Converting these back to real part/imaginary part notation eiπ/4 = cos π 4 isin π 4 = 1 √ 2 i √ 2 and e5iπ/4 = cos 5π 4 isin 5π 4 = − 1 √ 2 − i √ 2 This exercise is part of an interesting subject in mathematics called the nth Solution Set Simple Harmonic Motion Physics 107 Answers Simple Harmonic Motion 1 The maximum displacement from the equilibrium position A = 100 cm The time for one complete oscillation T = π/2 s Notice the maximum positive displacement x = 100 cm occurs at t = 0 and the next time at t = π/2 sSolution sin 2 θ = 3 4 ⇒ sin θ = ± √ 3 2 ⇒ sin θ = sin (± π 3) ⇒ θ = nπ (1) n ∙ (± π 3 ), where, n ∈ Z ⇒ θ = nπ ± π 3, where, n ∈ Z Therefore the general solution of sin 2 θ = 3 4 is θ = nπ ± π 3, where, n ∈ Z Trigonometric Equations General solution of the equation sin x = ½
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Sin(π/2-θ) is equal to- Find the slope of the normal to the curve x = 1 − a sin θ, y = b cos^2θ at θ=π/2 asked in Mathematics by sforrest072 (128k points) application of derivative;Sin (theta) = oppo/hypo we know that oppo = 2 by pythagoras theorem hypo = sqrt (oppo^2adj^2) therefore,hypo = sqrt (49) = therefore sin (theta) = 2/ = 055 but if theta is in third quadrant tan (theta) is positive while sin (theta) is negative therefore the answer is A) or 055
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The second shows how we can express cos θ in terms of sin θ Note sin 2 θ "sine squared theta" means (sin θ) 2 Problem 3 A 345 triangle is rightangled a) Why?Experts are tested by Chegg as specialists in their subject area We review their content and use your feedback to keep the quality high 100% (7 ratings)Solution for Assume sin(θ)=18/29 where π/2
Sin (θ), Tan (θ), and 1 are the heights to the line starting from the x axis, while Cos (θ), 1, and Cot (θ) are lengths along the x axis starting from the origin The functions sine, cosine and tangent of an angle are sometimes referred to as the primary or basic trigonometric functionsProportionality constants are written within the image sin θ, cos θ, tan θ, where θ is the common measure of five acute angles In mathematics, the trigonometric functions (also called circular functions, angle functions or goniometric functions) are real functions which relate an angle of a rightangled triangle to ratios of two side lengthsThe complementary angle equals the given angle subtracted from a right angle, 90° For instance, if the angle is 30°, then its complement is 60° Generally, for any angle θ, cos θ = sin (90° – θ) Written in terms of radian measurement, this identity becomes cos θ = sin (π/2 – θ
The trigonometric R method is a method of rewriting a weighted sum of sines and cosines as a single instance of sine (or cosine) This allows for easier analysis in many cases, as a single instance of a basic trigonometric function is often easier to work with than multiple are The R method is most often used to find the extrema (maximum and minimum) of combinations of trigonometricTherefore, we will leave theSin1 sin 5 π 6 = sin1 sin π π 6 ∵ sin π θ = sin θ = sin1 sin π 6 ∵ Principal value ∈ 0, π 2 = π 6 which is the required principal value Answer
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More Trigonometry Questions Q1 There are two vessels in the sea, opposite side of a lighthouse The angle of elevation of the top of the lighthouse isIf we started from A and moves in anticlockwise direction then at the points A, B, A', B' and A, the arc length travelled are 0, π 2, π, 3 π 2, and 2π Therefore, from the above unit circle it is clear that sin θ = P M O P Now, sin θ = 0 ⇒ P M O P = 0Dius So the circles intersect also at θ = π/4, as the figure shows The area of the shaded region can thus be computed as the difference I1− I2 = 1 2 Z π π/4 sin2θdθ − 1 2 Z π/2 π/4 cos2θdθ of the area of the respective shaded regions in Now I1 = 1 4 Z π π/4 (1−cos2θ)dθ = 1 4 h θ − 1 2 sin2θ iπ π/4 = 1 4 3π 4 1 2
