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Please refer to the MCQ Questions for Class 11 Trigonometric Functions Maths Chapter 3 with Answers The following Trigonometric Functions Class 11 Mathematics MCQ Questions have been designed based on the latest syllabus and examination pattern for Class 110 votes 1 answer Prove that (i) cos(30°45º), and α <
{∵ 0 ≤ θ ≤ π/2} sin θ 2 cos θ = 1 ⇒ sin 90°Proportionality 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 lengthsIf X A 1 Cos Thet)
4 Let P = a i j be a 3 ×The trigonometric functional values of angles coterminal with 0, π/2 , π, and 3π/2 are the same as those above, and the trigonometric functional values repeat themselves (eg, π and 3π are coterminal and sin (π) = sin (π 2π) = sin (3π) = 0) This illustrates the fact that the trigonometric functions are periodic We will discuss this in greater detail in the next sectionThe solution of the corresponding equal equation is obtained as sin x = 1 / 2 = sin π / 6 ⇒ x = π / 6 The sine function is positive in the first and second quarter Hence, the second angle between "0" and "2π" is ⇒ x = π − θ = π − π / 6 = 5π / 6 Both angles are less than "π" Thus, we do not need to convert the angle into an equivalent negative angle Further
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A 2 5 2 4 B − 2 5 2 4 C 1 8 1 3 D − 1 8 1 3 Medium Open in App Solution Verified by Toppr Correct option is B − 2 5 2 4 Was this answer helpful?Dividing Sin from both sides Or, θ = 90°÷Cos2 θ sin2 θ sin2 θ sin 2θ = 1 sin θ ie cot2 θ 1 = csc2 θ Summarising, cos 2θsin θ = 1 (6) 1tan 2θ = sec θ (7) cot 2θ1 = csc θ (8) Examples 1 Simplify the expression sec2 θ sec2 θ −1 sec2 θ sec2 θ−1 = sec2 θ tan2 θ = 1 cos2 θ sin2 θ cos2 θ = 1 sin2 θ = csc2 θ 2 Show that 1−2cos2 θ sinθcosθ = tanθ−
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👉If this answer fits with your question then feel free to UPVOTE👍 👇👇S i n Θ = − 5 3 A s p e r tFrom the tangent function definition it can also be seen that when the sin θ = cos θ, at π /4 radians (45°), the tan θ equals 1 Then, for the interval 0 ≤ θ <
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Sin2(θ)cos2(θ) = 1 (To the real showoffs try R dx sin2(ax) cos2(ax) = − 1 32a sin(4ax) x 8) As the exercise suggests, we first replace the sin and cos functions, then do some arithmetic sin2(θ)cos2(θ) = e iθ−e− iθ 2i 2 e e− 2 2 = 1 −4 eiθ −e−iθ 2 1 4 eiθ e−iθ 2 = 1 4 − eiθ −e−iθ 2Answer (1 of 4) Pi/2 can be written as 90 degrees Sin(90theta) lies in the 2nd quadrant where the Sines before conversion remain positive and since the value is changing through a multiple of 90 degrees, the trigonometric function will change Hence, Sin(90theta) = Cos(Theta) or, Sin(pi/Example 3 The value of sin 2 5 o sin 2 10 o sin 2 15 o sin 2 85 o sin 2 90 o is equal to Solution Given expression is sin 2 5 o sin 2 10 o sin 2 15 o sin 2 85 o sin 2 90 o We know that sin 90 o = 1 or sin 2 90 o = 1 Similarly, sin 45 o =1 / √2 or sin 2 45 o = 1 / 2 and the angles are in AP of 18 terms
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Using (9) with A = θ, B = π 2 we obtain cos θ − π 2 ≡ cosθ cos π 2 sinθ sin π 2 ≡ cosθ (0)sinθ (1) ie sinθ ≡ cos θ − π 2 ≡ cos π 2 −θ This result explains why the graph of sinθ has exactly the same shape as the graph of cosθ but it is shifted to the right by π 2The first shows how we can express sin θ in terms of cos θ;1 The area of the region 2 Let be a root of the quadratic equation, If , then arg is equal to 3 Let and be three vectors such that the projection vector of on , If is perpendicular to , then is equal to 4 Let and be points on the parabola,
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(1 − sin 2 θ ) (1 tan 2 θ ) is equal to Prove that cos 2 α cos 2 ( α β ) − 2 cos α cos β cos ( α β ) = sin 2 β cos 2 θ 3 − 4 sin 2 θ is equal to2 The coefficient of x 4 in the expansion of ( 1 − 2 x) 5 is equal to Binomial Theorem 3 The equation 5 x 2 y 2 y = 8 represents 4 The centre of the ellipse 4 x 2 y 2 − 8 x 4 y − 8 = 0 is 5 The area bounded by the curves y = − x 2 3 and y = 0 isSin( π 2) = 1 cos( π 2) = 0 So we have sin( π 2 x) = cos(x) Since this answer is very usefull for student here the full demonstration to obtain sin(a b) = sin(a)cos(b) cos(a)sin(b) (do not read this if you are not fan of math) a complex numbers can be written in trigonometrics form z = (cos(x) isin(x)) → (1)
