Example Question - pythagorean identity
Here are examples of questions we've helped users solve.
Trigonometric Ratios in a Right-Angled Triangle
The image shows a right-angled triangle, ΔABC, with angle A = 60°, angle B = 90°, and the hypotenuse AC = 20 ft. We are required to solve a series of questions related to this triangle. a) Using trigonometric ratios to find the perpendicular (opposite side to angle A, which is BC) and base (adjacent side to angle A, which is AB): We use the sine and cosine functions, which are defined as follows for a right-angled triangle: - Sine of an angle is the ratio of the length of the opposite side to the hypotenuse. - Cosine of an angle is the ratio of the length of the adjacent side to the hypotenuse. For angle A, sin(60°) = opposite/hypotenuse => sin(60°) = BC/AC BC = AC * sin(60°) BC = 20 * (√3/2) (since sin(60°) is √3/2) BC = 10√3 ft For the base (AB), cos(60°) = adjacent/hypotenuse => cos(60°) = AB/AC AB = AC * cos(60°) AB = 20 * (1/2) (since cos(60°) is 1/2) AB = 10 ft b) Using the perpendicular (BC) and base (AB) obtained above, find the value of tan(60°): Tangent of an angle is the ratio of the length of the opposite side to the adjacent side. tan(60°) = opposite/adjacent tan(60°) = BC/AB tan(60°) = (10√3)/10 tan(60°) = √3 c) Find the values of Sin C and Cos C: In a right triangle, the sine of one non-right angle is the cosine of the other, and vice versa. Since angle C is the 90-angle B (90° - 60° = 30°), we can find the sine and cosine of angle C by using their known values at 30°. sin(C) = sin(30°) = 1/2 cos(C) = cos(30°) = √3/2 d) Using the values of Sin C and Cos C, prove that Sin² C + Cos² C = 1: This is a well-known trigonometric identity known as the Pythagorean identity. Now let's substitute the values obtained for Sin C and Cos C: Sin² C + Cos² C = (1/2)² + (√3/2)² Sin² C + Cos² C = 1/4 + 3/4 Sin² C + Cos² C = 4/4 Sin² C + Cos² C = 1 This proves the Pythagorean identity for angle C.
Trigonometric Ratios in the Second Quadrant
The question gives us the cosine of an angle (θ) and the range for the angle, which is the second quadrant (\(\frac{\pi}{2} < \theta < \pi\)), and asks us to find the values of the remaining trigonometric ratios. Given: \[ \cos \theta = -\frac{\sqrt{3}}{2} \] Since the cosine of θ is negative and the angle is in the second quadrant, sine will be positive (as sine is positive in the second quadrant). Let's use the Pythagorean identity to find sine: \[ \sin^2 \theta + \cos^2 \theta = 1 \] Substitute the given value of cos θ: \[ \sin^2 \theta + \left(-\frac{\sqrt{3}}{2}\right)^2 = 1 \] \[ \sin^2 \theta + \frac{3}{4} = 1 \] Solving for \(\sin^2 \theta\): \[ \sin^2 \theta = 1 - \frac{3}{4} = \frac{1}{4} \] Taking the square root to find sin θ yields two possible values, +1/2 and -1/2. Since we are in the second quadrant and sine is positive: \[ \sin \theta = \frac{1}{2} \] The tangent of θ is the ratio of sine to cosine: \[ \tan \theta = \frac{\sin \theta}{\cos \theta} \] \[ \tan \theta = \frac{\frac{1}{2}}{-\frac{\sqrt{3}}{2}} \] \[ \tan \theta = -\frac{1}{\sqrt{3}} \] \[ \tan \theta = -\frac{\sqrt{3}}{3} \] For the reciprocals: \[ \csc \theta = \frac{1}{\sin \theta} = 2 \] \[ \sec \theta = \frac{1}{\cos \theta} = -\frac{2}{\sqrt{3}} = -\frac{2\sqrt{3}}{3} \] \[ \cot \theta = \frac{1}{\tan \theta} = -\sqrt{3} \] So, the six trigonometric ratios for the given angle θ in the second quadrant are: \[ \sin \theta = \frac{1}{2}, \cos \theta = -\frac{\sqrt{3}}{2}, \tan \theta = -\frac{\sqrt{3}}{3} \] \[ \csc \theta = 2, \sec \theta = -\frac{2\sqrt{3}}{3}, \cot \theta = -\sqrt{3} \]