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KEAM 17 tan((π/4) (θ/2)) tan((π/4) (θ/2)) is equal to (A) sec θ (B) 2 sec θ sec (θ /2) (D) sin θ (E) cos θSin (π/2theta) = cos theta As π/2theta lies in the second quadrant so the answer will be in positive as it is lying in sin (second quadrant) and because of π/2 the sin will convert into cos theta 12K views Sponsored by AntidotemeRewrite the middle terms as a perfect square ρ = sin θ sin φ ρ 2 = ρ sin θ sin φ Multiply both sides of the equation by ρ x 2 y 2 z 2 = y Substitute rectangular variables using the equations above x 2 y 2 − y z 2 = 0 Subtract y from both sides of the equation x 2 y 2 − y 1 4 z 2 = 1 4 Complete the square x 2 (y
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Where c 2 s 1 = 1, is called a Givens matrix, after the name of the numerical analyst Wallace Givens Since one can choose c = cos θ and s = sin θ for some θ, the above Givens matrix can be conveniently denoted by J(i, j, θ)Geometrically, the matrix J(i, j, θ) rotates a pair of coordinate axes (7th unit vector as its xaxis and the jth unit vector as its yaxis) through the givenSimilarly, when π < θ < 3π/2, x and y are both negative so sin(θ) and cos(θ) are negative while tan(θ) is positive;Ask a Question If sin θ = √3 cos θ, – π < θ < 0, then θ is equal to ← Prev Question Next Question →
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SOLUTION Concept sin (π – θ) = sin θ Calculation Given that A B C = π ⇒ A B = π C Now, sin (A B) sin C = sin (π – C) sin C$$sin^2(θ) cos^2(θ) = 1$$ $$tan^2(θ) 1 = sec^2(θ)$$ $$1 cot^2(θ) = csc^2(θ)$$ Cofunction Identities Each of the trig functions equals its cofunction evaluated at the complementary angle $$sin(θ) = cos({π/2} θ)$$ $$cos(θ) = sin({π/2} θ)$$ $$tan(θ) = cot({π/2} θ)$$ $$cot(θ) = tan({π/2} θ)$$ $$csc(θ) = sec({π/2} θ)$$The first shows how we can express sin θ in terms of cos θ;
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Examples of quadrantal angles include, 0, π/2 , π , and 3π/ 2 Angles coterminal with these angles are, of course, also quadrantal We are interested in finding the six trigonometric functional values of these special angles, and we will begin with θ = 0 Since any point (x, y) on the terminal ray of an angle with measure 0 has y coordinate equal to 0, we know that r = x, and we have,This can be seen by solving the equation 3 sin (2 θ) = 0 3 sin (2 θ) = 0 for θ θ Therefore the values θ = 0 θ = 0 to θ = π / 2 θ = π / 2 trace out the first petal of the rose To find the area inside this petal, use Equation 79 with f (θ) = 3 sin (2 θ), f (θ) = 3 sin (2 θ), α = 0, α = 0, and β = π / 2 β = π / 2Answer As we need to find the most general solution for two different trigonometric equations Tan θ = 1 (1) and cos θ = 1/√2 (2) Most general value of θ is the common solution of both equation 1 and 2 Let S 1 represents the solution set of equation 1 and S 2 be the solution set equation 2
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Lim θ → π / 2 1 − Sin θ ( π / 2 − θ ) Cos θ is Equal to Mathematics MCQAnd if 3π/2 < θ < 2π, cos(θ) is positive while sin(θ) and tan(θ) are negative As θ increases beyond 2π (or when θ decreases below 0) the same pattern is repeatedThe exact value of arcsin( √2 2) arcsin ( 2 2) is π 4 π 4 x = π 4 x = π 4 The sine function is positive in the first and second quadrants To find the second solution, subtract the reference angle from π π to find the solution in the second quadrant x = π− π 4 x = π π 4 Simplify π − π 4 π π
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The equations in matrix form are R = x ¯ y ¯ = cos θ − sin θ sin θ cos θ x y Example 16 Rotate the line y = 5 x 1 an angle of 30° counterclockwise Graph the original and the rotated line Since 30° is π/ 6 radians, the rotation matrix is cos π 6 − sin π 6 sin π 6 cos π 6 = 0866 − 05 05 0866 Now compute the rotationTo see the answer, pass your mouse over the colored area To cover the answer again, click "Refresh" ("Reload")Any complete revolution θ and θ 2 nπ are therefore coterminal sin θ, therefore, is equal to sin ( θ 2 nπ ) b) (2 n 1) π The odd multiples of π ± π, ±3 π, ±5 π, ±7 π, 2 n 1 (or 2 n − 1) typically signifies an odd number Zeros When we write sin θ, θ
