IB Mathematics: Analysis and Approaches Quiz: Trig Ratios And Identities
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Trig Ratios And IdentitiesQuestion 1 of 20

If cos⁡(2α)=725\cos(2\alpha) = \frac{7}{25}, where 2α2\alpha is an acute angle, find the value of cos⁡4(α)−sin⁡4(α)\cos^4(\alpha) - \sin^4(\alpha).

725\frac{7}{25}
2425\frac{24}{25}
49625\frac{49}{625}
75\frac{\sqrt{7}}{5}
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IB Mathematics: Analysis and Approaches Quiz

IB Mathematics: Analysis and Approaches Quiz: Trig Ratios And Identities

Practice Trig Ratios And Identities in IB Mathematics: Analysis and Approaches with focused quiz questions that help you check what you know, review explanations, and build confidence with test-style prompts.

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This quiz focuses on Trig Ratios And Identities, giving you a quick way to practice the rules, question types, and explanations that matter most for IB Mathematics: Analysis and Approaches.

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Question 1

If cos⁡(2α)=725\cos(2\alpha) = \frac{7}{25}, where 2α2\alpha is an acute angle, find the value of cos⁡4(α)−sin⁡4(α)\cos^4(\alpha) - \sin^4(\alpha).

  1. 725\frac{7}{25} (correct answer)
  2. 2425\frac{24}{25}
  3. 49625\frac{49}{625}
  4. 75\frac{\sqrt{7}}{5}
Explanation: The expression cos⁡4(α)−sin⁡4(α)\cos^4(\alpha) - \sin^4(\alpha) is a difference of two squares. It can be factored as (cos⁡2(α)−sin⁡2(α))(cos⁡2(α)+sin⁡2(α))(\cos^2(\alpha) - \sin^2(\alpha))(\cos^2(\alpha) + \sin^2(\alpha)). Using the Pythagorean identity, cos⁡2(α)+sin⁡2(α)=1\cos^2(\alpha) + \sin^2(\alpha) = 1. The expression simplifies to cos⁡2(α)−sin⁡2(α)\cos^2(\alpha) - \sin^2(\alpha). This is the double angle identity for cosine, cos⁡(2α)\cos(2\alpha). Since we are given that cos⁡(2α)=725\cos(2\alpha) = \frac{7}{25}, the value of the expression is 725\frac{7}{25}.

Question 2

For what value of kk is the equation cos⁡2(x)+3sin⁡(x)=k\cos^2(x) + 3\sin(x) = k true for x=π6x = \frac{\pi}{6}?

  1. 32\frac{3}{2}
  2. 74\frac{7}{4}
  3. 94\frac{9}{4} (correct answer)
  4. 3+32\frac{3+\sqrt{3}}{2}
Explanation: Substitute x=π6x = \frac{\pi}{6} into the expression. We know that sin⁡(π6)=12\sin(\frac{\pi}{6}) = \frac{1}{2} and cos⁡(π6)=32\cos(\frac{\pi}{6}) = \frac{\sqrt{3}}{2}. The expression becomes k=cos⁡2(π6)+3sin⁡(π6)=(32)2+3(12)=34+32k = \cos^2(\frac{\pi}{6}) + 3\sin(\frac{\pi}{6}) = (\frac{\sqrt{3}}{2})^2 + 3(\frac{1}{2}) = \frac{3}{4} + \frac{3}{2}. To add these fractions, find a common denominator: 34+64=94\frac{3}{4} + \frac{6}{4} = \frac{9}{4}.

Question 3

If cos⁡(2α)=725\cos(2\alpha) = \frac{7}{25}, where 2α2\alpha is an acute angle, find the value of cos⁡4(α)−sin⁡4(α)\cos^4(\alpha) - \sin^4(\alpha).

