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Geometry and Trigonometry - Trigonometric Identities and Equations

Grade 12IB

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

🔑Concepts

The Unit Circle: The unit circle is a circle with a radius of 11 centered at the origin (0,0)(0,0) on the Cartesian plane. For any point P(x,y)P(x, y) on the circumference that forms an angle θ\theta with the positive x-axis, the coordinates are defined as x=cos(θ)x = \cos(\theta) and y=sin(θ)y = \sin(\theta). Visually, this creates a right-angled triangle where the hypotenuse is the radius (11), the adjacent side is xx, and the opposite side is yy. This relationship ensures that trigonometric functions are periodic, repeating every 2π2\pi radians.

The Pythagorean Identity: Derived directly from the unit circle and the Pythagorean theorem (a2+b2=c2a^2 + b^2 = c^2), we establish that sin2(θ)+cos2(θ)=1\sin^2(\theta) + \cos^2(\theta) = 1. This identity is crucial for converting between sine and cosine in equations. Visually, no matter where the point PP is on the circle, the square of the horizontal distance plus the square of the vertical distance always equals the square of the radius (121^2).

Double Angle Identities: These identities express functions of 2θ2\theta in terms of θ\theta. For example, sin(2θ)=2sin(θ)cos(θ)\sin(2\theta) = 2\sin(\theta)\cos(\theta). Visually, if you look at the graph of y=sin(2x)y = \sin(2x), it appears 'compressed' compared to y=sin(x)y = \sin(x), having half the period and completing two full oscillations in the same interval (2π2\pi) where the original sine graph completes only one.

Solving Trigonometric Equations: To solve equations such as sin(x)=k\sin(x) = k, we look for all values of xx within a specified domain (like [0,2π][0, 2\pi]). Visually, this corresponds to finding the x-coordinates where the horizontal line y=ky = k intersects the trigonometric graph. Due to symmetry, there are usually two solutions within one period (e.g., in the first and second quadrants for a positive sine value).

Reciprocal and Quotient Identities: The tangent function is defined as the ratio of sine to cosine: tan(θ)=sin(θ)cos(θ)\tan(\theta) = \frac{\sin(\theta)}{\cos(\theta)}. Reciprocal functions include cosecant (csc(θ)=1sin(θ)\csc(\theta) = \frac{1}{\sin(\theta)}), secant (sec(θ)=1cos(θ)\sec(\theta) = \frac{1}{\cos(\theta)}), and cotangent (cot(θ)=1tan(θ)\cot(\theta) = \frac{1}{\tan(\theta)}). Visually, tan(θ)\tan(\theta) represents the slope of the line passing through the origin and the point PP on the unit circle.

Quadratic Trigonometric Equations: Some equations appear in the form asin2(x)+bsin(x)+c=0a\sin^2(x) + b\sin(x) + c = 0. These can be solved using algebraic factoring or the quadratic formula by substituting u=sin(x)u = \sin(x). Once uu is found, we solve for xx by identifying the corresponding angles on the unit circle. Visually, these problems involve finding multiple intersection points across several 'waves' of the trigonometric function graph.

Compound Angle Identities: These formulas describe the sine, cosine, or tangent of the sum or difference of two angles, such as cos(AB)=cos(A)cos(B)+sin(A)sin(B)\cos(A - B) = \cos(A)\cos(B) + \sin(A)\sin(B). Geometrically, these represent the rotation of a vector by one angle and then further adjusting it by a second angle, helping to find exact values for non-standard angles like 1515^{\circ} or 7575^{\circ}.

📐Formulae

sin2(θ)+cos2(θ)=1\sin^2(\theta) + \cos^2(\theta) = 1

tan(θ)=sin(θ)cos(θ)\tan(\theta) = \frac{\sin(\theta)}{\cos(\theta)}

sin(2θ)=2sin(θ)cos(θ)\sin(2\theta) = 2\sin(\theta)\cos(\theta)

cos(2θ)=cos2(θ)sin2(θ)=2cos2(θ)1=12sin2(θ)\cos(2\theta) = \cos^2(\theta) - \sin^2(\theta) = 2\cos^2(\theta) - 1 = 1 - 2\sin^2(\theta)

