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Trigonometry - Trigonometrical Identities

Grade 10ICSE

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

🔑Concepts

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Trigonometric Ratios in a Right Triangle: Imagine a right-angled triangle where θ\theta is one of the acute angles. The side opposite the 90∘90^{\circ} angle is the hypotenuse, the side opposite θ\theta is the perpendicular (opposite side), and the side adjacent to θ\theta is the base. The fundamental ratios are defined as sin⁡θ=PerpendicularHypotenuse\sin \theta = \frac{Perpendicular}{Hypotenuse}, cos⁡θ=BaseHypotenuse\cos \theta = \frac{Base}{Hypotenuse}, and tan⁡θ=PerpendicularBase\tan \theta = \frac{Perpendicular}{Base}.

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Reciprocal Identities: These identities show that the three primary ratios have corresponding reciprocal ratios. Visually, if you invert the fraction of a ratio, you get its reciprocal: sin⁡θ\sin \theta and csc⁡θ\csc \theta are reciprocals, cos⁡θ\cos \theta and sec⁡θ\sec \theta are reciprocals, and tan⁡θ\tan \theta and cot⁡θ\cot \theta are reciprocals.

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Quotient Identities: These express tangent and cotangent in terms of sine and cosine. Think of tan⁡θ\tan \theta as the 'slope' of a line in a coordinate system, representing the ratio of vertical change (sin⁡θ\sin \theta) to horizontal change (cos⁡θ\cos \theta). Therefore, tan⁡θ=sin⁡θcos⁡θ\tan \theta = \frac{\sin \theta}{\cos \theta} and cot⁡θ=cos⁡θsin⁡θ\cot \theta = \frac{\cos \theta}{\sin \theta}.

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Pythagorean Identities: These are the most critical identities derived from the Pythagorean theorem (a2+b2=c2a^2 + b^2 = c^2). In a unit circle context, the coordinates of a point are (cos⁡θ,sin⁡θ)(\cos \theta, \sin \theta). Applying the theorem to this circle results in the identity sin⁡2θ+cos⁡2θ=1\sin^2 \theta + \cos^2 \theta = 1. From this, two other identities are derived: 1+tan⁡2θ=sec⁡2θ1 + \tan^2 \theta = \sec^2 \theta and 1+cot⁡2θ=csc⁡2θ1 + \cot^2 \theta = \csc^2 \theta.

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Complementary Angle Relations: Since the two acute angles in a right triangle sum to 90∘90^{\circ}, the sine of one angle is the cosine of its complement. Visually, the 'opposite' side for one angle becomes the 'adjacent' side for the other. This gives us relations like sin⁡(90∘−θ)=cos⁡θ\sin(90^{\circ} - \theta) = \cos \theta and tan⁡(90∘−θ)=cot⁡θ\tan(90^{\circ} - \theta) = \cot \theta.

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Proving Identities - Strategy: When proving that the Left Hand Side (LHS) equals the Right Hand Side (RHS), the best approach is to start with the more complex-looking side. Common visual patterns to look for include fractions that can be combined using a common denominator, terms that can be factored out, or expressions that resemble the difference of two squares (a2−b2)(a^2 - b^2).

📐Formulae

sin⁡θ=1csc⁡θ  ⟹  sin⁡θ⋅csc⁡θ=1\sin \theta = \frac{1}{\csc \theta} \implies \sin \theta \cdot \csc \theta = 1

cos⁡θ=1sec⁡θ  ⟹  cos⁡θ⋅sec⁡θ=1\cos \theta = \frac{1}{\sec \theta} \implies \cos \theta \cdot \sec \theta = 1

tan⁡θ=1cot⁡θ  ⟹  tan⁡θ⋅cot⁡θ=1\tan \theta = \frac{1}{\cot \theta} \implies \tan \theta \cdot \cot \theta = 1

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

cot⁡θ=cos⁡θsin⁡θ\cot \theta = \frac{\cos \theta}{\sin \theta}

sin⁡2θ+cos⁡2θ=1\sin^2 \theta + \cos^2 \theta = 1

sin⁡2θ=1−cos⁡2θ\sin^2 \theta = 1 - \cos^2 \theta

cos⁡2θ=1−sin⁡2θ\cos^2 \theta = 1 - \sin^2 \theta

1+tan⁡2θ=sec⁡2θ  ⟹  sec⁡2θ−tan⁡2θ=11 + \tan^2 \theta = \sec^2 \theta \implies \sec^2 \theta - \tan^2 \theta = 1

1+cot⁡2θ=csc⁡2θ  ⟹  csc⁡2θ−cot⁡2θ=11 + \cot^2 \theta = \csc^2 \theta \implies \csc^2 \theta - \cot^2 \theta = 1

💡Examples

Problem 1:

Prove that sin⁡A1+cos⁡A+1+cos⁡Asin⁡A=2csc⁡A\frac{\sin A}{1 + \cos A} + \frac{1 + \cos A}{\sin A} = 2\csc A

Solution:

Step 1: Take the LHS and find the common denominator. LHS=sin⁡2A+(1+cos⁡A)2sin⁡A(1+cos⁡A)LHS = \frac{\sin^2 A + (1 + \cos A)^2}{\sin A(1 + \cos A)} Step 2: Expand the term (1+cos⁡A)2(1 + \cos A)^2 using the identity (a+b)2=a2+2ab+b2(a+b)^2 = a^2 + 2ab + b^2. LHS=sin⁡2A+1+2cos⁡A+cos⁡2Asin⁡A(1+cos⁡A)LHS = \frac{\sin^2 A + 1 + 2\cos A + \cos^2 A}{\sin A(1 + \cos A)} Step 3: Group sin⁡2A\sin^2 A and cos⁡2A\cos^2 A together. LHS=(sin⁡2A+cos⁡2A)+1+2cos⁡Asin⁡A(1+cos⁡A)LHS = \frac{(\sin^2 A + \cos^2 A) + 1 + 2\cos A}{\sin A(1 + \cos A)} Step 4: Use the identity sin⁡2A+cos⁡2A=1\sin^2 A + \cos^2 A = 1. LHS=1+1+2cos⁡Asin⁡A(1+cos⁡A)=2+2cos⁡Asin⁡A(1+cos⁡A)LHS = \frac{1 + 1 + 2\cos A}{\sin A(1 + \cos A)} = \frac{2 + 2\cos A}{\sin A(1 + \cos A)} Step 5: Factor out 2 from the numerator. LHS=2(1+cos⁡A)sin⁡A(1+cos⁡A)LHS = \frac{2(1 + \cos A)}{\sin A(1 + \cos A)} Step 6: Cancel the common factor (1+cos⁡A)(1 + \cos A). LHS=2sin⁡A=2csc⁡A=RHSLHS = \frac{2}{\sin A} = 2\csc A = RHS

Explanation:

This problem is solved by using the algebraic technique of finding a common denominator and then applying the fundamental Pythagorean identity sin⁡2θ+cos⁡2θ=1\sin^2 \theta + \cos^2 \theta = 1 to simplify the numerator.

Problem 2:

Prove that 1−cos⁡θ1+cos⁡θ=csc⁡θ−cot⁡θ\sqrt{\frac{1 - \cos \theta}{1 + \cos \theta}} = \csc \theta - \cot \theta

Solution:

Step 1: Start with the LHS and rationalize the denominator by multiplying the numerator and denominator by (1−cos⁡θ)(1 - \cos \theta) inside the square root. LHS=(1−cos⁡θ)(1−cos⁡θ)(1+cos⁡θ)(1−cos⁡θ)LHS = \sqrt{\frac{(1 - \cos \theta)(1 - \cos \theta)}{(1 + \cos \theta)(1 - \cos \theta)}} Step 2: Simplify the numerator and denominator. LHS=(1−cos⁡θ)21−cos⁡2θLHS = \sqrt{\frac{(1 - \cos \theta)^2}{1 - \cos^2 \theta}} Step 3: Use the identity 1−cos⁡2θ=sin⁡2θ1 - \cos^2 \theta = \sin^2 \theta. LHS=(1−cos⁡θ)2sin⁡2θLHS = \sqrt{\frac{(1 - \cos \theta)^2}{\sin^2 \theta}} Step 4: Remove the square root. LHS=1−cos⁡θsin⁡θLHS = \frac{1 - \cos \theta}{\sin \theta} Step 5: Split the fraction. LHS=1sin⁡θ−cos⁡θsin⁡θLHS = \frac{1}{\sin \theta} - \frac{\cos \theta}{\sin \theta} Step 6: Apply reciprocal and quotient identities. LHS=csc⁡θ−cot⁡θ=RHSLHS = \csc \theta - \cot \theta = RHS

Explanation:

The key strategy here is 'rationalizing' the expression under the square root to create perfect squares in both the numerator and denominator, allowing the root to be removed.