Trigonometric Identity Proof with Justifications
The image contains two steps of a proof of a trigonometric identity, and there is a question asking which option best justifies the equation for each step. Step 1: cos^2(θ) + [csc(θ) - sin(θ)]^2 = cos^2(θ) + [csc^2(θ) - 2csc(θ)sin(θ) + sin^2(θ)]. The justification for Step 1 is: **Option C: Using a form of the Pythagorean Identity.** This is because the term [csc(θ) - sin(θ)]^2 is expanded using the distributive property (a - b)^2 = a^2 - 2ab + b^2, and csc^2(θ) is the reciprocal identity of sin^2(θ) stated by the Pythagorean identity csc^2(θ) = 1/sin^2(θ). Step 2: cos^2(θ) + [csc^2(θ) - 2csc(θ)sin(θ) + sin^2(θ)] = cos^2(θ) + [1 + sin^2(θ) - 2csc(θ)sin(θ)]. The justification for Step 2 is: **Option B: Rearranging terms.** The terms have been reorganized to group the 1 (which comes from the identity csc^2(θ) = 1/sin^2(θ)) with sin^2(θ) to show how they come together due to the use of the Pythagorean identity. In this case, the "Rearranging terms" option best describes the action taken to move the 1 next to sin^2(θ) to later apply the identity sin^2(θ) + 1 = csc^2(θ). Please note that although the Pythagorean identity is used again in Step 2, the rearrangement of terms to show the identity is the focus of this step, which is why Option B is chosen over A.
Solving Complex Number Division
Certainly! To find \( \frac{z_1}{z_2} \) given that \( z_1 = \cos \theta + j \sin \theta \) and \( z_2 = \cos \theta - j \sin \theta \), we can directly divide the complex numbers as follows: \[ \frac{z_1}{z_2} = \frac{\cos \theta + j \sin \theta}{\cos \theta - j \sin \theta} \] To divide these two complex numbers, we can multiply the numerator and the denominator by the conjugate of the denominator to remove the imaginary parts from the denominator. The conjugate of \( z_2 \) is \( \cos \theta + j \sin \theta \), which is actually the same as \( z_1 \). So, we multiply both top and bottom by this conjugate: \[ \frac{z_1}{z_2} = \frac{(\cos \theta + j \sin \theta)(\cos \theta + j \sin \theta)}{(\cos \theta - j \sin \theta)(\cos \theta + j \sin \theta)} \] Multiplying out the numerator, we get: Numerator: \( (\cos^2 \theta + 2j \cos \theta \sin \theta - \sin^2 \theta) \) (using \( j^2 = -1 \)) Multiplying out the denominator, we get: Denominator: \( (\cos^2 \theta - (\sin^2 \theta)) \) Simplify the denominator using \( \cos^2 \theta - \sin^2 \theta = \cos(2\theta) \): Denominator: \( \cos(2\theta) \) However, the denominator simplifies even further because \( \cos^2 \theta - \sin^2 \theta = 1 \) (from the Pythagorean identity \( \cos^2 \theta + \sin^2 \theta = 1 \)). Therefore, the denominator actually is: Denominator: \( 1 \) Returning to the numerator, we use the identity \( \cos^2 \theta + \sin^2 \theta = 1 \) again: Numerator: \( 1 + 2j \cos \theta \sin \theta \) The numerator simplifies using the double angle identity for sine: \( 2\sin \theta \cos \theta = \sin(2\theta) \): Numerator: \( 1 + j \sin(2\theta) \) Putting it all together, we have: \[ \frac{z_1}{z_2} = 1 + j \sin(2\theta) \] This is the final simplified form of \( \frac{z_1}{z_2} \).
