Q: Sin 15 plus cos 105 equals?

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The result is variant, therefore uncertain. The only sure thing is that sin(x)2 + cos(x)2 = 1.

cos*cot + sin = cos*cos/sin + sin = cos2/sin + sin = (cos2 + sin2)/sin = 1/sin = cosec

sec + tan = cos /(1 + sin) sec and tan are defined so cos is non-zero. 1/cos + sin/cos = cos/(1 + sin) (1 + sin)/cos = cos/(1 + sin) cross-multiplying, (1 + sin)2 = cos2 (1 + sin)2 = 1 - sin2 1 + 2sin + sin2 = 1 - sin2 2sin2 + 2sin = 0 sin2 + sin = 0 sin(sin + 1) = 0 so sin = 0 or sin = -1 But sin = -1 implies that cos = 0 and cos is non-zero. Therefore sin = 0 or the solutions are k*pi radians where k is an integer.

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

Related questions

[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,

There is a hint to how to solve this in what is required to be shown: a and b are both squared.Ifa cos θ + b sin θ = 8a sin θ - b cos θ = 5then square both sides of each to get:a² cos² θ + 2ab cos θ sin θ + b² sin² θ = 64a² sin² θ - 2ab sin θ cos θ + b² cos² θ = 25Now add the two together:a² cos² θ + a² sin² θ + b² sin² θ + b² cos² θ = 89→ a²(cos² θ + sin² θ) + b² (sin² θ + cos² θ) = 89using cos² θ + sin² θ = 1→ a² + b² = 89

leonhard euler

No. sin(0) = 0 So cos(0)*sin(0) = 0 so the left hand side = 1

The derivative of cos(x) equals -sin(x); therefore, the anti-derivative of -sin(x) equals cos(x).

The result is variant, therefore uncertain. The only sure thing is that sin(x)2 + cos(x)2 = 1.

cos*cot + sin = cos*cos/sin + sin = cos2/sin + sin = (cos2 + sin2)/sin = 1/sin = cosec

sin cubed + cos cubed (sin + cos)( sin squared - sin.cos + cos squared) (sin + cos)(1 + sin.cos)

sec + tan = cos /(1 + sin) sec and tan are defined so cos is non-zero. 1/cos + sin/cos = cos/(1 + sin) (1 + sin)/cos = cos/(1 + sin) cross-multiplying, (1 + sin)2 = cos2 (1 + sin)2 = 1 - sin2 1 + 2sin + sin2 = 1 - sin2 2sin2 + 2sin = 0 sin2 + sin = 0 sin(sin + 1) = 0 so sin = 0 or sin = -1 But sin = -1 implies that cos = 0 and cos is non-zero. Therefore sin = 0 or the solutions are k*pi radians where k is an integer.

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)

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.

To show that (cos tan = sin) ??? Remember that tan = (sin/cos) When you substitute it for tan, cos tan = cos (sin/cos) = sin QED