  1. 725\frac{7}{25} (correct answer)
  2. 2425\frac{24}{25}
  3. 49625\frac{49}{625}
  4. 75\frac{\sqrt{7}}{5}
Explanation: The expression cos⁡4(α)−sin⁡4(α)\cos^4(\alpha) - \sin^4(\alpha) is a difference of two squares. It can be factored as (cos⁡2(α)−sin⁡2(α))(cos⁡2(α)+sin⁡2(α))(\cos^2(\alpha) - \sin^2(\alpha))(\cos^2(\alpha) + \sin^2(\alpha)). Using the Pythagorean identity, cos⁡2(α)+sin⁡2(α)=1\cos^2(\alpha) + \sin^2(\alpha) = 1. The expression simplifies to cos⁡2(α)−sin⁡2(α)\cos^2(\alpha) - \sin^2(\alpha). This is the double angle identity for cosine, cos⁡(2α)\cos(2\alpha). Since we are given that cos⁡(2α)=725\cos(2\alpha) = \frac{7}{25}, the value of the expression is 725\frac{7}{25}.

Question 4

Given that sin⁡(x)=p\sin(x) = p where π2<x<π\frac{\pi}{2} < x < \pi, express tan⁡(x)\tan(x) in terms of pp.

  1. −1−p2p-\frac{\sqrt{1-p^2}}{p}
  2. −p1−p2-\frac{p}{\sqrt{1-p^2}} (correct answer)
  3. p1−p2\frac{p}{\sqrt{1-p^2}}
  4. 1−p2p\frac{\sqrt{1-p^2}}{p}
Explanation: First, find cos⁡(x)\cos(x) using sin⁡2(x)+cos⁡2(x)=1\sin^2(x) + \cos^2(x) = 1. We have cos⁡2(x)=1−sin⁡2(x)=1−p2\cos^2(x) = 1 - \sin^2(x) = 1 - p^2, so cos⁡(x)=±1−p2\cos(x) = \pm\sqrt{1-p^2}. The given domain π2<x<π\frac{\pi}{2} < x < \pi is the second quadrant, where cosine is negative. Therefore, cos⁡(x)=−1−p2\cos(x) = -\sqrt{1-p^2}. Now, use the identity tan⁡(x)=sin⁡(x)cos⁡(x)\tan(x) = \frac{\sin(x)}{\cos(x)}. Substituting the known values gives tan⁡(x)=p−1−p2=−p1−p2\tan(x) = \frac{p}{-\sqrt{1-p^2}} = -\frac{p}{\sqrt{1-p^2}}.

Question 5

The expression (tan⁡(θ)+cot⁡(θ))sin⁡(θ)cos⁡(θ)(\tan(\theta) + \cot(\theta))\sin(\theta)\cos(\theta) simplifies to:

  1. 1 (correct answer)
  2. 2
  3. sin⁡(θ)+cos⁡(θ)\sin(\theta) + \cos(\theta)
  4. tan⁡2(θ)\tan^2(\theta)
Explanation: First, express tan⁡(θ)\tan(\theta) and cot⁡(θ)\cot(\theta) in terms of sin⁡(θ)\sin(\theta) and cos⁡(θ)\cos(\theta). The expression becomes (sin⁡(θ)cos⁡(θ)+cos⁡(θ)sin⁡(θ))sin⁡(θ)cos⁡(θ)\left(\frac{\sin(\theta)}{\cos(\theta)} + \frac{\cos(\theta)}{\sin(\theta)}\right)\sin(\theta)\cos(\theta). Combine the terms in the parenthesis with a common denominator of sin⁡(θ)cos⁡(θ)\sin(\theta)\cos(\theta): (sin⁡2(θ)+cos⁡2(θ)sin⁡(θ)cos⁡(θ))sin⁡(θ)cos⁡(θ)\left(\frac{\sin^2(\theta) + \cos^2(\theta)}{\sin(\theta)\cos(\theta)}\right)\sin(\theta)\cos(\theta). Using the identity sin⁡2(θ)+cos⁡2(θ)=1\sin^2(\theta) + \cos^2(\theta) = 1, this simplifies to (1sin⁡(θ)cos⁡(θ))sin⁡(θ)cos⁡(θ)\left(\frac{1}{\sin(\theta)\cos(\theta)}\right)\sin(\theta)\cos(\theta). The sin⁡(θ)cos⁡(θ)\sin(\theta)\cos(\theta) terms cancel, leaving a result of 1.

