There is nothing in the question to justify an assumption that the angle is a vertex of a triangle. Even if it were, there is no justification for assuming that it is a right angled triangle. So the use of ratios of sides of a triangle are totally inappropriate.
In the circumstances, the only realistic way is to use a calculator. Make sure that the calculator's mode is set to the correct measurement unit for the angle - degrees or radians - and use the cosine or sine function.
Spreadsheets such as Excel only use angles measured in radians. So, if given the angle in degrees, multiply the value by pi/180 and then use the appropriate trigonometric function.
There is no reason at all. For most angles sin plus cos do not equal one.
To simplify the expression sin(30°) cos(90°) sin(90°) cos(30°), we first evaluate the trigonometric functions at the given angles. sin(30°) = 1/2, cos(90°) = 0, sin(90°) = 1, and cos(30°) = √3/2. Substituting these values into the expression, we get (1/2) * 0 * 1 * (√3/2) = 0. Therefore, the final result of sin(30°) cos(90°) sin(90°) cos(30°) is 0.
I would start by looking up the formulae for multiple angles, and convert that to simgle angles. In this case, sin 2x = 2 sin x cos x, so your equation becomes:2 sin x cos x sin x = cos x2 sin2x cos x = cos xNext divide both sides by cos x; note that you must consider the possibility that cos x = 0 (this may give additional solutions to the equation).
Like normal expansion of brackets, along with: cos(A + B) = cos A cos B - sin A sin B sin(A + B) = sin A cos B + cos A sin B 5(cos 20 + i sin 20) × 8(cos 15 + i sin 15) = 5×8 × (cos 20 + i sin 20)(cos 15 + i sin 15) = 40(cos 20 cos 15 + i sin 15 cos 20 + i cos 15 sin 20 + i² sin 20 sin 15) = 40(cos 20 cos 15 - sin 20 cos 15 + i(sin 15 cos 20 + cos 15 sin 20)) = 40(cos(20 +15) + i sin(15 + 20)) = 40(cos 35 + i sin 35)
There can be no significant simplicfication if some of the angles are theta and others are x, so assume that all angles are x. [csc(x) - cot(x)]*[cos(x) + 1] =[1/sin(x) - cos(x)/sin(x)]*[cos(x) + 1] =1/sin(x)*[1 - cos(x)]*[cos(x) + 1] =1/sin(x)*[1 - cos2(x)] =1/sin(x)*[sin2(x)] = sin(x)
There is no reason at all. For most angles sin plus cos do not equal one.
sin θ = cos (90° - θ) cos θ = sin (90° - θ)
Assuming that the angles are all stated in degrees: sin(45) = cos(45) = 1/2 sqrt(2) sin(45) cos(45) = (1/2)2 x (2) = 1/2 sin(230) = - 0.7660444 sin(45) cos(45) - sin(230) = 0.5 + 0.7660444 = 1.2660444 (rounded)
To simplify the expression sin(30°) cos(90°) sin(90°) cos(30°), we first evaluate the trigonometric functions at the given angles. sin(30°) = 1/2, cos(90°) = 0, sin(90°) = 1, and cos(30°) = √3/2. Substituting these values into the expression, we get (1/2) * 0 * 1 * (√3/2) = 0. Therefore, the final result of sin(30°) cos(90°) sin(90°) cos(30°) is 0.
sin(x) = cos(90° - x) cos(x) = sin(90° - x)
I would start by looking up the formulae for multiple angles, and convert that to simgle angles. In this case, sin 2x = 2 sin x cos x, so your equation becomes:2 sin x cos x sin x = cos x2 sin2x cos x = cos xNext divide both sides by cos x; note that you must consider the possibility that cos x = 0 (this may give additional solutions to the equation).
sin cos tan -soh cah toa
Like normal expansion of brackets, along with: cos(A + B) = cos A cos B - sin A sin B sin(A + B) = sin A cos B + cos A sin B 5(cos 20 + i sin 20) × 8(cos 15 + i sin 15) = 5×8 × (cos 20 + i sin 20)(cos 15 + i sin 15) = 40(cos 20 cos 15 + i sin 15 cos 20 + i cos 15 sin 20 + i² sin 20 sin 15) = 40(cos 20 cos 15 - sin 20 cos 15 + i(sin 15 cos 20 + cos 15 sin 20)) = 40(cos(20 +15) + i sin(15 + 20)) = 40(cos 35 + i sin 35)
There can be no significant simplicfication if some of the angles are theta and others are x, so assume that all angles are x. [csc(x) - cot(x)]*[cos(x) + 1] =[1/sin(x) - cos(x)/sin(x)]*[cos(x) + 1] =1/sin(x)*[1 - cos(x)]*[cos(x) + 1] =1/sin(x)*[1 - cos2(x)] =1/sin(x)*[sin2(x)] = sin(x)
There are many. For example, if A and B are the two acute angles, then A + B = 90 degrees or sin(A) = cos(B) or cos(A) = sin(B) or tan(A) = 1/tan(B)
[sin - cos + 1]/[sin + cos - 1] = [sin + 1]/cosiff [sin - cos + 1]*cos = [sin + 1]*[sin + cos - 1]iff sin*cos - cos^2 + cos = sin^2 + sin*cos - sin + sin + cos - 1iff -cos^2 = sin^2 - 11 = sin^2 + cos^2, which is true,
sin(3A) = sin(2A + A) = sin(2A)*cos(A) + cos(2A)*sin(A)= sin(A+A)*cos(A) + cos(A+A)*sin(A) = 2*sin(A)*cos(A)*cos(A) + {cos^2(A) - sin^2(A)}*sin(A) = 2*sin(A)*cos^2(A) + sin(a)*cos^2(A) - sin^3(A) = 3*sin(A)*cos^2(A) - sin^3(A)