No, but cos(-x) = cos(x), because the cosine function is an even function.
Sin 15 + cos 105 = -1.9045
3cos
(sin x + cos x) / cosx = sin x / cos x + cosx / cos x = tan x + 1
No, (sinx)^2 + (cosx)^2=1 is though
No, but cos(-x) = cos(x), because the cosine function is an even function.
Sin 15 + cos 105 = -1.9045
1
2 cos * cos * -1 = 2cos(square) * -1 =cos(square) + cos(square) *-1 =1- sin(square) +cos(square) * -1 1 - 1 * -1 =0
3cos
[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,
leonhard euler
(sin x + cos x) / cosx = sin x / cos x + cosx / cos x = tan x + 1
No, (sinx)^2 + (cosx)^2=1 is though
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, expand the brackets, collect [useful] terms together, apply A+B formula above (backwards) and apply even nature of cos function: (cos A + sin A)(cos 2A + sin 2A) = cos A cos 2A + cos A sin 2A + sin A cos 2A + sin A sin 2A = (cos A cos 2A + sin A sin 2A) + (cos A sin 2A + sin A cos 2A) = cos(A - 2A) + sin(A + 2A) = cos(-A) + sin 3A = cos A + sin 3A which is the right hand side as required.
The result is variant, therefore uncertain. The only sure thing is that sin(x)2 + cos(x)2 = 1.
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.