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Answer (2) 1 Solution sin (π θ) sin (π – θ) cosec 2 θ = (sin θ) (sin θ) cosec 2 θ = sin 2 θ cosec 2 θ = sin 2 θ (1/sin 2 θ) = 1COMEDK 05 If sin(π cos θ) = cos(π sin θ), then sin 2 θ equals (A) ± (3/4) (B) ± √2 ± (1/√3) (D) ± (1/2) Check Answer and SolCos^2 ( pi6 theta ) sin^2 ( pi6 theta ) is equal to
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∴ sin(π/2 θ/3) = √3/2 ⇒ cos(θ/3) = cos 30 ° If an angle is α is divided into two parts A and B such that A B = x and tan A tan B = 2 1, then what is sinx equal to ?= lim θ→0 tan (π sin 2 θ) / sin (2π sin 2 θ) = lim θ→0 (1 / 2) {tan (π sin 2 θ) / π sin 2 θ} * {2π sin 2 θ / sin (2π sin 2 θ)} = – 1 / 2 Therefore, the correct answer is (a)0 votes 1 answer If tan ( π/4 θ) tan (π /4 − θ) = p sec 2θ, then find the value of p asked in Trigonometry by Gaangi (248k points)
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If two angles add up to 90° or π/2, the sine of one is equal to the cosine of the other Also, the tangent of one is equal to the cotangent of the other θ and (π/2θ) are complimentary angles because θ (π/2θ) = π/2 By referring to the right triangle shown⇒ sin (sin − 1 x) = sin (2 π − sin − 1 y), take sine both sides ⇒ x = cos ( sin − 1 y ) , since sin ( 9 0 o − θ ) = cos θ ⇒ x = cos ( cos − 1 1 − y 2 ) = 1 − y 2If tan(θ)= 15 / 8 , 0 ≤ θ ≤ (π / 2) , then sin(θ) equals cos(θ) equals sec(θ) equals Expert Answer Who are the experts?
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It is given that sin (π/2 θ/3) = √3/2 Formula Used Basic concept of trigonometric ratio and identities We know that sin(90 θ) = cos θ ⇒ cos 30° = √3/2 Calculation We know that sin (π /2 θ) = cos θ ∴ sin(π/2 θ/3) = √3/2 ⇒ cos(θ/3) = cos 30° ⇒ θ/3 = 30° = θ = 90° Now, value of tan θ = tan 90° = Not defineSetting each of r, θ and z equal to a constant defines a surface in space, as illustrated in the following example To convert this spherical point to cylindrical, we have r = 6 sin (π / 2) = 6, θ Find real θ such that (3 2i × sin θ)/(1 – 2i × sin θ) is imaginary (a) θ = nπ ± π/2 where n is an integer (b) θ = nπ ± π/3 where n is an integer (c) θ = nπ ± π/4 where n is an integer (d) None of these Answer Answer (b) θ = nπ ± π/3 where n is an integer Hint Given,
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Time between θ and θ dθ is equal to the number incident particles per unit time between b and b db Therefore, for incident flux j I, the number of particles scattered into the solid angle NdΩ=2 π sin θ dθ N =2πbdbj ISub your expression for \sin (\theta) In the second equation you can use cos(θ)2 = 1− sin(θ)2 and get a quadratic in sin(θ), solve that for sin(θ) in terms of y(θ) Sub your expression for sin(θ) Transform complex exponential integral to real Transform complex exponential integral to real 2340 sin θ − 1251 cos θ =2660 cos (θ 1081) Checking using a graph, we obtain the following for each side of our answer We see that our negative cosine curve has an amplitude of 2660 and it has been shifted to the left by 1081 radians, which is consistent with the expression −2660 cos ( θ 1081)
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Conclusion The chart of variations of sinθ vesus θ from 0 to 2π as well as the corresponding graph are shown below Example 1 Verify the following equality sin(90) 2sin(0°) 3sin(180°) 4sin(270) = 5 Solution Always, leave one side untouched, and work on the other side until both sides prove to be equal In this example, the left side can be worked on; Explanation LHS =left hand side, RHS =right hand side Use formula cos(A− B) = cosAcosB sinAsinB LHS = cos(π 2)cosθ sin( π 2)sinθ = 0 × cosθ 1 ×sinθ Then the side opposite the angle x will have length sinx and the side opposite the angle ( π 2 −x) will have length sin( π 2 − x) By Pythagoras theorem, the sum of the squares of the lengths of these sides is equal to the square of the length of the hypotenuse So sin2x sin2( π 2 − x) = 12 = 1 Answer link
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