Question 6

If tan⁡(θ)=3\tan(\theta) = 3 and cos⁡(θ)<0\cos(\theta) < 0, find the exact value of sin⁡(θ)\sin(\theta).

  1. −310-\frac{3}{\sqrt{10}} (correct answer)
  2. −110-\frac{1}{\sqrt{10}}
  3. 110\frac{1}{\sqrt{10}}
  4. 310\frac{3}{\sqrt{10}}
Explanation: Since tan⁡(θ)=sin⁡(θ)cos⁡(θ)>0\tan(\theta) = \frac{\sin(\theta)}{\cos(\theta)} > 0 and cos⁡(θ)<0\cos(\theta) < 0, it must be that sin⁡(θ)<0\sin(\theta) < 0 as well. This places θ\theta in the third quadrant. We can construct a right-angled triangle with opposite side 3 and adjacent side 1. The hypotenuse would be 32+12=10\sqrt{3^2 + 1^2} = \sqrt{10}. From this reference triangle, ∣sin⁡(θ)∣=310|\sin(\theta)| = \frac{3}{\sqrt{10}}. Since θ\theta is in the third quadrant, sin⁡(θ)\sin(\theta) is negative. Thus, sin⁡(θ)=−310\sin(\theta) = -\frac{3}{\sqrt{10}}.

Question 7

The expression 1−sin⁡2(x)1−cos⁡2(x)\frac{1 - \sin^2(x)}{1 - \cos^2(x)} is equivalent to:

  1. tan⁡2(x)\tan^2(x)
  2. cot⁡2(x)\cot^2(x) (correct answer)
  3. −1-1
  4. 1
Explanation: Using the Pythagorean identity sin⁡2(x)+cos⁡2(x)=1\sin^2(x) + \cos^2(x) = 1, we can rewrite the numerator and the denominator. The numerator 1−sin⁡2(x)1 - \sin^2(x) is equal to cos⁡2(x)\cos^2(x). The denominator 1−cos⁡2(x)1 - \cos^2(x) is equal to sin⁡2(x)\sin^2(x). So the expression becomes cos⁡2(x)sin⁡2(x)\frac{\cos^2(x)}{\sin^2(x)}. This is equal to (cos⁡(x)sin⁡(x))2\left(\frac{\cos(x)}{\sin(x)}\right)^2. Since cot⁡(x)=cos⁡(x)sin⁡(x)\cot(x) = \frac{\cos(x)}{\sin(x)}, the expression simplifies to cot⁡2(x)\cot^2(x).

Question 8

The equation tan⁡2(x)−2tan⁡(x)−3=0\tan^2(x) - 2\tan(x) - 3 = 0 holds for an acute angle xx. Which of the following is the value of sin⁡(x)\sin(x)?

  1. −12-\frac{1}{\sqrt{2}}
  2. 110\frac{1}{\sqrt{10}}
  3. 310\frac{3}{\sqrt{10}} (correct answer)
  4. 12\frac{1}{\sqrt{2}}
Explanation: Let u=tan⁡(x)u = \tan(x). The equation becomes a quadratic: u2−2u−3=0u^2 - 2u - 3 = 0. Factoring gives (u−3)(u+1)=0(u-3)(u+1) = 0, so u=3u=3 or u=−1u=-1. This means tan⁡(x)=3\tan(x) = 3 or tan⁡(x)=−1\tan(x) = -1. Since xx is an acute angle (in the first quadrant), its tangent must be positive. Thus, we must have tan⁡(x)=3\tan(x) = 3. To find sin⁡(x)\sin(x), we can visualize a right-angled triangle where the opposite side is 3 and the adjacent side is 1. The hypotenuse is 32+12=10\sqrt{3^2 + 1^2} = \sqrt{10}. Therefore, sin⁡(x)=oppositehypotenuse=310\sin(x) = \frac{\text{opposite}}{\text{hypotenuse}} = \frac{3}{\sqrt{10}}.

