krit.club logo

Trigonometry - Trigonometric Identities of Sum and Difference of Angles

Grade 11ICSE

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

🔑Concepts

•

Compound Angles: Compound angles are the algebraic sum or difference of two or more simple angles, typically represented as (A+B)(A+B) or (A−B)(A-B). Visually, if a ray rotates by angle AA and then further rotates by angle BB, the final position of the ray from the initial axis is the compound angle A+BA+B.

•

Geometric Construction of Sine Sum: The identity sin⁡(A+B)=sin⁡Acos⁡B+cos⁡Asin⁡B\sin(A+B) = \sin A \cos B + \cos A \sin B can be visualized using a unit circle or by stacking two right-angled triangles. If triangle 1 has angle AA and its hypotenuse serves as the base for triangle 2 with angle BB, the total vertical height of the combined structure represents the sine of the sum.

•

The Cosine Sum Logic: In the expansion cos⁡(A+B)=cos⁡Acos⁡B−sin⁡Asin⁡B\cos(A+B) = \cos A \cos B - \sin A \sin B, the negative sign occurs because as the angle increases (moving from AA to A+BA+B), the horizontal projection (cosine) on the x-axis generally decreases or moves in the opposite direction on the unit circle.

•

Tangent of Compound Angles: The identity for tan⁡(A±B)\tan(A \pm B) relates the slopes of two lines. Visually, if two lines have slopes m1=tan⁡Am_1 = \tan A and m2=tan⁡Bm_2 = \tan B, the tangent of the angle between them or their sum is a rational function of their individual slopes, specifically tan⁡A±tan⁡B1∓tan⁡Atan⁡B\frac{\tan A \pm \tan B}{1 \mp \tan A \tan B}.

•

Transformation of Standard Angles: Sum and difference identities are primarily used to find the trigonometric values of non-standard angles by breaking them down into standard angles (30∘,45∘,60∘,90∘30^\circ, 45^\circ, 60^\circ, 90^\circ). For example, 75∘75^\circ is visualized as 45∘+30∘45^\circ + 30^\circ and 15∘15^\circ as 45∘−30∘45^\circ - 30^\circ.

•

Co-function and Quadrant Relationships: These identities are consistent with quadrant rules (ASTC). For instance, sin⁡(90∘+θ)=sin⁡90∘cos⁡θ+cos⁡90∘sin⁡θ=cos⁡θ\sin(90^\circ + \theta) = \sin 90^\circ \cos \theta + \cos 90^\circ \sin \theta = \cos \theta. Visually, this represents a 90∘90^\circ rotation on the Cartesian plane where the y-coordinate of the new point corresponds to the x-coordinate of the original point.

•

Product of Sum and Difference: The identity sin⁡(A+B)sin⁡(A−B)=sin⁡2A−sin⁡2B\sin(A+B)\sin(A-B) = \sin^2 A - \sin^2 B is a powerful simplification tool. It shows that the product of sines of compound angles is equivalent to the difference of the squares of the sines of the individual angles.

📐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}

sin⁡(A+B)sin⁡(A−B)=sin⁡2A−sin⁡2B=cos⁡2B−cos⁡2A\sin(A + B)\sin(A - B) = \sin^2 A - \sin^2 B = \cos^2 B - \cos^2 A

cos⁡(A+B)cos⁡(A−B)=cos⁡2A−sin⁡2B=cos⁡2B−sin⁡2A\cos(A + B)\cos(A - B) = \cos^2 A - \sin^2 B = \cos^2 B - \sin^2 A

💡Examples

Problem 1:

Find the exact value of sin⁡75∘\sin 75^\circ using sum and difference identities.

Solution:

  1. Express 75∘75^\circ as a sum of two standard angles: 75∘=45∘+30∘75^\circ = 45^\circ + 30^\circ
  2. Apply the sine sum identity sin⁡(A+B)=sin⁡Acos⁡B+cos⁡Asin⁡B\sin(A+B) = \sin A \cos B + \cos A \sin B: sin⁡(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
  3. Substitute the standard values: sin⁡75∘=(12)(32)+(12)(12)\sin 75^\circ = \left(\frac{1}{\sqrt{2}}\right) \left(\frac{\sqrt{3}}{2}\right) + \left(\frac{1}{\sqrt{2}}\right) \left(\frac{1}{2}\right)
  4. Simplify the expression: sin⁡75∘=322+122=3+122\sin 75^\circ = \frac{\sqrt{3}}{2\sqrt{2}} + \frac{1}{2\sqrt{2}} = \frac{\sqrt{3} + 1}{2\sqrt{2}}

Explanation:

To solve for an angle not on the standard unit circle, we decompose it into the sum of two angles whose sine and cosine values are known (45∘45^\circ and 30∘30^\circ). We then apply the sine sum formula and simplify the fractions.

Problem 2:

Prove that cos⁡8∘−sin⁡8∘cos⁡8∘+sin⁡8∘=tan⁡37∘\frac{\cos 8^\circ - \sin 8^\circ}{\cos 8^\circ + \sin 8^\circ} = \tan 37^\circ.

Solution:

  1. Start with the Left Hand Side (LHS) and divide both numerator and denominator by cos⁡8∘\cos 8^\circ: LHS=cos⁡8∘cos⁡8∘−sin⁡8∘cos⁡8∘cos⁡8∘cos⁡8∘+sin⁡8∘cos⁡8∘=1−tan⁡8∘1+tan⁡8∘LHS = \frac{\frac{\cos 8^\circ}{\cos 8^\circ} - \frac{\sin 8^\circ}{\cos 8^\circ}}{\frac{\cos 8^\circ}{\cos 8^\circ} + \frac{\sin 8^\circ}{\cos 8^\circ}} = \frac{1 - \tan 8^\circ}{1 + \tan 8^\circ}
  2. Recognize that 1=tan⁡45∘1 = \tan 45^\circ. Substitute this into the formula: LHS=tan⁡45∘−tan⁡8∘1+(tan⁡45∘)(tan⁡8∘)LHS = \frac{\tan 45^\circ - \tan 8^\circ}{1 + (\tan 45^\circ)(\tan 8^\circ)}
  3. This expression matches the structure of the identity tan⁡(A−B)=tan⁡A−tan⁡B1+tan⁡Atan⁡B\tan(A-B) = \frac{\tan A - \tan B}{1 + \tan A \tan B}: LHS=tan⁡(45∘−8∘)LHS = \tan(45^\circ - 8^\circ)
  4. Simplify the subtraction: LHS=tan⁡37∘=RHSLHS = \tan 37^\circ = RHS

Explanation:

This is a common algebraic manipulation in trigonometry. By dividing by cos⁡θ\cos \theta, we convert a sine-cosine expression into a tangent expression, which allows us to use the tan⁡(A−B)\tan(A-B) identity by recognizing that 11 is tan⁡45∘\tan 45^\circ.