Trigonometric Identities - Math 3

If $\cos \psi = -\frac{3}{5}$ and $\psi$ is in the second quadrant, what is the value of $\frac{\sin \psi}{1 + \cos \psi}$ ?
Answer: $2$ When you encounter trigonometric problems with given values and quadrant information, your first step is to find the missing trigonometric function using the Pythagorean identity, then carefully track signs based on the quadrant. Since $\cos \psi = -\frac{3}{5}$ and $\psi$ is in the second quadrant, you need to find $\sin \psi$ . Using $\sin^2 \psi + \cos^2 \psi = 1$ : $\sin^2 \psi = 1 - \cos^2 \psi = 1 - \left(-\frac{3}{5}\right)^2 = 1 - \frac{9}{25} = \frac{16}{25}$ So $\sin \psi = \pm\frac{4}{5}$ . Since $\psi$ is in the second quadrant where sine is positive, $\sin \psi = \frac{4}{5}$ . Now you can evaluate the expression: $\frac{\sin \psi}{1 + \cos \psi} = \frac{\frac{4}{5}}{1 + \left(-\frac{3}{5}\right)} = \frac{\frac{4}{5}}{\frac{2}{5}} = \frac{4}{5} \cdot \frac{5}{2} = 2$ The correct answer is C) $2$ . Choice A) $\frac{1}{2}$ results from incorrectly using $\sin \psi = \frac{1}{5}$ instead of the correct value. Choice B) $-2$ comes from using the wrong sign for sine (treating it as negative in the second quadrant). Choice D) $-\frac{1}{2}$ combines both errors: wrong sine value and wrong sign. Remember: in quadrant problems, always determine signs carefully using "All Students Take Calculus" (All positive in I, Sine positive in II, Tangent positive in III, Cosine positive in IV).
Which of the following is equivalent to $\frac{\sin^2 \beta}{1 - \cos \beta}$ for $\cos \beta \neq 1$ ?
Answer: $1 + \cos \beta$ Using $\sin^2 \beta = 1 - \cos^2 \beta = (1 - \cos \beta)(1 + \cos \beta)$ , we can substitute: $\frac{\sin^2 \beta}{1 - \cos \beta} = \frac{(1 - \cos \beta)(1 + \cos \beta)}{1 - \cos \beta} = 1 + \cos \beta$ (provided $\cos \beta \neq 1$ ). Choice B is the denominator. Choice C would result from incorrectly inverting the entire expression. Choice D would result from incorrectly canceling terms.
For what value of $\cos \phi$ does the expression $\frac{\sin^2 \phi + \cos^2 \phi}{\cos \phi}$ equal $\frac{4}{3}$ ?
Answer: $\cos \phi = \frac{3}{4}$ When you encounter trigonometric expressions with both sine and cosine terms, always look for opportunities to apply fundamental identities. The key insight here is recognizing that $\sin^2 \phi + \cos^2 \phi = 1$ by the Pythagorean identity. Let's simplify the given expression first. Since $\sin^2 \phi + \cos^2 \phi = 1$ , we can rewrite: $\frac{\sin^2 \phi + \cos^2 \phi}{\cos \phi} = \frac{1}{\cos \phi}$ Now we need to solve: $\frac{1}{\cos \phi} = \frac{4}{3}$ Cross-multiplying gives us: $3 = 4\cos \phi$ Therefore: $\cos \phi = \frac{3}{4}$ This confirms that answer C is correct. Let's examine why the other options fail. If we substitute each into our equation $\frac{1}{\cos \phi} = \frac{4}{3}$ : For A) $\cos \phi = \frac{1}{3}$ : This gives $\frac{1}{\frac{1}{3}} = 3 \neq \frac{4}{3}$ For B) $\cos \phi = \frac{4}{3}$ : This gives $\frac{1}{\frac{4}{3}} = \frac{3}{4} \neq \frac{4}{3}$ . Also, cosine values must be between -1 and 1, so $\frac{4}{3}$ is impossible. For D) $\cos \phi = \frac{1}{4}$ : This gives $\frac{1}{\frac{1}{4}} = 4 \neq \frac{4}{3}$ Study tip: Always simplify trigonometric expressions using basic identities before solving equations. The Pythagorean identity $\sin^2 \phi + \cos^2 \phi = 1$ is especially powerful for eliminating complex terms and revealing simpler relationships.
