Manipulate normally, noting:
(1 + cot x)² - 2 cot x = 1² + 2 cot x + cot² x - 2 cot x
= 1 + cot² x
= 1 + (cos x / sin x)²
= 1 + cos² x / sin² x
= 1 + cos² x / (1 - cos² x)
= ((1 - cos² x) + cos² x)/(1 - cos² x)
= 1/(1² - cos² x)
= 1/((1 + cos x)(1 - cos x))
= 1/(1 - cos x)/(1 + cos x)
QED.
3cos
Until an "equals" sign shows up somewhere in the expression, there's nothing to prove.
No, but cos(-x) = cos(x), because the cosine function is an even function.
Sin 15 + cos 105 = -1.9045
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)
3cos
Until an "equals" sign shows up somewhere in the expression, there's nothing to prove.
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.
[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 + Cos Does NOT equal '1'. However, Sin^(2) + Cos^(2) = 1 = ( 1^(2) It is by Pythagoras. Remember in a right angled triangle H^(2) = o^(2) + a^(2) Assume ' h = 1' Then 1^(2) = 1 = o^(2) + a^)2) But o = hSin & a = hCos Substituting 1^(2) = (hSin)^(2) + (hCos(^(2)) However, we assume h = 1 Hence 1^(2) = Sin^(2) + Cos^(2)
No, but cos(-x) = cos(x), because the cosine function is an even function.
Not correct. sin2alpha + cos2alpha = 1
When tan A = 815, sin A = 0.9999992 and cos A = 0.0012270 so that sin A + cos A*cos A*(1-cos A) = 1.00000075, approx.
sin cubed + cos cubed (sin + cos)( sin squared - sin.cos + cos squared) (sin + cos)(1 + sin.cos)
Sin 15 + cos 105 = -1.9045
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)
The expression (\cos(x + y) \cos x + \cos y) does not simplify to a standard identity. Instead, it can be rewritten using the angle addition formula for cosine: (\cos(x + y) = \cos x \cos y - \sin x \sin y). Therefore, the original expression is not generally true, and its simplification would depend on specific values of (x) and (y).