Question 9

The expression 11−sin⁡(x)+11+sin⁡(x)\frac{1}{1 - \sin(x)} + \frac{1}{1 + \sin(x)} is equivalent to:

  1. 2
  2. 2sin⁡2(x)\frac{2}{\sin^2(x)}
  3. 2cos⁡(x)\frac{2}{\cos(x)}
  4. 2cos⁡2(x)\frac{2}{\cos^2(x)} (correct answer)
Explanation: To add the fractions, find a common denominator, which is (1−sin⁡(x))(1+sin⁡(x))(1 - \sin(x))(1 + \sin(x)). This product simplifies to 1−sin⁡2(x)1 - \sin^2(x), which by the Pythagorean identity is equal to cos⁡2(x)\cos^2(x). The numerator becomes (1+sin⁡(x))+(1−sin⁡(x))=2(1 + \sin(x)) + (1 - \sin(x)) = 2. Therefore, the expression is 2cos⁡2(x)\frac{2}{\cos^2(x)}.

Question 10

The expression (tan⁡(θ)+cot⁡(θ))sin⁡(θ)cos⁡(θ)(\tan(\theta) + \cot(\theta))\sin(\theta)\cos(\theta) simplifies to:

  1. 1 (correct answer)
  2. 2
  3. sin⁡(θ)+cos⁡(θ)\sin(\theta) + \cos(\theta)
  4. tan⁡2(θ)\tan^2(\theta)
Explanation: First, express tan⁡(θ)\tan(\theta) and cot⁡(θ)\cot(\theta) in terms of sin⁡(θ)\sin(\theta) and cos⁡(θ)\cos(\theta). The expression becomes (sin⁡(θ)cos⁡(θ)+cos⁡(θ)sin⁡(θ))sin⁡(θ)cos⁡(θ)\left(\frac{\sin(\theta)}{\cos(\theta)} + \frac{\cos(\theta)}{\sin(\theta)}\right)\sin(\theta)\cos(\theta). Combine the terms in the parenthesis with a common denominator of sin⁡(θ)cos⁡(θ)\sin(\theta)\cos(\theta): (sin⁡2(θ)+cos⁡2(θ)sin⁡(θ)cos⁡(θ))sin⁡(θ)cos⁡(θ)\left(\frac{\sin^2(\theta) + \cos^2(\theta)}{\sin(\theta)\cos(\theta)}\right)\sin(\theta)\cos(\theta). Using the identity sin⁡2(θ)+cos⁡2(θ)=1\sin^2(\theta) + \cos^2(\theta) = 1, this simplifies to (1sin⁡(θ)cos⁡(θ))sin⁡(θ)cos⁡(θ)\left(\frac{1}{\sin(\theta)\cos(\theta)}\right)\sin(\theta)\cos(\theta). The sin⁡(θ)cos⁡(θ)\sin(\theta)\cos(\theta) terms cancel, leaving a result of 1.

Question 11

Given that tan⁡(θ)=k\tan(\theta) = k for an acute angle θ\theta, which expression represents the product sin⁡(θ)cos⁡(θ)\sin(\theta)\cos(\theta)?

