If I understand you correctly, yes. Normally we write y=sin(x) or y=cos(x), and because we make our x-axis run horizontally and our y-axis run vertically, this gives us a wave running left-to-right. If instead we wrote x=sin(y) or x=cos(y), the waves would run bottom-to-top. However, notice that in the first case, y is a function of x, but in the second case, x is a function of y. If we wished to make x=sin(y) or x=cos(y) into a function of x, we need to restrict the values of y in the domain so that it will pass the vertical line test. These "restricted" functions have a name, namely arcsin and arccos, and they are referred to as the inverse trigonometric functions.
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cos(2x) = 1 - 2(sin(x)^2), so sin(x)^2 = 1/2 - 1/2*cos(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)
Like normal expansion of brackets, along with: cos(A + B) = cos A cos B - sin A sin B sin(A + B) = sin A cos B + cos A sin B 5(cos 20 + i sin 20) × 8(cos 15 + i sin 15) = 5×8 × (cos 20 + i sin 20)(cos 15 + i sin 15) = 40(cos 20 cos 15 + i sin 15 cos 20 + i cos 15 sin 20 + i² sin 20 sin 15) = 40(cos 20 cos 15 - sin 20 cos 15 + i(sin 15 cos 20 + cos 15 sin 20)) = 40(cos(20 +15) + i sin(15 + 20)) = 40(cos 35 + i sin 35)
cos*cot + sin = cos*cos/sin + sin = cos2/sin + sin = (cos2 + sin2)/sin = 1/sin = cosec
To show that (cos tan = sin) ??? Remember that tan = (sin/cos) When you substitute it for tan, cos tan = cos (sin/cos) = sin QED