin⁡Asin⁡B\cos(A + B) = \cos A \cos B - \sin A \sin B

cos⁡(A−B)=cos⁡Acos⁡B+sin⁡Asin⁡B\cos(A - B) = \cos A \cos B + \sin A \sin B

tan⁡(A+B)=tan⁡A+tan⁡B1−tan⁡Atan⁡B\tan(A + B) = \frac{\tan A + \tan B}{1 - \tan A \tan B}

tan⁡(A−B)=tan⁡A−tan⁡B1+tan⁡Atan⁡B\tan(A - B) = \frac{\tan A - \tan B}{1 + \tan A \tan B}

cot⁡(A+B)=cot⁡Acot⁡B−1cot⁡B+cot⁡A\cot(A + B) = \frac{\cot A \cot B - 1}{\cot B + \cot A}

cot⁡(A−B)=cot⁡Acot⁡B+1cot⁡B−cot⁡A\cot(A - B) = \frac{\cot A \cot B + 1}{\cot B - \cot A}

💡Examples

Problem 1:

Find the exact value of sin⁡15∘\sin 15^\circ.

Solution:

Step 1: Express 15∘15^\circ as a difference of two standard angles: 15∘=45∘−30∘15^\circ = 45^\circ - 30^\circ. \nStep 2: Use the formula sin⁡(A−B)=sin⁡Acos⁡B−cos⁡Asin⁡B\sin(A - B) = \sin A \cos B - \cos A \sin B. \nStep 3: Substitute A=45∘A = 45^\circ and B=30∘B = 30^\circ: \nsin⁡(45∘−30∘)=sin⁡45∘cos⁡30∘−cos⁡45∘sin⁡30∘\sin(45^\circ - 30^\circ) = \sin 45^\circ \cos 30^\circ - \cos 45^\circ \sin 30^\circ \nStep 4: Plug in the standard values: \nsin⁡15∘=(12)(32)−(12)(12)\sin 15^\circ = \left(\frac{1}{\sqrt{2}}\right) \left(\frac{\sqrt{3}}{2}\right) - \left(\frac{1}{\sqrt{2}}\right) \left(\frac{1}{2}\right) \nStep 5: Simplify the expression: \nsin⁡15∘=322−122=3−122\sin 15^\circ = \frac{\sqrt{3}}{2\sqrt{2}} - \frac{1}{2\sqrt{2}} = \frac{\sqrt{3} - 1}{2\sqrt{2}}

Explanation:

This approach decomposes a non-standard angle into standard angles whose trigonometric values are known from the unit circle, then applies the sine difference identity.

Problem 2:

Prove that cos⁡11∘+sin⁡11∘cos⁡11∘−sin⁡11∘=tan⁡56∘\frac{\cos 11^\circ + \sin 11^\circ}{\cos 11^\circ - \sin 11^\circ} = \tan 56^\circ.

Solution:

Step 1: Take the Left Hand Side (LHS) and divide both numerator and denominator by cos⁡11∘\cos 11^\circ: \nLHS=cos⁡11∘cos⁡11∘+sin⁡11∘cos⁡11∘cos⁡11∘cos⁡11∘−sin⁡11∘cos⁡11∘\text{LHS} = \frac{\frac{\cos 11^\circ}{\cos 11^\circ} + \frac{\sin 11^\circ}{\cos 11^\circ}}{\frac{\cos 11^\circ}{\cos 11^\circ} - \frac{\sin 11^\circ}{\cos 11^\circ}} \nStep 2: Simplify using tan⁡θ=sin⁡θcos⁡θ\tan \theta = \frac{\sin \theta}{\cos \theta}: \nLHS=1+tan⁡11∘1−tan⁡11∘\text{LHS} = \frac{1 + \tan 11^\circ}{1 - \tan 11^\circ} \nStep 3: Recognize that 1=tan⁡45∘1 = \tan 45^\circ. Substitute this into the expression: \nLHS=tan⁡45∘+tan⁡11∘1−(tan⁡45∘)(tan⁡11∘)\text{LHS} = \frac{\tan 45^\circ + \tan 11^\circ}{1 - (\tan 45^\circ)(\tan 11^\circ)} \nStep 4: Observe that this matches the form tan⁡A+tan⁡B1−tan⁡Atan⁡B=tan⁡(A+B)\frac{\tan A + \tan B}{1 - \tan A \tan B} = \tan(A + B). \nStep 5: Therefore, LHS=tan⁡(45∘+11∘)=tan⁡56∘\text{LHS} = \tan(45^\circ + 11^\circ) = \tan 56^\circ. \nStep 6: LHS = RHS. Proved.

Explanation:

This example uses the tangent sum formula in reverse. By dividing by cos⁡A\cos A, we transform a sine-cosine fraction into a tangent expression, which is a common technique in trigonometric proofs.