  1. 1k2+1\frac{1}{k^2+1}
  2. kk2−1\frac{k}{k^2-1}
  3. kk2+1\frac{k}{\sqrt{k^2+1}}
  4. kk2+1\frac{k}{k^2+1} (correct answer)
Explanation: When you encounter trigonometric problems involving one ratio and need to find another, the key strategy is to use the Pythagorean identity and the fundamental relationships between sine, cosine, and tangent. Given that tan⁡(θ)=k\tan(\theta) = k where θ\theta is acute, you can construct a right triangle where the opposite side is kk and the adjacent side is 11. Using the Pythagorean theorem, the hypotenuse equals k2+1\sqrt{k^2 + 1}. From this triangle: sin⁡(θ)=kk2+1\sin(\theta) = \frac{k}{\sqrt{k^2+1}} and cos⁡(θ)=1k2+1\cos(\theta) = \frac{1}{\sqrt{k^2+1}} Therefore: sin⁡(θ)cos⁡(θ)=kk2+1⋅1k2+1=kk2+1\sin(\theta)\cos(\theta) = \frac{k}{\sqrt{k^2+1}} \cdot \frac{1}{\sqrt{k^2+1}} = \frac{k}{k^2+1} This confirms answer D is correct. Answer A, 1k2+1\frac{1}{k^2+1}, represents cos⁡2(θ)\cos^2(\theta), not the product sin⁡(θ)cos⁡(θ)\sin(\theta)\cos(\theta). This is a common error when students confuse which trigonometric expression they're calculating. Answer B, kk2−1\frac{k}{k^2-1}, uses subtraction instead of addition in the denominator. This mistake often occurs when students misremember the Pythagorean identity or confuse it with other algebraic identities. Answer C, kk2+1\frac{k}{\sqrt{k^2+1}}, represents sin⁡(θ)\sin(\theta) alone, not the product. Students might select this if they forget to multiply by the cosine term. Remember: when working with trigonometric ratios, always draw a reference triangle to visualize the relationships. This prevents algebraic errors and helps you verify your final expression makes geometric sense.

Question 12

Which of the following expressions is equivalent to 1−cos⁡(θ)sin⁡(θ)\frac{1 - \cos(\theta)}{\sin(\theta)}?

  1. cos⁡(θ)1+sin⁡(θ)\frac{\cos(\theta)}{1 + \sin(\theta)}
  2. 1cos⁡(θ)\frac{1}{\cos(\theta)}
  3. sin⁡(θ)1−cos⁡(θ)\frac{\sin(\theta)}{1 - \cos(\theta)}
  4. sin⁡(θ)1+cos⁡(θ)\frac{\sin(\theta)}{1 + \cos(\theta)} (correct answer)
Explanation: To find an equivalent expression, we can multiply the numerator and denominator by the conjugate of the numerator, which is 1+cos⁡(θ)1 + \cos(\theta). This gives 1−cos⁡(θ)sin⁡(θ)×1+cos⁡(θ)1+cos⁡(θ)=1−cos⁡2(θ)sin⁡(θ)(1+cos⁡(θ))\frac{1 - \cos(\theta)}{\sin(\theta)} \times \frac{1 + \cos(\theta)}{1 + \cos(\theta)} = \frac{1 - \cos^2(\theta)}{\sin(\theta)(1 + \cos(\theta))}. Using the Pythagorean identity sin⁡2(θ)+cos⁡2(θ)=1\sin^2(\theta) + \cos^2(\theta) = 1, we know that 1−cos⁡2(θ)=sin⁡2(θ)1 - \cos^2(\theta) = \sin^2(\theta). The expression becomes sin⁡2(θ)sin⁡(θ)(1+cos⁡(θ))\frac{\sin^2(\theta)}{\sin(\theta)(1 + \cos(\theta))}. Canceling one factor of sin⁡(θ)\sin(\theta) from the numerator and denominator yields sin⁡(θ)1+cos⁡(θ)\frac{\sin(\theta)}{1 + \cos(\theta)}.

Question 13

If xx is an angle in the second quadrant such that tan⁡(x)=−43\tan(x) = -\frac{4}{3}, what is the value of 5(sin⁡(x)+cos⁡(x))5(\sin(x) + \cos(x))?

