r = 2 + cos 0 r = 4+ 3 sin 0

Q: Establish each identity.

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A: Solution 1: Given 5∠π42 Here…

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Q: Sin 0/1 + co² B) 2.

A: sinθ1+cot2θ                             =sinθ1+cos2θsin2θ…

Q: cos 2x-1 i) x sin 2x 2 tan 30–20 ii) tan 40

A: follow next step.

Q: 3. [cos 6° + i sin 6°] 30 = COS + i sin %3D

A: follow next step

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A: Use trigonometric ratio to solve

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Q: cos 2x + 2 sin 2 x = _______

A: This question can be solved by applying the formula of cos2x cos2x=cos2x-sin2x

Q: - 4 cos 0-2 sec 0 = 0. %3D

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Q: 3- y 4 cos 3t - 2 sin 3t

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Q: 4 sin 0 1 – 3 cos 0

A: Using Pythagorean identity we write the equation with sin, then solve for sin theta.

Q: 2 - ƒ (t) = sin 0.5t + ²⁄t – 3 8 cos 32t

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Q: -1 COS (cos (뚜)) COs 4 (sin (프)) -1 tan 2

A: We can make it easier for you

Q: 1+ cos(0) sin(0) sin(8) 1 + cos(0) = 2 csc(0)

A: Given: 1+cosθsinθ+sinθ1+cosθ=2cosecθ

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Q: sin(2x) 2+ cos(2z) da ++

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Q: 1/6 2 +3 sin 0 cos 20 1/6 2 +3 sin 0 (T %3D COS Gos 20

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Q: (1 + cos α)(1 − cos α) = sin2 α (1 + cos α)(1 − cos α) = 1 −(     )=sin2α

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Q: 3 sin(0) + 5 cos(0)

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Q: 1 sin x + 2 cos x + 3

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Q: Solve for the missing variable. 1. cos(x) = sin(18) %3D 2. cos(37) = sin(x + 3)

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Q: 19. TT/2 -π/2 sin 0 de 20. * sec² 0 de

A: 19 ∫-π/2π/2 sinθdθ [-cosθ ]-π/2π/2 0 - 0 =0 other way sin is odd function so also integration is = 0…

Q: 8. 2 sin? x- V3sinx = 0 9. 3 csc(2x) +6 = 0 %3D 10. 2 cos? x = 2+ sinx %3D

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Q: Cos 2x %3D 1-sin 2x

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A: We have to find the value of they ha in given region

Q: 2 2. (sin cos x) = sin 2x +1 2х +

A: The given identity

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A: Using fundamental identity of trigonometry.

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A: Integration is the process of finding the anti-derivative function of the given function.…

Q: 2. Convert r = -8 cos 0 into Cartesian equation.

A: r2=x2+y2 x= r cos theta y= r sin theta

Q: 2 Convert r = to rectangular form. 7 sin 0 COS

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Q: 3. sin(2x - 6) = cos(4x + 18) %3D

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Q: 3 cos(0) sin(0) + 2 cos(0) = 0

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Q: 3) [I - cos х х

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Q: π/2 0 sin³ cos Ode 0

A: This question is based on substitution method.

Q: •π 2 (sin³r)√1 (sin³ x)√1 + cos x dx

A: Given problem:-

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Q: Find all solutions of the equation in the interval [0, 2n). (2 sin 0 + /2)(2 cos 0 + /3)=0

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Q: (A[cos(6) - 3 cos(3) + 2] + B[– sin(6) +3 sin(3)] = 2, A[sin(6) + 3 sin(3)] + B[cos(6) – 3 cos(3) +…

A: Given that, Acos6-3cos3+2+B-sin6+3sin3sin3k+Asin6-3sin3+Bcos6-3cos3+2cos3k=2sin3k

Q: Q.23) cos Q.24) cos(sin ¹x) = 24 sin11x) √1-1 d)√1-x² b) (sinx+1)+c 11

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Q: sin 0 35. r = cos 0 – 1

A: since you have asked multiple questions in single request so we will be answering only first…

Q: 0. (sin 0 + cos 0)² = 1+ sin 20 %3D

A: We’ll answer the first question since the exact one wasn’t specified. Please submit a new question…

Q: cos –1 (-V)] 2 sec OS

A: Simplification

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Q: 2 + cos 0 = 1 + 2 sin² 0

A: The given equation is  2+cosθ=1+2sin2θ The given set of possible answers is  -π3,-π4,π6,π4,2π3,π,3π2…

Q: sin 2 2) = (9z + 4), h(3) (2) (1) = sin(cos r), f"(1) %3D

A: