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Trigonometry - Trigonometric Identities of Multiple and Sub-multiple Angles

Grade 11ICSE

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

🔑Concepts

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Multiple Angles Definition: Multiple angles are integral multiples of an angle AA, represented as nAnA (e.g., 2A,3A,4A2A, 3A, 4A). Visually, if angle AA is represented by a rotation of a terminal ray in the Cartesian plane, 2A2A represents a rotation twice as large, potentially moving the ray into a different quadrant.

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Double Angle Relationships: These identities relate trigonometric functions of 2A2A to functions of AA. They are derived from the addition formulas by setting both angles equal (i.e., A+AA + A). For example, the sine of a double angle is twice the product of the sine and cosine of the original angle.

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Sub-multiple Angles: These are fractions of an angle, commonly A2\frac{A}{2} or A3\frac{A}{3}. Formulas for these are derived by replacing AA with A2\frac{A}{2} in the double-angle identities. Graphically, functions like sin⁡(x2)\sin(\frac{x}{2}) appear 'stretched' horizontally, having a period of 4π4\pi instead of the standard 2π2\pi.

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Power Reduction and Squared Identities: The cos⁡2A\cos 2A formula is unique because it can be expressed in three ways, allowing us to convert sin⁡2A\sin^2 A and cos⁡2A\cos^2 A into linear expressions of cos⁡2A\cos 2A. This is essential for simplifying expressions where the square of a trigonometric ratio needs to be removed.

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Tangent Half-Angle Substitution: The functions sin⁡A\sin A, cos⁡A\cos A, and tan⁡A\tan A can all be expressed as rational functions of t=tan⁡(A2)t = \tan(\frac{A}{2}). This creates a bridge between trigonometry and algebra, often used to solve complex trigonometric equations by converting them into algebraic ones.

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Triple Angle Identities: These relate sin⁡3A\sin 3A, cos⁡3A\cos 3A, and tan⁡3A\tan 3A to the functions of angle AA. They are derived using the sum formulas for (2A+A)(2A + A) and then substituting the double angle results. For instance, sin⁡3A\sin 3A depends only on the value of sin⁡A\sin A and its cube.

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Quadrant Sign Rules for Half-Angles: When calculating values for sub-multiple angles like sin⁡(A2)=±1−cos⁡A2\sin(\frac{A}{2}) = \pm \sqrt{\frac{1 - \cos A}{2}}, the choice of the positive or negative sign depends entirely on which quadrant the half-angle A2\frac{A}{2} falls in, which may be different from the quadrant of AA.

📐Formulae

sin⁡2A=2sin⁡Acos⁡A=2tan⁡A1+tan⁡2A\sin 2A = 2 \sin A \cos A = \frac{2 \tan A}{1 + \tan^2 A}

cos⁡2A=cos⁡2A−sin⁡2A=2cos⁡2A−1=1−2sin⁡2A=1−tan⁡2A1+tan⁡2A\cos 2A = \cos^2 A - \sin^2 A = 2 \cos^2 A - 1 = 1 - 2 \sin^2 A = \frac{1 - \tan^2 A}{1 + \tan^2 A}

tan⁡2A=2tan⁡A1−tan⁡2A\tan 2A = \frac{2 \tan A}{1 - \tan^2 A}

sin⁡3A=3sin⁡A−4sin⁡3A\sin 3A = 3 \sin A - 4 \sin^3 A

cos⁡3A=4cos⁡3A−3cos⁡A\cos 3A = 4 \cos^3 A - 3 \cos A

tan⁡3A=3tan⁡A−tan⁡3A1−3tan⁡2A\tan 3A = \frac{3 \tan A - \tan^3 A}{1 - 3 \tan^2 A}

sin⁡A=2sin⁡A2cos⁡A2=2tan⁡A21+tan⁡2A2\sin A = 2 \sin \frac{A}{2} \cos \frac{A}{2} = \frac{2 \tan \frac{A}{2}}{1 + \tan^2 \frac{A}{2}}

cos⁡A=cos⁡2A2−sin⁡2A2=1−tan⁡2A21+tan⁡2A2\cos A = \cos^2 \frac{A}{2} - \sin^2 \frac{A}{2} = \frac{1 - \tan^2 \frac{A}{2}}{1 + \tan^2 \frac{A}{2}}

1−cos⁡A=2sin⁡2A2  ⟹  sin⁡A2=±1−cos⁡A21 - \cos A = 2 \sin^2 \frac{A}{2} \implies \sin \frac{A}{2} = \pm \sqrt{\frac{1 - \cos A}{2}}

1+cos⁡A=2cos⁡2A2  ⟹  cos⁡A2=±1+cos⁡A21 + \cos A = 2 \cos^2 \frac{A}{2} \implies \cos \frac{A}{2} = \pm \sqrt{\frac{1 + \cos A}{2}}

💡Examples

Problem 1:

Prove that 1−cos⁡2θsin⁡2θ=tan⁡θ\frac{1 - \cos 2\theta}{\sin 2\theta} = \tan \theta.

Solution:

LHS: 1−cos⁡2θsin⁡2θ\frac{1 - \cos 2\theta}{\sin 2\theta}

Step 1: Substitute 1−cos⁡2θ1 - \cos 2\theta with 2sin⁡2θ2 \sin^2 \theta and sin⁡2θ\sin 2\theta with 2sin⁡θcos⁡θ2 \sin \theta \cos \theta.

=2sin⁡2θ2sin⁡θcos⁡θ= \frac{2 \sin^2 \theta}{2 \sin \theta \cos \theta}

Step 2: Cancel the common terms 22 and sin⁡θ\sin \theta.

=sin⁡θcos⁡θ= \frac{\sin \theta}{\cos \theta}

Step 3: Use the identity tan⁡θ=sin⁡θcos⁡θ\tan \theta = \frac{\sin \theta}{\cos \theta}.

=tan⁡θ=RHS= \tan \theta = RHS.

Explanation:

This solution uses double angle identities for cosine (specifically the power reduction form) and sine to simplify the fraction into a single trigonometric ratio.

Problem 2:

If sin⁡A=35\sin A = \frac{3}{5} and AA is in the first quadrant, find the value of sin⁡3A\sin 3A.

Solution:

Step 1: Write down the triple angle formula for sine.

sin⁡3A=3sin⁡A−4sin⁡3A\sin 3A = 3 \sin A - 4 \sin^3 A

Step 2: Substitute the given value sin⁡A=35\sin A = \frac{3}{5} into the formula.

sin⁡3A=3(35)−4(35)3\sin 3A = 3 \left(\frac{3}{5}\right) - 4 \left(\frac{3}{5}\right)^3

Step 3: Perform the arithmetic calculations.

sin⁡3A=95−4(27125)\sin 3A = \frac{9}{5} - 4 \left(\frac{27}{125}\right)

sin⁡3A=95−108125\sin 3A = \frac{9}{5} - \frac{108}{125}

Step 4: Find a common denominator (125125).

sin⁡3A=9×25125−108125=225−108125=117125\sin 3A = \frac{9 \times 25}{125} - \frac{108}{125} = \frac{225 - 108}{125} = \frac{117}{125}

Explanation:

This problem demonstrates the direct application of the triple angle identity. Since the result is positive and AA is in the first quadrant (0<A<90∘0 < A < 90^{\circ}), 3A3A will fall between 00 and 270∘270^{\circ}. The value 117125\frac{117}{125} (approx 0.9360.936) is a valid sine value.