  1. -7
  2. -1
  3. 1 (correct answer)
  4. 7
Explanation: In the second quadrant, sin⁡(x)\sin(x) is positive and cos⁡(x)\cos(x) is negative. From tan⁡(x)=−43\tan(x) = -\frac{4}{3}, we can model a right-angled triangle with an opposite side of 4 and an adjacent side of 3. The hypotenuse is 42+32=25=5\sqrt{4^2 + 3^2} = \sqrt{25} = 5. Using the quadrant information for the signs, we have sin⁡(x)=45\sin(x) = \frac{4}{5} and cos⁡(x)=−35\cos(x) = -\frac{3}{5}. Now substitute these values into the expression: 5(sin⁡(x)+cos⁡(x))=5(45+(−35))=5(45−35)=5(15)=15(\sin(x) + \cos(x)) = 5(\frac{4}{5} + (-\frac{3}{5})) = 5(\frac{4}{5} - \frac{3}{5}) = 5(\frac{1}{5}) = 1.

Question 14

Given that 3sin⁡2(θ)+7cos⁡(θ)=53\sin^2(\theta) + 7\cos(\theta) = 5 for 0<θ<π20 < \theta < \frac{\pi}{2}, find the value of cos⁡(θ)\cos(\theta).

  1. 13\frac{1}{3} (correct answer)
  2. 12\frac{1}{2}
  3. 23\frac{2}{3}
  4. 2
Explanation: Use the identity sin⁡2(θ)=1−cos⁡2(θ)\sin^2(\theta) = 1 - \cos^2(\theta) to write the equation solely in terms of cos⁡(θ)\cos(\theta). 3(1−cos⁡2(θ))+7cos⁡(θ)=53(1 - \cos^2(\theta)) + 7\cos(\theta) = 5. Expand and rearrange: 3−3cos⁡2(θ)+7cos⁡(θ)−5=03 - 3\cos^2(\theta) + 7\cos(\theta) - 5 = 0, which simplifies to −3cos⁡2(θ)+7cos⁡(θ)−2=0-3\cos^2(\theta) + 7\cos(\theta) - 2 = 0. Multiply by -1 to get 3cos⁡2(θ)−7cos⁡(θ)+2=03\cos^2(\theta) - 7\cos(\theta) + 2 = 0. Let c=cos⁡(θ)c = \cos(\theta). The quadratic is 3c2−7c+2=03c^2 - 7c + 2 = 0. Factoring gives (3c−1)(c−2)=0(3c - 1)(c - 2) = 0. The solutions are c=13c = \frac{1}{3} or c=2c = 2. Since the range of cos⁡(θ)\cos(\theta) is [−1,1][-1, 1], the solution cos⁡(θ)=2\cos(\theta) = 2 is impossible. Therefore, cos⁡(θ)=13\cos(\theta) = \frac{1}{3}. This is a valid solution as it is positive, consistent with θ\theta being in the first quadrant.

Question 15

The expression 11−sin⁡(x)+11+sin⁡(x)\frac{1}{1 - \sin(x)} + \frac{1}{1 + \sin(x)} is equivalent to:

  1. 2
  2. 2sin⁡2(x)\frac{2}{\sin^2(x)}
  3. 2cos⁡(x)\frac{2}{\cos(x)}
  4. 2cos⁡2(x)\frac{2}{\cos^2(x)} (correct answer)
Explanation: To add the fractions, find a common denominator, which is (1−sin⁡(x))(1+sin⁡(x))(1 - \sin(x))(1 + \sin(x)). This product simplifies to 1−sin⁡2(x)1 - \sin^2(x), which by the Pythagorean identity is equal to cos⁡2(x)\cos^2(x). The numerator becomes (1+sin⁡(x))+(1−sin⁡(x))=2(1 + \sin(x)) + (1 - \sin(x)) = 2. Therefore, the expression is 2cos⁡2(x)\frac{2}{\cos^2(x)}.

