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(sin x + cos x) / cosx = sin x / cos x + cosx / cos x = tan x + 1
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.
sin2(1) = 1 - cos2(1) = 1 - [cos(1)]2
Try to write everything in terms of sines and cosines:1 / cos B - cos B = (sin B / cos B) sin B1 / cos B - cos B = sin2B / cos BMultiply by the common denominator, cos B:1 - cos2B = sin2BUse the pithagorean identity on the left side:sin2B + cos2B - cos2B = sin2Bsin2B = sin2B
tan = sin/cos Now cos2 = 1 - sin2 so cos = +/- sqrt(1 - sin2) In the second quadrant, cos is negative, so cos = - sqrt(1 - sin2) So that tan = sin/[-sqrt(1 - sin2)] or -sin/sqrt(1 - sin2)
the Alamo
cos = sqrt(1 - sin^2)
(sin x + cos x) / cosx = sin x / cos x + cosx / cos x = tan x + 1
Provided that any denominator is non-zero, sin = sqrt(1 - cos^2)tan = sqrt(1 - cos^2)/cos sec = 1/cos cosec = 1/sqrt(1 - cos^2) cot = cos/sqrt(1 - cos^2)
That is just not true. Sin and cos terms are used for many other purposes : for example the components of a force along orthogonal axes.
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.
He was on the Mexican side, he was captured by the Texan army, but was later released after surrender.
(1 - cos(2x))/2, where x is the variable. And/Or, 1 - cos(x)^2, where x is the variable.
sin2(1) = 1 - cos2(1) = 1 - [cos(1)]2
Try to write everything in terms of sines and cosines:1 / cos B - cos B = (sin B / cos B) sin B1 / cos B - cos B = sin2B / cos BMultiply by the common denominator, cos B:1 - cos2B = sin2BUse the pithagorean identity on the left side:sin2B + cos2B - cos2B = sin2Bsin2B = sin2B
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.
tan θ = sin θ / cos θ sec θ = 1 / cos θ sin ² θ + cos² θ = 1 → sin² θ - 1 = - cos² θ → tan² θ - sec² θ = (sin θ / cos θ)² - (1 / cos θ)² = sin² θ / cos² θ - 1 / cos² θ = (sin² θ - 1) / cos² θ = - cos² θ / cos² θ = -1