Given that $\cos \lambda \neq 0$ and $\sin \lambda \neq 0$ , which of the following trigonometric identities must be true?
Answer: All of the above These are fundamental trigonometric identities derived from $\sin^2 \lambda + \cos^2 \lambda = 1$ . Identity A: Dividing by $\cos^2 \lambda$ gives $\tan^2 \lambda + 1 = \sec^2 \lambda$ . Identity B: Dividing by $\sin^2 \lambda$ gives $1 + \cot^2 \lambda = \csc^2 \lambda$ . Identity C: This is equivalent to identity A. Since $\cos \lambda \neq 0$ and $\sin \lambda \neq 0$ , all three identities are valid.
If $\sec \alpha + \tan \alpha = 3$ , what is the value of $\sec \alpha - \tan \alpha$ ?
Answer: $\frac{1}{3}$ We use the identity $(\sec \alpha + \tan \alpha)(\sec \alpha - \tan \alpha) = \sec^2 \alpha - \tan^2 \alpha = 1$ (from the Pythagorean identity $\sec^2 \alpha - \tan^2 \alpha = 1$ ). Given that $\sec \alpha + \tan \alpha = 3$ , we have $3 \cdot (\sec \alpha - \tan \alpha) = 1$ . Therefore, $\sec \alpha - \tan \alpha = \frac{1}{3}$ . Choice B results from incorrectly squaring the given equation. Choice C assumes the two expressions are equal. Choice D uses the wrong sign from a calculation error.
If $\cos \alpha = \frac{5}{13}$ and $\sin \alpha > 0$ , what is the value of $\frac{1 + \tan^2 \alpha}{\sec \alpha}$ ?
Answer: $\frac{13}{5}$ First, we use the identity $1 + \tan^2 \alpha = \sec^2 \alpha$ . So the expression becomes $\frac{\sec^2 \alpha}{\sec \alpha} = \sec \alpha$ . Since $\cos \alpha = \frac{5}{13}$ , we have $\sec \alpha = \frac{1}{\cos \alpha} = \frac{13}{5}$ . Choice B gives $\cos \alpha$ instead of $\sec \alpha$ . Choice C would result from squaring $\sec \alpha$ . Choice D would result from squaring $\cos \alpha$ .
Which of the following expressions is NOT equivalent to $\frac{\sin x}{1 + \cos x}$ ?
Answer: $\cot\left(\frac{x}{2}\right) - \csc\left(\frac{x}{2}\right)$ We can verify each choice. For choice A: multiply numerator and denominator of $\frac{\sin x}{1 + \cos x}$ by $(1 - \cos x)$ to get $\frac{\sin x(1 - \cos x)}{(1 + \cos x)(1 - \cos x)} = \frac{\sin x(1 - \cos x)}{1 - \cos^2 x} = \frac{\sin x(1 - \cos x)}{\sin^2 x} = \frac{1 - \cos x}{\sin x}$ . So A is equivalent. For choice C: using half-angle identities, $\frac{\sin x}{1 + \cos x} = \frac{2\sin(x/2)\cos(x/2)}{1 + (2\cos^2(x/2) - 1)} = \frac{2\sin(x/2)\cos(x/2)}{2\cos^2(x/2)} = \frac{\sin(x/2)}{\cos(x/2)} = \tan(x/2)$ . So C is equivalent. Choice D simplifies to $\frac{2\sin(x/2)\cos(x/2)}{2\cos^2(x/2)} = \frac{\sin(x/2)}{\cos(x/2)} = \tan(x/2)$ , which we showed equals the original expression. For choice B: $\cot(x/2) - \csc(x/2) = \frac{\cos(x/2)}{\sin(x/2)} - \frac{1}{\sin(x/2)} = \frac{\cos(x/2) - 1}{\sin(x/2)}$ . This is the negative of what we want, so B is NOT equivalent.
If $\cos \alpha = \frac{5}{13}$ and $\alpha$ is in the fourth quadrant, what is the value of $\sin(2\alpha)$ ?