Question 16

The expression cos⁡(x)+sin⁡(x)tan⁡(x)\cos(x) + \sin(x)\tan(x) is equivalent to:

  1. 1
  2. 1sin⁡(x)\frac{1}{\sin(x)}
  3. 1cos⁡(x)\frac{1}{\cos(x)} (correct answer)
  4. 2cos⁡(x)2\cos(x)
Explanation: Substitute tan⁡(x)=sin⁡(x)cos⁡(x)\tan(x) = \frac{\sin(x)}{\cos(x)} into the expression: cos⁡(x)+sin⁡(x)(sin⁡(x)cos⁡(x))=cos⁡(x)+sin⁡2(x)cos⁡(x)\cos(x) + \sin(x) \left( \frac{\sin(x)}{\cos(x)} \right) = \cos(x) + \frac{\sin^2(x)}{\cos(x)}. To add these terms, find a common denominator: cos⁡2(x)cos⁡(x)+sin⁡2(x)cos⁡(x)=cos⁡2(x)+sin⁡2(x)cos⁡(x)\frac{\cos^2(x)}{\cos(x)} + \frac{\sin^2(x)}{\cos(x)} = \frac{\cos^2(x) + \sin^2(x)}{\cos(x)}. Using the Pythagorean identity sin⁡2(x)+cos⁡2(x)=1\sin^2(x) + \cos^2(x) = 1, the expression simplifies to 1cos⁡(x)\frac{1}{\cos(x)}.

Question 17

Simplify the expression sin⁡4(θ)−cos⁡4(θ)sin⁡2(θ)−cos⁡2(θ)\frac{\sin^4(\theta) - \cos^4(\theta)}{\sin^2(\theta) - \cos^2(\theta)}, assuming the denominator is not zero.

  1. -1
  2. 0
  3. 1 (correct answer)
  4. 2
Explanation: The numerator sin⁡4(θ)−cos⁡4(θ)\sin^4(\theta) - \cos^4(\theta) is a difference of two squares, which can be factored as (sin⁡2(θ)−cos⁡2(θ))(sin⁡2(θ)+cos⁡2(θ))(\sin^2(\theta) - \cos^2(\theta))(\sin^2(\theta) + \cos^2(\theta)). The expression becomes (sin⁡2(θ)−cos⁡2(θ))(sin⁡2(θ)+cos⁡2(θ))sin⁡2(θ)−cos⁡2(θ)\frac{(\sin^2(\theta) - \cos^2(\theta))(\sin^2(\theta) + \cos^2(\theta))}{\sin^2(\theta) - \cos^2(\theta)}. The term (sin⁡2(θ)−cos⁡2(θ))(\sin^2(\theta) - \cos^2(\theta)) cancels out, leaving sin⁡2(θ)+cos⁡2(θ)\sin^2(\theta) + \cos^2(\theta). By the Pythagorean identity, this is equal to 1.

Question 18

Given that sin⁡(x)=p\sin(x) = p where π2<x<π\frac{\pi}{2} < x < \pi, express tan⁡(x)\tan(x) in terms of pp.

  1. −1−p2p-\frac{\sqrt{1-p^2}}{p}
  2. −p1−p2-\frac{p}{\sqrt{1-p^2}} (correct answer)
  3. p1−p2\frac{p}{\sqrt{1-p^2}}
  4. 1−p2p\frac{\sqrt{1-p^2}}{p}
Explanation: First, find cos⁡(x)\cos(x) using sin⁡2(x)+cos⁡2(x)=1\sin^2(x) + \cos^2(x) = 1. We have cos⁡2(x)=1−sin⁡2(x)=1−p2\cos^2(x) = 1 - \sin^2(x) = 1 - p^2, so cos⁡(x)=±1−p2\cos(x) = \pm\sqrt{1-p^2}. The given domain π2<x<π\frac{\pi}{2} < x < \pi is the second quadrant, where cosine is negative. Therefore, cos⁡(x)=−1−p2\cos(x) = -\sqrt{1-p^2}. Now, use the identity tan⁡(x)=sin⁡(x)cos⁡(x)\tan(x) = \frac{\sin(x)}{\cos(x)}. Substituting the known values gives tan⁡(x)=p−1−p2=−p1−p2\tan(x) = \frac{p}{-\sqrt{1-p^2}} = -\frac{p}{\sqrt{1-p^2}}.

