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Geometry and Trigonometry - Trigonometric identities and equations

Grade 11IB_AA

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

🔑Concepts

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The Pythagorean Identity sin⁡2θ+cos⁡2θ=1\sin^2\theta + \cos^2\theta = 1 is derived from the Unit Circle, where any point (x,y)(x, y) on the circle satisfies x2+y2=1x^2 + y^2 = 1 with x=cos⁡θx = \cos\theta and y=sin⁡θy = \sin\theta.

Unit circle demonstrating the geometric origin of Pythagorean identities.
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Trigonometric equations of the form sin⁡(kx)=a\sin(kx) = a or cos⁡(kx)=a\cos(kx) = a often result in multiple solutions within a given domain. These can be visualized as intersections between the horizontal line y=ay = a and the periodic trigonometric function.

Graph of sin(x) intersected by a constant line to show multiple solutions.
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Double angle identities allow for the simplification of expressions where the argument is doubled (2θ2\theta). cos⁡(2θ)\cos(2\theta) has three useful forms which can be selected based on the other terms in the equation to facilitate factoring.

Flowchart showing the three forms of the cosine double angle identity.
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The ASTC (All Students Take Calculus) rule or CAST diagram identifies the quadrants where sine, cosine, and tangent are positive, helping to find principal and secondary solutions for θ\theta in the range [0,2π][0, 2\pi].

📐Formulae

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

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

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

cos⁡(2θ)=cos⁡2θ−sin⁡2θ\cos(2\theta) = \cos^2\theta - \sin^2\theta

cos⁡(2θ)=2cos⁡2θ−1\cos(2\theta) = 2\cos^2\theta - 1

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

tan⁡2θ+1=sec⁡2θ\tan^2\theta + 1 = \sec^2\theta

1+cot⁡2θ=csc⁡2θ1 + \cot^2\theta = \csc^2\theta

💡Examples

Problem 1:

Solve the equation 2cos⁡2x+3sin⁡x−3=02\cos^2x + 3\sin x - 3 = 0 for 0≤x≤2π0 \leq x \leq 2\pi.

Solution:

  1. Use the identity cos⁡2x=1−sin⁡2x\cos^2x = 1 - \sin^2x: 2(1−sin⁡2x)+3sin⁡x−3=02(1 - \sin^2x) + 3\sin x - 3 = 0
  2. Expand and simplify: 2−2sin⁡2x+3sin⁡x−3=02 - 2\sin^2x + 3\sin x - 3 = 0 −2sin⁡2x+3sin⁡x−1=0-2\sin^2x + 3\sin x - 1 = 0 2sin⁡2x−3sin⁡x+1=02\sin^2x - 3\sin x + 1 = 0
  3. Factor the quadratic: (2sin⁡x−1)(sin⁡x−1)=0(2\sin x - 1)(\sin x - 1) = 0
  4. Solve for sin⁡x\sin x: sin⁡x=12 or sin⁡x=1\sin x = \frac{1}{2} \text{ or } \sin x = 1
  5. Find xx in the domain [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}

Explanation:

We first convert the equation into a single trigonometric ratio (sine) using the Pythagorean identity. Then, we treat it as a quadratic equation to find the values of sin⁡x\sin x before identifying the specific angles.

Problem 2:

Solve sin⁡(2x)=cos⁡x\sin(2x) = \cos x for 0≤x≤2π0 \leq x \leq 2\pi.

Solution:

  1. Use the double angle identity for sin⁡(2x)\sin(2x): 2sin⁡xcos⁡x=cos⁡x2\sin x \cos x = \cos x
  2. Rearrange the equation to one side: 2sin⁡xcos⁡x−cos⁡x=02\sin x \cos x - \cos x = 0
  3. Factor out cos⁡x\cos x: cos⁡x(2sin⁡x−1)=0\cos x (2\sin x - 1) = 0
  4. Set each factor to zero: cos⁡x=0  ⟹  x=π2,3π2\cos x = 0 \implies x = \frac{\pi}{2}, \frac{3\pi}{2} 2sin⁡x−1=0  ⟹  sin⁡x=12  ⟹  x=π6,5π62\sin x - 1 = 0 \implies \sin x = \frac{1}{2} \implies x = \frac{\pi}{6}, \frac{5\pi}{6}
  5. The solution set is x∈{π6,π2,5π6,3π2}x \in \{\frac{\pi}{6}, \frac{\pi}{2}, \frac{5\pi}{6}, \frac{3\pi}{2}\}.

