y=sin x
y=cos x cos x
sin x = cos^2 x
sin x = 1-sin^2 x
sin x -1 + sin^2 x = 0
sin^2 x + sin x -1 = 0
Let y=sin x
y^2+y-1 = 0
This equation is of form ay^2+by+c=0
a = 1 b = 1 c = -1
y=[-b+/-sqrt(b^2-4ac)]/2a]
y=[-1 +/-sqrt(1^2-4(1)(-1)]/(2)(1)
discriminant is b^2-4ac =5
y=[-1 +√(5)] / 2
y=[-1 -√(5)] / 2
sin x = [-1 +√(5)] / 2
x = sin^-1 [-1 +√(5)] / 2] = 0.6662394 radians
x = sin^-1 [-1 -√(5)] / 2] = sin^-1 (-1.618) -- has no solution
When x = 0.6662394 radians, sin x and cos x times cos x are equal.
The derivative of cos(x) equals -sin(x); therefore, the anti-derivative of -sin(x) equals cos(x).
2 x cosine squared x -1 which also equals cos (2x)
2
Remember that tan = sin/cos. So your expression is sin/cos times cos. That's sin(theta).
If tan 3a is equal to sin cos 45 plus sin 30, then the value of a = 0.4.
The derivative of cos(x) equals -sin(x); therefore, the anti-derivative of -sin(x) equals cos(x).
cos(125) = cos(180 - 55) = cos(180)*cos(55) + sin(180)*sin(55) = -cos(55) since cos(180) = -1, and sin(180) = 0 So A = 55 degrees.
Sin(x) cos(x) = 1/2 of sin(2x)
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)
To show that (cos tan = sin) ??? Remember that tan = (sin/cos) When you substitute it for tan, cos tan = cos (sin/cos) = sin QED
sin2 + cos2 = 1 So, (1 - 2*cos2)/(sin*cos) = (sin2 + cos2 - 2*cos2)/(sin*cos) = (sin2 - cos2)/(sin*cos) = sin2/(sin*cos) - cos2/(sin*cos) = sin/cos - cos-sin = tan - cot
Rewrite sec x as 1/cos x. Then, sec x sin x = (1/cos x)(sin x) = sin x/cos x. By definition, this is equal to tan x.
[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 15 + cos 105 = -1.9045
cos*cot + sin = cos*cos/sin + sin = cos2/sin + sin = (cos2 + sin2)/sin = 1/sin = cosec
2 x cosine squared x -1 which also equals cos (2x)
There is no reason at all. For most angles sin plus cos do not equal one.