If tan y equals to x where y is acute find 3 cos y in terms of x?
3cos(y) = 3/(sqrt(1+x^2)
To solve this equation, you must first put the equation in terms of cosx. 3cosx=2 cosx = 2/3 Next, you find the reference angle (α) by finding the cos inverse of 2/3. α=cos-1(2/3) = 48.19 Degrees (approximately) find the distance from from 48.19 and 90 you then add the difference to 270 giving you the two answers which are ... 48.19 and 311.81 (approximately)
Start on the left-hand side. cos(x) + tan(x)sin(x) Put tan(x) in terms of sin(x) and cos(x). cos(x) + [sin(x)/cos(x)]sin(x) Multiply. cos(x) + sin2(x)/cos(x) Make the denominators equal. cos2(x)/cos(x) + sin2(x)/cos(x) Add. [cos2(x) + sin2(x)]/cos(x) Use the Pythagorean Theorem to simplify. 1/cos(x) Since 1/cos(x) is the same as sec(x)- the right-hand side- the proof is complete.
Find dy over dx in terms of x and y if cos to the power 2 braket 6yclose bracket plus sin to the power to open bracket 6y close bracket equals y plus 14?
Note : sin² Φ + cos² Φ = 1 for all real Φ. __________________________________ Now, given that : cos² (6y) + sin² (6y) = y + 14 ∴ 1 = y + 14 ∴differentiating w.r.t. x, ... 0 = (dy/dx) + 0 ∴ dy/dx = 0 ........... Ans. ___________________________________ Happy To Help ! ___________________________________
You find the value of the integral of the difference cos x - sin x between the limits of the shaded area. Note that if some of the area is below the x-axis when both cos x and sin x are less than 0, then the area will be negative and could result in a total area of zero (for example if the limits are 0 to 2π which results in a complete cycle of…
How would you prove left cosA plus sinA right times left cos2A plus sin2A right equals cosA plus sin3A?
You need to make use of the formulae for sin(A+B) and cos(A+B), and that cos is an even function: sin(A+B) = cos A sin B + sin A cos B cos(A+B) = cos A cos B - sin A sin B cos even fn → cos(-x) = cos(x) To prove: (cos A + sin A)(cos 2A + sin 2A) = cos A + sin 3A The steps are to work with the left hand side…
How do you prove that the sin over one minus the cosine minus one plus the cosine over the sine equals zero?
Multiply both sides by sin(1-cos) and you lose the denominators and get (sin squared) minus 1+cos times 1-cos. Then multiply out (i.e. expand) 1+cos times 1-cos, which will of course give the difference of two squares: 1 - (cos squared). (because the cross terms cancel out.) (This is diff of 2 squares because 1 is the square of 1.) And so you get (sin squared) - (1 - (cos squared)) = (sin squared) + (cos…
tan(x) = sin(x)/cos(x) Therefore, all trigonometric ratios can be expressed in terms of sin and cos. So the identity can be rewritten in terms of sin and cos. Then there are only two "tools": sin^2(x) + cos^2(x) = 1 and sin(x) = cos(pi/2 - x) Suitable use of these will enable you to prove the identity.