Answer: $-\frac{120}{169}$ First, we find $\sin \alpha$ using $\sin^2 \alpha + \cos^2 \alpha = 1$ . We have $\sin^2 \alpha = 1 - \frac{25}{169} = \frac{144}{169}$ , so $\sin \alpha = \pm\frac{12}{13}$ . Since $\alpha$ is in quadrant IV, $\sin \alpha = -\frac{12}{13}$ . Using the double angle identity $\sin(2\alpha) = 2\sin \alpha \cos \alpha$ , we get $\sin(2\alpha) = 2 \cdot (-\frac{12}{13}) \cdot \frac{5}{13} = -\frac{120}{169}$ . Choice B makes an arithmetic error in the calculation. Choice C uses the wrong sign for $\sin \alpha$ . Choice D makes both sign and calculation errors.
For which value of $k$ does the equation $\sin^2 \theta + k\cos^2 \theta = 1$ hold for all values of $\theta$ ?
Answer: $k = 1$ Using the fundamental identity $\sin^2 \theta + \cos^2 \theta = 1$ , we can substitute to get $\sin^2 \theta + k\cos^2 \theta = 1$ . For this to equal the fundamental identity, we need $k\cos^2 \theta = \cos^2 \theta$ , which means $k = 1$ . Choice A would give $\sin^2 \theta = 1$ , which is not true for all $\theta$ . Choice C would give $\sin^2 \theta - \cos^2 \theta = 1$ , which is not always true. Choice D would give $\sin^2 \theta + 2\cos^2 \theta = 1$ , which contradicts the fundamental identity.
Which expression is equivalent to $\frac{\cos^2 x - \sin^2 x}{\cos^2 x + \sin^2 x}$ for all values of $x$ where the expression is defined?
Answer: $\cos 2x$ Using the fundamental identity $\sin^2 x + \cos^2 x = 1$ , the denominator becomes $\cos^2 x + \sin^2 x = 1$ . The numerator $\cos^2 x - \sin^2 x$ is the double angle identity for cosine: $\cos 2x = \cos^2 x - \sin^2 x$ . Therefore, the expression simplifies to $\frac{\cos 2x}{1} = \cos 2x$ . Choice B is wrong because $\sin 2x = 2\sin x \cos x$ . Choice C is wrong because $\tan^2 x = \frac{\sin^2 x}{\cos^2 x}$ . Choice D is wrong because $\sec^2 x = \frac{1}{\cos^2 x}$ .
If $\tan \theta = -\frac{4}{3}$ and $\sin \theta > 0$ , what is the value of $\sin \theta - \cos \theta$ ?
Answer: $\frac{7}{5}$ Since $\tan \theta = -\frac{4}{3} < 0$ and $\sin \theta > 0$ , we must have $\cos \theta < 0$ , so $\theta$ is in quadrant II. Using $\tan \theta = \frac{\sin \theta}{\cos \theta} = -\frac{4}{3}$ and $\sin^2 \theta + \cos^2 \theta = 1$ , we can write $\sin \theta = -\frac{4}{3}\cos \theta$ . Substituting: $(-\frac{4}{3}\cos \theta)^2 + \cos^2 \theta = 1$ , so $\frac{16}{9}\cos^2 \theta + \cos^2 \theta = 1$ , giving $\frac{25}{9}\cos^2 \theta = 1$ . Thus $\cos^2 \theta = \frac{9}{25}$ , so $\cos \theta = \pm\frac{3}{5}$ . Since we're in quadrant II, $\cos \theta = -\frac{3}{5}$ and $\sin \theta = -\frac{4}{3} \cdot (-\frac{3}{5}) = \frac{4}{5}$ . Therefore, $\sin \theta - \cos \theta = \frac{4}{5} - (-\frac{3}{5}) = \frac{4}{5} + \frac{3}{5} = \frac{7}{5}$ . Choice B subtracts incorrectly. Choice C uses wrong signs. Choice D uses both terms negative.
Given that $\sin A = \frac{5}{13}$ and $\cos A = \frac{12}{13}$ , what is the value of $\sin^2 A + \cos^2 A + \tan^2 A$ ?
