WebbThis function has period 2π, so we can use the Fourier series formula to find its representation as a sum of sine and cosine functions: f(x) = a0/2 + Σ(ancos(nx) + bnsin(nx)) where a0 = 1/π ∫[−π,π] f(x) dx = 1/π ∫[−π,π] x dx = 0 an = 1/π ∫[−π,π] f(x) cos(nx) dx = 1/π ∫[−π,π] x cos(nx) dx = 0 bn = 1/π ∫[−π,π] f(x) sin(nx) dx = 1/π ∫[−π,π] x sin(nx) dx ... WebbNow, cos −1(4x 3−3x)=cos −1(4cos 3θ−3cosθ) =cos −1(cos3θ) Since θϵ[0, 3π], Hence, 3θ =3cos −1(x) =a+bcos −1(x) Hence, a=0 and b=3 Now, lim y→abcosy =bcos(0) =b =3 Solve any question of Inverse Trigonometric Functions with:- Patterns of problems > Was this answer helpful? Similar questions If cos −1ax+cos −1ay=α,then Hard View solution >
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subtracted, and the resulting vector is called the DO ENTIRE...
WebbProve the following:cos 4 x+cos 3 x+cos 2 x/sin 4 x+sin 3 x+sin 2 x= 3 x. Login. Study Materials. ... cos 4x+ cos 3x + cos 2x sin 4x + sin 3x + sin 2x = cot 3 x. Open in App. Solution. We have . ... = 2 cos 4 x + 2 x 2 cos 4 x − 2 x 2 + cos 3x 2 sin 4 x + 2 x 2 cos 4 x ... Webb21.Evaluate f(x)=−a3+6a−7 at a = - 1 and state the remainder. -14. To evaluate the function f(x) = -a^3 + 6a - 7 at a = -1, we substitute -1 for a in the expression and simplify: f(-1) = -(-1)^3 + 6(-1) - 7 = 1 - 6 - 7 = -12. Therefore, the value of the function at a = -1 is -12. To find the remainder, we can use synthetic division or long ... Webb10 apr. 2024 · Very Long Questions [5 Marks Questions]. Ques. By applying the binomial theorem, represent that 6 n – 5n always leaves behind remainder 1 after it is divided by 25. Ans. Consider that for any two given numbers, assume x and y, the numbers q and r can be determined such that x = yq + r.After that, it can be said that b divides x with q as the … side effect of corticosteroid therapy