Explanation:

It is critical not to divide both sides by cos⁡x\cos x, as this would result in losing solutions where cos⁡x=0\cos x = 0. Instead, factor the equation to find all possible roots.

Problem 3:

Given cos⁡θ=35\cos\theta = \frac{3}{5} and 3π2<θ<2π\frac{3\pi}{2} < \theta < 2\pi, find the value of sin⁡(2θ)\sin(2\theta).

Solution:

  1. Identify the quadrant: 3π2<θ<2π\frac{3\pi}{2} < \theta < 2\pi is the 4th quadrant, where sin⁡θ\sin\theta is negative.
  2. Find sin⁡θ\sin\theta using sin⁡2θ+cos⁡2θ=1\sin^2\theta + \cos^2\theta = 1: sin⁡2θ+(35)2=1\sin^2\theta + \left(\frac{3}{5}\right)^2 = 1 sin⁡2θ=1−925=1625\sin^2\theta = 1 - \frac{9}{25} = \frac{16}{25} sin⁡θ=−1625=−45\sin\theta = -\sqrt{\frac{16}{25}} = -\frac{4}{5}
  3. Use the double angle formula: sin⁡(2θ)=2sin⁡θcos⁡θ\sin(2\theta) = 2\sin\theta\cos\theta sin⁡(2θ)=2(−45)(35)=−2425\sin(2\theta) = 2\left(-\frac{4}{5}\right)\left(\frac{3}{5}\right) = -\frac{24}{25}

Explanation:

The quadrant is used to determine the sign of the sine value. Once both sin⁡θ\sin\theta and cos⁡θ\cos\theta are known, the double angle identity provides the final result.

Problem 4:

Solve the equation tan⁡2x−3=0\tan^2 x - 3 = 0 for 0≤x≤π0 \leq x \leq \pi.

Graph showing tan(x) and horizontal lines at positive and negative square root of 3.

Solution:

  1. Isolate the trigonometric term: tan⁡2x=3\tan^2 x = 3
  2. Take the square root of both sides: tan⁡x=±3\tan x = \pm \sqrt{3}
  3. Case 1: tan⁡x=3⇒x=π3\tan x = \sqrt{3} \Rightarrow x = \frac{\pi}{3}
  4. Case 2: tan⁡x=−3⇒x=π−π3=2π3\tan x = -\sqrt{3} \Rightarrow x = \pi - \frac{\pi}{3} = \frac{2\pi}{3}
  5. The solutions in the interval [0,π][0, \pi] are x=π3,2π3x = \frac{\pi}{3}, \frac{2\pi}{3}.

Explanation:

Since the equation involves tan⁡2x\tan^2 x, we obtain two possible values for tan⁡x\tan x. We then find the angles in the first and second quadrants that correspond to these values within the restricted domain.

Problem 5:

Show that 1−cos⁡(2x)sin⁡(2x)=tan⁡x\frac{1 - \cos(2x)}{\sin(2x)} = \tan x.

Diagram showing the transformation of Left Hand Side to Right Hand Side.

Solution:

  1. Substitute the double angle identities: 1−cos⁡(2x)=1−(1−2sin⁡2x)=2sin⁡2x1 - \cos(2x) = 1 - (1 - 2\sin^2 x) = 2\sin^2 x
  2. Substitute the denominator: sin⁡(2x)=2sin⁡xcos⁡x\sin(2x) = 2\sin x \cos x
  3. Form the fraction: 2sin⁡2x2sin⁡xcos⁡x\frac{2\sin^2 x}{2\sin x \cos x}
  4. Simplify by cancelling 2sin⁡x2\sin x: sin⁡xcos⁡x\frac{\sin x}{\cos x}
  5. Use the identity tan⁡x=sin⁡xcos⁡x\tan x = \frac{\sin x}{\cos x} to conclude the proof.

Explanation:

By choosing the form of cos⁡(2x)\cos(2x) that eliminates the constant 11, the expression simplifies into a single trigonometric ratio.