Question 19

Given that tan⁡(θ)=k\tan(\theta) = k for an acute angle θ\theta, which expression represents the product sin⁡(θ)cos⁡(θ)\sin(\theta)\cos(\theta)?

  1. 1k2+1\frac{1}{k^2+1}
  2. kk2−1\frac{k}{k^2-1}
  3. kk2+1\frac{k}{\sqrt{k^2+1}}
  4. kk2+1\frac{k}{k^2+1} (correct answer)
Explanation: When you encounter trigonometric problems involving one ratio and need to find another, the key strategy is to use the Pythagorean identity and the fundamental relationships between sine, cosine, and tangent. Given that tan⁡(θ)=k\tan(\theta) = k where θ\theta is acute, you can construct a right triangle where the opposite side is kk and the adjacent side is 11. Using the Pythagorean theorem, the hypotenuse equals k2+1\sqrt{k^2 + 1}. From this triangle: sin⁡(θ)=kk2+1\sin(\theta) = \frac{k}{\sqrt{k^2+1}} and cos⁡(θ)=1k2+1\cos(\theta) = \frac{1}{\sqrt{k^2+1}} Therefore: sin⁡(θ)cos⁡(θ)=kk2+1⋅1k2+1=kk2+1\sin(\theta)\cos(\theta) = \frac{k}{\sqrt{k^2+1}} \cdot \frac{1}{\sqrt{k^2+1}} = \frac{k}{k^2+1} This confirms answer D is correct. Answer A, 1k2+1\frac{1}{k^2+1}, represents cos⁡2(θ)\cos^2(\theta), not the product sin⁡(θ)cos⁡(θ)\sin(\theta)\cos(\theta). This is a common error when students confuse which trigonometric expression they're calculating. Answer B, kk2−1\frac{k}{k^2-1}, uses subtraction instead of addition in the denominator. This mistake often occurs when students misremember the Pythagorean identity or confuse it with other algebraic identities. Answer C, kk2+1\frac{k}{\sqrt{k^2+1}}, represents sin⁡(θ)\sin(\theta) alone, not the product. Students might select this if they forget to multiply by the cosine term. Remember: when working with trigonometric ratios, always draw a reference triangle to visualize the relationships. This prevents algebraic errors and helps you verify your final expression makes geometric sense.

Question 20

The expression 1−sin⁡2(x)1−cos⁡2(x)\frac{1 - \sin^2(x)}{1 - \cos^2(x)} is equivalent to:

  1. tan⁡2(x)\tan^2(x)
  2. cot⁡2(x)\cot^2(x) (correct answer)
  3. −1-1
  4. 1
Explanation: Using the Pythagorean identity sin⁡2(x)+cos⁡2(x)=1\sin^2(x) + \cos^2(x) = 1, we can rewrite the numerator and the denominator. The numerator 1−sin⁡2(x)1 - \sin^2(x) is equal to cos⁡2(x)\cos^2(x). The denominator 1−cos⁡2(x)1 - \cos^2(x) is equal to sin⁡2(x)\sin^2(x). So the expression becomes cos⁡2(x)sin⁡2(x)\frac{\cos^2(x)}{\sin^2(x)}. This is equal to (cos⁡(x)sin⁡(x))2\left(\frac{\cos(x)}{\sin(x)}\right)^2. Since cot⁡(x)=cos⁡(x)sin⁡(x)\cot(x) = \frac{\cos(x)}{\sin(x)}, the expression simplifies to cot⁡2(x)\cot^2(x).