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To calculate: The value of the trigonometric function
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Answer to Problem 25E
The value of the trigonometric function
Explanation of Solution
Given information:
The trigonometric function
Formula used:
Coordinate plane is divided into four quadrants.
In the first quadrant all trigonometric functions that is
In the second quadrant only sine and cosecant trigonometric functions that is
In the third quadrant only tangent and cotangent trigonometric functions that is
In the fourth quadrant only cosine and secant trigonometric functions that is
The reference angle
Calculation:
Consider the provided trigonometric function
Denote,
First estimate the reference angle
From the figure provided below,
The provided angle has terminal side on negative side of
The reference angle is,
Recall that coordinate plane is divided into four quadrants.
In the third quadrant only tangent and cotangent trigonometric functions that is
In the fourth quadrant only cosine and secant trigonometric functions that is
So, sine trigonometric function,
Therefore,
Thus, the value of the trigonometric function
Chapter 6 Solutions
Precalculus: Mathematics for Calculus - 6th Edition
- -b±√√b2-4ac 2a @4x²-12x+9=0 27 de febrero de 2025 -b±√√b2-4ac 2a ⑥2x²-4x-1=0 a = 4 b=-12 c=9 a = 2 b = 9 c = \ x=-42±√(2-4 (4) (9) 2(4)) X = (12) ±√44)-(360) 2(108) x = ±√ X = =±√√²-4(2) (1) 2() X = ±√ + X = X = + X₁ = = X₁ = X₁ = + X₁ = = =arrow_forward3.9 (A/B). A beam ABCDE, with A on the left, is 7 m long and is simply supported at Band E. The lengths of the various portions are AB 1-5m, BC = 1-5m, CD = 1 m and DE : 3 m. There is a uniformly distributed load of 15kN/m between B and a point 2m to the right of B and concentrated loads of 20 KN act at 4 and 0 with one of 50 KN at C. (a) Draw the S.F. diagrams and hence determine the position from A at which the S.F. is zero. (b) Determine the value of the B.M. at this point. (c) Sketch the B.M. diagram approximately to scale, quoting the principal values. [3.32 m, 69.8 KNm, 0, 30, 69.1, 68.1, 0 kNm.]arrow_forward4. Verify that V X (aẢ) = (Va) XẢ + aV X Ả where Ả = xyz(x + y + 2) A and a = 3xy + 4zx by carrying out the detailed differentiations.arrow_forward
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