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∴ LHS = sin θ cos(π/4) sin(π/4)cos θ cos θ cos(π/4) sin(π/4)sin θ ∵ sin(π/4)=cos(π/4) = 1/√2 ⇒ LHS = ⇒ LHS = ⇒ LHS = √2 cos θ = RHS Hence, sinα cosα = √2 cos θ Question 16 If sin θ cos θ = 1, then find the general value of θ Answer Given, sin θ cos θ = 1 We need to solve the above equation If we can convert this to a single trigonometricThe 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?Also, the tangent of one is equal to the cotangent of the other θ and sin(π/2θ) = cosθ cos π/2θ) = sinθ tan(π/2θ) = cotθ cot(π/2θ) = tanθ Proof The Trigonometric ratios of angle (θ) Thinking of θ as an acute angle, (θ) ends in the 4th Quadrant where only cosine is positive (θ) means starting from A, the origin on the unit circle, and just going clockwise
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If sin − 1 x sin − 1 y = 2 π , then x 2 is equal to A 1 − y 2 B y 2 C 0 D 1 − y Medium Open in App Solution Verified by Toppr Correct option is A 1 − y 2 We have, sin − 1 x sin − 1 y = 2 π ⇒ sin − 1 x = 2 π − sin − 1 y ⇒ 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) = cos2340 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)If α is a root of 2 5 cos 2 θ 5 cos θ − 1 2 = 0, 2 π <
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Sin Pi Plus Theta Sin Pi Minus Theta Cosec 2 Theta Equal To sin(π θ) sin(π θ) cosec 2 θ = 1) 1 2) 1 3) sin θ 4) sin θ 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) = 1 Was this answer helpful?If sin(π cos θ) = cos(π sin θ), then sin 2 θ equals Rs 10,000 Worth of NEET &π/2 Then ∫ (sinn θ sinθ)^1/n cosθ/(sin^n1θ) dθ is equal to (where C is a constant of integration) asked in Mathematics by Niharika (756k points) jee mains 19;
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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 functionsProblem Answer θ is equal to 30 °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 given
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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 ordiate of all points on the curve 2 s i n 2 x 3 c o s 2 x 1 where the tangent is horizontal isThe Trigonometric ratios of angle π/2 θ Thinking of θ as an acute angle (that ends in the 1st Quadrant), (π/2θ) or (90°θ) ends in the 2nd Quadrant where only sine of the angle is positive The (π/2θ) formulas are similar to the (π/2θ) formulas except only sine is positive because (π/2θ) ends in the 2nd Quadrant sin(π/2θ) = cosθCosθ1 sinθ = sinπ2 θ1 cosπ2 θ = 2sinπ4
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Solve for θ in the following equation sin 2θ = cosθ A 30°Tan Theta Sin Pi 2 Plus Theta Cos Pi 2 Minus Theta Equals To tan θ sin(π/2 θ) cos(π/2 θ) = 1) 1 2) 0 3) 1/√2 4) None of these Answer (4) None of these Solution We know that, sin(π/2 θ) = cos θ cos(π/2 – θ) = sin θ Now, tan θ sin(π/2 θ) cos(π/2 – θ) = tan θ (cos θ) (sin θ) = (sin θ/cos θ) (sin θ)(cos θ) = sin 2 θ Was this answer helpful?Videos 581 Syllabus Advertisement
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Medium Open in App Solution Verified by Toppr 2 c o s Θ = 1 − s i n Θ o n s q u a r i n g b o t h t h e s i d e s ⇒ 4 c o s 2 Θ = s i n 2 Θ − 2 s i n θ 1 ⇒ 4 (1 − s i n 2 Θ) = s i n 2 Θ − 2 s i n Θ 1 ⇒ 5 s i n 2 Θ − 2 s i n Θ − 3 = 0 ⇒ s i n Θ = 1;3π/2, x and y are both negative so sin(θ) and cos(θ) are negative while tan(θ) is positive;π /4 the tangent is less than 1 and for the interval π /4 <
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View Solution Latest Problem Solving in Plane Trigonometry Problems More Questions in Plane Trigonometry Problems Online Questions and Answers in Plane Trigonometry Problems MCQ in Plane Trigonometry Problems DOWNLOAD PDF / PRINTLet a tangent be drawn to the ellipse x^2/27 y^2 at (3√3 cos θ, sin θ) where θ ∈ (0, π/2 ) asked Mar 27 in Mathematics by Yaad ( 355k points) jeeTo see the answer, pass your mouse over the colored area To cover the answer again, click Refresh (Reload)
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X) = (√3 cos x sin x)/2 (ii) cos(π θ) = cos θ (iii) sin(π θ) = sin θ asked in Trigonometry by Anjali01 (476k points) trigonometry;Lim θ → π / 2 1 − Sin θ ( π / 2 − θ ) Cos θ is Equal to Department of PreUniversity Education, Karnataka PUC Karnataka Science Class 11 Textbook Solutions 9192 Important Solutions 3 Question Bank Solutions 53 Concept Notes &Assume θ = 90°
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If tan (π cos θ) = cot (π sin θ) than a value of cos θ − 4 π ) among the following is A 2 2 1 B 2 1 C 2 1 D 4 1 Medium Open in App Solution Verified by Toppr Correct option is A 2 2 1 tan π cos θ = cot π sin θ tan π cos θ = tan 2 π − π sin θ π cos θ = 2 π − π sin θ cos θ sin θ = 2 1 2 1 cos θ 2 1 sin θ = 2 2 1 cos θ −If sin (θ α)/cos(θ α) = (1 m/1 m), then tan(π/4 θ) tan(π/4 α) is equal to ← Prev Question Next Question → 1 vote 252k views asked in Mathematics by AnjaliVarma (294k points) closed May 15 by Vikash Kumar If sin (θ α)/cos(θ α) = (1 m/1 m), then tan(π/4 θ) tan(π/4 α) is equal to (a) 1/m (b) m (c) 2/m (d) 2m jee;Similarly, when π <
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