tan(2θ)=2tan(θ)1tan2(θ)\tan(2\theta) = \frac{2\tan(\theta)}{1 - \tan^2(\theta)}

sin(A±B)=sin(A)cos(B)±cos(A)sin(B)\sin(A \pm B) = \sin(A)\cos(B) \pm \cos(A)\sin(B)

cos(A±B)=cos(A)cos(B)sin(A)sin(B)\cos(A \pm B) = \cos(A)\cos(B) \mp \sin(A)\sin(B)

tan(A±B)=tan(A)±tan(B)1tan(A)tan(B)\tan(A \pm B) = \frac{\tan(A) \pm \tan(B)}{1 \mp \tan(A)\tan(B)}

💡Examples

Problem 1:

Solve the equation 2cos2(x)+3sin(x)3=02\cos^2(x) + 3\sin(x) - 3 = 0 for 0x2π0 \leq x \leq 2\pi.

Solution:

Step 1: Use the Pythagorean identity cos2(x)=1sin2(x)\cos^2(x) = 1 - \sin^2(x) to get the equation in terms of sine only. 2(1sin2(x))+3sin(x)3=02(1 - \sin^2(x)) + 3\sin(x) - 3 = 0 Step 2: Expand and simplify. 22sin2(x)+3sin(x)3=0    2sin2(x)+3sin(x)1=02 - 2\sin^2(x) + 3\sin(x) - 3 = 0 \implies -2\sin^2(x) + 3\sin(x) - 1 = 0 Step 3: Multiply by 1-1 to make the leading coefficient positive. 2sin2(x)3sin(x)+1=02\sin^2(x) - 3\sin(x) + 1 = 0 Step 4: Factor the quadratic expression. (2sin(x)1)(sin(x)1)=0(2\sin(x) - 1)(\sin(x) - 1) = 0 Step 5: Solve for sin(x)\sin(x). sin(x)=12\sin(x) = \frac{1}{2} or sin(x)=1\sin(x) = 1 Step 6: Find the values of xx in the range [0,2π][0, 2\pi]. For sin(x)=12\sin(x) = \frac{1}{2}, x=π6,5π6x = \frac{\pi}{6}, \frac{5\pi}{6}. For sin(x)=1\sin(x) = 1, x=π2x = \frac{\pi}{2}. Final solution set: x{π6,π2,5π6}x \in \{\frac{\pi}{6}, \frac{\pi}{2}, \frac{5\pi}{6}\}.

Explanation:

This approach uses the Pythagorean identity to create a single-variable quadratic equation. We factor the quadratic and then find all angles within one full rotation of the unit circle that satisfy the resulting sine values.

Problem 2:

Prove the identity: sin(2θ)1+cos(2θ)=tan(θ)\frac{\sin(2\theta)}{1 + \cos(2\theta)} = \tan(\theta).

Solution:

Step 1: Start with the Left Hand Side (LHS) and apply the double angle identities. LHS = sin(2θ)1+cos(2θ)\frac{\sin(2\theta)}{1 + \cos(2\theta)} Step 2: Substitute sin(2θ)=2sin(θ)cos(θ)\sin(2\theta) = 2\sin(\theta)\cos(\theta) and cos(2θ)=2cos2(θ)1\cos(2\theta) = 2\cos^2(\theta) - 1 (choosing this form of the cosine identity helps cancel the 11 in the denominator). LHS = 2sin(θ)cos(θ)1+(2cos2(θ)1)\frac{2\sin(\theta)\cos(\theta)}{1 + (2\cos^2(\theta) - 1)} Step 3: Simplify the denominator. LHS = 2sin(θ)cos(θ)2cos2(θ)\frac{2\sin(\theta)\cos(\theta)}{2\cos^2(\theta)} Step 4: Cancel common factors (22 and cos(θ)\cos(\theta)). LHS = sin(θ)cos(θ)\frac{\sin(\theta)}{\cos(\theta)} Step 5: Recognize the quotient identity. LHS = tan(θ)=RHS\tan(\theta) = \text{RHS}.

Explanation:

This proof relies on choosing the correct form of the cos(2θ)\cos(2\theta) identity to simplify the expression efficiently. By canceling the constant term in the denominator, the fraction simplifies directly to the tangent definition.