Simplifying Trigonometric Expression
To solve the expression, we need to simplify it by using trigonometric identities where possible. The expression is: \[ \frac{\csc(x)}{1 - \cos^2(x)} - \csc^3(x) \] We can recognize that \(\csc(x) = \frac{1}{\sin(x)}\), and using the Pythagorean identity \(1 - \cos^2(x) = \sin^2(x)\), we can rewrite the first term: \[ \frac{\csc(x)}{1 - \cos^2(x)} = \frac{\frac{1}{\sin(x)}}{\sin^2(x)} = \frac{1}{\sin^3(x)} \] Now we rewrite \(\csc^3(x)\) as: \[ \csc^3(x) = \left(\frac{1}{\sin(x)}\right)^3 = \frac{1}{\sin^3(x)} \] With these substitutions, the expression simplifies to: \[ \frac{1}{\sin^3(x)} - \frac{1}{\sin^3(x)} \] Now it is clear that both terms are the same and they cancel each other out. Thus, the simplified expression equals zero. \[ \frac{\csc(x)}{1 - \cos^2(x)} - \csc^3(x) = 0 \]
Trigonometry Problem Solution: Finding the Exact Value of sin(θ)
The image contains a trigonometry problem. It states: "Question 27 You are told that cos(θ) = 8/17. a) If θ is in the first quadrant, then the exact value of sin(θ) is ______. Note: In this question we require you to input your answer without decimals and without entering the words sin, cos, tan or cot. For example, if your answer is 3/17, then enter sqrt(17^2-3^2)/17)" To solve this, we will use the Pythagorean identity which relates the sine and cosine of an angle: sin^2(θ) + cos^2(θ) = 1. We are given that cos(θ) = 8/17. Squaring both sides we get: cos^2(θ) = (8/17)^2 = 64/289. Now we can find sin^2(θ): sin^2(θ) = 1 - cos^2(θ) = 1 - 64/289 = 289/289 - 64/289 = 225/289. Since we are in the first quadrant, sin(θ) will be positive, so: sin(θ) = √(225/289) = 15/17. Therefore, the exact value of sin(θ) is 15/17.
Trigonometry Question: Finding Cosine Value Given Sine
The image shows a trigonometry question which reads: "Question 26 You are told that sin θ = \(\frac{7}{\sqrt{389}}\). a) If θ is in the first quadrant, then the exact value of cos θ is __________. Note: In this question we require you input your answer without decimals and without entering the words sin, cos or tan. For example, if your answer is \(\frac{\sqrt{17}}{17}\), then enter sqrt(5)/sqrt(17)" To solve for cos θ when given sin θ, we can use the Pythagorean identity, which states that sin² θ + cos² θ = 1. Since we're given sin θ = \(\frac{7}{\sqrt{389}}\), we can square this to find sin² θ: (sin θ)² = \(\left(\frac{7}{\sqrt{389}}\right)^2 = \frac{49}{389}\) Next we'll use the identity to solve for cos² θ: cos² θ = 1 - sin² θ = 1 - \(\frac{49}{389}\) To find cos² θ, we need to subtract \(\frac{49}{389}\) from 1. Since 1 can be written as \(\frac{389}{389}\), we get: cos² θ = \(\frac{389}{389} - \frac{49}{389} = \frac{389 - 49}{389} = \frac{340}{389}\) So, cos² θ = \(\frac{340}{389}\). We are now looking for the positive square root since θ is in the first quadrant where cosine values are positive: cos θ = \(\sqrt{\frac{340}{389}}\) The square root of a fraction is the square root of the numerator over the square root of the denominator: cos θ = \(\frac{\sqrt{340}}{\sqrt{389}}\) In the most simplified fractional form, this is the exact value for cos θ. However, note that \(\sqrt{340}\) can be further simplified since 340 has a square factor, which is 4 (2²). \(\sqrt{340} = \sqrt{4 \cdot 85} = 2\sqrt{85}\) Therefore, the final simplified answer for cosθ is: cos θ = \(\frac{2\sqrt{85}}{\sqrt{389}}\) And in the answer format requested in the question: 2sqrt(85)/sqrt(389)
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