Answer: $\frac{169}{144}$ When you encounter trigonometric expressions involving multiple functions, start by identifying what fundamental relationships you can use. This question tests your knowledge of the Pythagorean identity and how to find tangent from sine and cosine. To find $\sin^2 A + \cos^2 A + \tan^2 A$ , begin with what you know. The Pythagorean identity tells us that $\sin^2 A + \cos^2 A = 1$ for any angle. This is always true regardless of the specific values. Next, find $\tan^2 A$ . Since $\tan A = \frac{\sin A}{\cos A}$ , we have: $\tan A = \frac{5/13}{12/13} = \frac{5}{12}$ Therefore: $\tan^2 A = \left(\frac{5}{12}\right)^2 = \frac{25}{144}$ The complete expression equals: $\sin^2 A + \cos^2 A + \tan^2 A = 1 + \frac{25}{144} = \frac{144}{144} + \frac{25}{144} = \frac{169}{144}$ Answer A ( $\frac{394}{169}$ ) likely results from incorrectly calculating $\sin^2 A + \cos^2 A$ as $\left(\frac{5}{13}\right)^2 + \left(\frac{12}{13}\right)^2 = \frac{169}{169}$ and making arithmetic errors. Answer B ( $\frac{313}{169}$ ) comes from similar calculation mistakes with the denominators. Answer D ( $\frac{25}{144}$ ) represents finding only $\tan^2 A$ and forgetting to add the Pythagorean identity term. Remember: $\sin^2 A + \cos^2 A$ always equals 1, so focus your calculations on the additional trigonometric terms in these types of problems.
If $\sec^2 \delta - \tan^2 \delta = k$ for all values of $\delta$ where the expression is defined, what is the value of $k$ ?
Answer: $k = 1$ This is a fundamental trigonometric identity: $\sec^2 \delta - \tan^2 \delta = 1$ . This can be derived from $\sin^2 \delta + \cos^2 \delta = 1$ by dividing both sides by $\cos^2 \delta$ : $\frac{\sin^2 \delta}{\cos^2 \delta} + \frac{\cos^2 \delta}{\cos^2 \delta} = \frac{1}{\cos^2 \delta}$ , which gives $\tan^2 \delta + 1 = \sec^2 \delta$ , or $\sec^2 \delta - \tan^2 \delta = 1$ . Choice A would contradict the fundamental identity. Choice C represents $\tan^2 \delta - \sec^2 \delta$ . Choice D would be $\sec^2 \delta + \tan^2 \delta$ minus something.
For which expression is $\sin^4 \xi + \cos^4 \xi$ equivalent?
Answer: $1 - 2\sin^2 \xi \cos^2 \xi$ We can use the identity $(\sin^2 \xi + \cos^2 \xi)^2 = 1$ and expand: $\sin^4 \xi + 2\sin^2 \xi \cos^2 \xi + \cos^4 \xi = 1$ . Rearranging: $\sin^4 \xi + \cos^4 \xi = 1 - 2\sin^2 \xi \cos^2 \xi$ . Choice A has the wrong coefficient. Choice C has the wrong sign. Choice D has both wrong sign and coefficient.
If $\cot \gamma = -\frac{7}{24}$ and $\sin \gamma > 0$ , what is the value of $\csc \gamma$ ?
Answer: $\frac{25}{24}$ When you encounter trigonometric problems with given ratios and quadrant constraints, you need to use the Pythagorean identity and determine signs based on which quadrant the angle lies in. Since $\cot \gamma = -\frac{7}{24}$ and $\sin \gamma > 0$ , you can determine that $\gamma$ is in Quadrant II. Here's why: cotangent equals $\frac{\cos \gamma}{\sin \gamma}$ , so if cotangent is negative and sine is positive, then cosine must be negative—which only occurs in Quadrant II. To find $\csc \gamma = \frac{1}{\sin \gamma}$ , you first need to find $\sin \gamma$ . Using the identity $\cot^2 \gamma + 1 = \csc^2 \gamma$ : $\left(-\frac{7}{24}\right)^2 + 1 = \csc^2 \gamma$ $\frac{49}{576} + 1 = \frac{49 + 576}{576} = \frac{625}{576} = \csc^2 \gamma$ Therefore, $\csc \gamma = \pm\frac{25}{24}$ . Since $\sin \gamma > 0$ in Quadrant II, $\csc \gamma = \frac{1}{\sin \gamma} > 0$ , so $\csc \gamma = \frac{25}{24}$ . Choice A ( $-\frac{25}{7}$ ) and choice B ( $\frac{25}{7}$ ) incorrectly use 7 in the denominator, likely from confusing cotangent's numerator with the final answer's denominator. Choice C ( $-\frac{25}{24}$ ) has the correct magnitude but wrong sign—this would be correct if the angle were in Quadrant III where both sine and cosecant are negative. Remember: always check quadrant signs after calculating trigonometric ratios. The given conditions about positive or negative values tell you which quadrant you're in, which determines the final sign.

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